2
$\begingroup$

The probability of a number being square-free (i.e., the number has no divisor that is a square, cube, etc.) is $6/\pi^2$. I have seen many appearances of $\pi$, and this is also similar to them. All of them can be explained intuitively why pi appears in them. But I don't see any connection between square-free numbers and $\pi$. So my question is:

Why $\pi$ appears here?

Note: I don't want rigorous proofs, I want intuitive explanations.

$\endgroup$
7
  • 1
    $\begingroup$ What do you mean by "The probability of a number being square free"? What is the mathematically rigorous definiition of this statement? $\endgroup$
    – 5xum
    Commented Dec 18, 2020 at 8:26
  • $\begingroup$ @5xum Given any random positive number, what is the chance that it is square free? I don't know much about probability, but I think a better definition is this: Let the probability of a positive integer less than some positive integer $n$ being square free be $P(n)$. What is $\lim_{n\to\infty}P(n)$? $\endgroup$ Commented Dec 18, 2020 at 8:30
  • $\begingroup$ You're asking essentially for an intuitive explanation why $\pi$ appears in $\zeta(2)$ $\endgroup$ Commented Dec 18, 2020 at 8:36
  • 4
    $\begingroup$ I can meet you halfway: if we provide an intuitive explanation of squares' connection to $\pi$ to motivate $\zeta(2)=\pi^2/6$, we can finish with $\prod_{p\in\Bbb P}(1-p^{-2})=1/\zeta(2)=6/\pi^2$. $\endgroup$
    – J.G.
    Commented Dec 18, 2020 at 8:37
  • 1
    $\begingroup$ @bof there are many probability problems where pi comes up. Almost all can be explained. Mathematics is what explains these. And for the reason why pi comes in a standard normal distribution, see this. $\endgroup$ Commented Dec 18, 2020 at 9:34

2 Answers 2

0
$\begingroup$
  • $6/\pi^2$ appears in the density of square-free numbers because $\zeta(2)=\pi^2/6$ and every integer is uniquely of the form $k n^2$ with $k$ square-free.

  • Then $\zeta(2)=\pi^2/6$ because $$\frac{\pi^2}{\sin^2(\pi z)} = \sum_n \frac1{(z-n)^2}$$ This latter formula is magic: the LHS minus the RHS is a bounded entire function, thus constant, which is the great achievement of complex analysis.

  • If you don't like it then the video mentioned in the comment is saying that $$\sum_{n=0}^{2^k} \frac1{|2^k (e^{2i \pi (2n+1)/2^{k+1}}-1)|^2}=1/4$$ and that $$\lim_{k\to \infty} \sum_{n=0}^{2^k} \frac1{|2^k (e^{2i \pi (2n+1)/2^{k+1}}-1)|^2}=\sum_{n=-\infty}^\infty \lim_{k\to \infty} \frac1{|2^k (e^{2i \pi (2n+1)/2^{k+1}}-1)|^2}$$ $$ = \sum_{n=-\infty}^\infty \frac1{ |i\pi (2n+1)|^2} = \frac1{\pi^2}2(1-1/4)\zeta(2)$$

$\endgroup$
0
$\begingroup$

The most intuitive thing I can think of would be the Riemann Zeta function. You are working with squares, and you will be adding a bunch of squares. In a similar way, if you wanted cube free numbers, you would get Apéry's constant.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .