What are the connections between pi and prime numbers? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T18:52:21Z http://mathoverflow.net/feeds/question/16991 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16991/what-are-the-connections-between-pi-and-prime-numbers What are the connections between pi and prime numbers? muad 2010-03-03T18:51:00Z 2013-03-15T21:27:07Z <p>I watched a video that said the probability for Gaussian integers to be relatively prime is an expression in $\pi$, and I also know about $\zeta(2) = \pi^2/6$ but I am wondering what are more connections between $\pi$ and prime numbers?</p> http://mathoverflow.net/questions/16991/what-are-the-connections-between-pi-and-prime-numbers/17003#17003 Answer by Richard Stanley for What are the connections between pi and prime numbers? Richard Stanley 2010-03-03T20:19:26Z 2010-03-03T20:19:26Z <p>The probability that two Gaussian integers are relatively prime is $6/(\pi^2 K) = 0.66370080461385348\cdots$, where $K= 1 - \frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{7^2}+\cdots$ (Catalan's constant). There is no known simple expression for $K$ in terms of $\pi$. See <a href="http://www.springerlink.com/content/y826m64747254t87" rel="nofollow">http://www.springerlink.com/content/y826m64747254t87</a>.</p> http://mathoverflow.net/questions/16991/what-are-the-connections-between-pi-and-prime-numbers/17008#17008 Answer by Gjergji Zaimi for What are the connections between pi and prime numbers? Gjergji Zaimi 2010-03-03T20:47:45Z 2010-03-03T20:47:45Z <p>Well, first of all, $\pi$ is not just a random real number. Almost every real number is transcendental so how can we make the notion "$\pi$ is special" (in a number-theoretical sense) more precise?</p> <p>Start by noticing that $$\pi=\int_{-\infty}^{\infty}\frac{dx}{1+x^2}$$ This already tells us that $\pi$ has something to do with rational numbers. It can be expressed as "a complex number whose real and imaginary parts are values of absolutely convergent integrals of rational functions with rational coefficients, over domains in $\mathbb{R}^n$ given by polynomial inequalities with rational coefficients." Such numbers are called <a href="http://en.wikipedia.org/wiki/Period_(number)" rel="nofollow">periods</a>. Coming back to the identity $$\zeta(2)=\frac{\pi^2}{6}$$ There is a very nice proof of this (that at first seems very unnatural) due to Calabi, it shows that $$\frac{3\zeta(2)}{4}=\int_0^1\int_0^1\frac{dxdy}{1-x^2y^2}$$ by expanding the corresponding geometric series, and then evaluates the integral to $\pi^2/8$ (So yes, $\pi^2$ and all other powers of $\pi$ are periods.) But the story doesn't end here as it is believed that there are truly deep connections between values of zeta functions (or <a href="http://en.wikipedia.org/wiki/L-function" rel="nofollow">L-functions</a>) and certain evaluations involving periods, such as $\pi$. Another famous problem about primes is Sylvester's problem of which primes can be written as a sum of two rational cubes. So one studies the <a href="http://en.wikipedia.org/wiki/Elliptic_curve" rel="nofollow">Elliptic curve</a> $$E_p: p=x^3+y^3$$ and one wants to know if there is one rational solution, the central value of the corresponding L-function will again involve $\pi$ up to some integer factor and some Gamma factor. Next, periods are also values of multiple zeta functions: $$\zeta(s_1,s_2,\dots,s_k)=\sum_{n_1>n_2>\cdots>n_k\geq 1}\frac{1}{n_1^{s_1}\cdots n_k^{s_k}}$$ And they also appear in other very important conjectures such as the <a href="http://en.wikipedia.org/wiki/Birch_and_Swinnerton-Dyer_conjecture" rel="nofollow">Birch and Swinnerton-Dyer conjecture</a>. But of course all of this is really hard to explain without using appropriate terminology, the language of <a href="http://en.wikipedia.org/wiki/Motive_(algebraic_geometry)" rel="nofollow">motives</a> etc. So, though, this answer doesn't mean much, it's trying to show that there is an answer to your question out there, and if you study a lot of modern number theory, it might just be satisfactory :-).</p> http://mathoverflow.net/questions/16991/what-are-the-connections-between-pi-and-prime-numbers/17041#17041 Answer by Gerry Myerson for What are the connections between pi and prime numbers? Gerry Myerson 2010-03-04T02:02:30Z 2010-03-04T02:02:30Z <p>There are a few formulas relating $\pi$ to arithmetic functions. For example, if $\sigma(n)$ is the sum of the divisors of $n$, then $\sum_1^n\sigma(n)=\pi^2n^2/12+O(n\log n)$. If $d(n)$ is the number of divisors of $n$, then $\sum_1^{\infty}n^{-2}d(n)=\pi^4/36$. If $\phi(n)$ is the Euler phi-function, then $\sum_1^n\phi(n)=3n^2\pi^{-2}+O(n\log n)$. These all appear in Section 3.5 of Eymard and Lafon, The Number $\pi$. </p> http://mathoverflow.net/questions/16991/what-are-the-connections-between-pi-and-prime-numbers/19501#19501 Answer by Wadim Zudilin for What are the connections between pi and prime numbers? Wadim Zudilin 2010-03-27T10:21:51Z 2010-03-27T10:21:51Z <p>There is a nice story, initiated by L. Van Hamme, which relates several Ramanujan's formulas for $\pi$ to supercongruences modulo powers of primes. The simplest way to witness this route is to make a look at (my) <a href="http://arxiv.org/abs/0805.2788" rel="nofollow">http://arxiv.org/abs/0805.2788</a> .</p> http://mathoverflow.net/questions/16991/what-are-the-connections-between-pi-and-prime-numbers/19502#19502 Answer by dke for What are the connections between pi and prime numbers? dke 2010-03-27T12:23:02Z 2010-03-27T12:23:02Z <p>A formula that surely belongs here linking $\pi$ and the primes is <code>$$2.3.5.7...=4\pi^2.$$</code> This is obtained via a zeta regularization in a similar way to the more well-known $\infty!=\sqrt{2\pi}$ (see e.g. <a href="http://www.google.co.uk/url?sa=t&amp;source=web&amp;ct=res&amp;cd=5&amp;ved=0CC8QFjAE&amp;url=http%3A%2F%2Fkatlas.math.toronto.edu%2Fdrorbn%2FMathBlog%2F2008-11%2Fone%2FGillet%40FI-_What_is_infinity_factorial_(and_why_might_we_care)Q.pdf&amp;rct=j&amp;q=infinity+factorial&amp;ei=kvWtS7TfEoS80gTqg5iPDg&amp;usg=AFQjCNHnBSldUsYexgXp29EpMFMtMWKh8g" rel="nofollow">here</a> for a short discussion of this). However, to find the product of the primes, one uses the prime zeta function <code>$$\sum_{p\; prime} \frac{1}{p^s}$$</code> which has the unfortunate property of having infinitely many singularities between 0 and 1 which breaks the standard regularization procedure. <a href="http://www.springerlink.com/content/n64k047n5r5j1881/" rel="nofollow">E. Muñoz García and R. Pérez Marco</a> circumvent this problem (literally) by adding in an extra variable and taking the limit from a different direction.</p> http://mathoverflow.net/questions/16991/what-are-the-connections-between-pi-and-prime-numbers/28754#28754 Answer by Robert Bruner for What are the connections between pi and prime numbers? Robert Bruner 2010-06-19T14:40:07Z 2010-06-19T14:40:07Z <p>More elementary: the probability that two positive integers have GCD=1 is $6/\pi^2 = 1/\zeta(2)$ because the probability that a prime $p$ divides the GCD is 1/p^2 by considering each p by p block of pairs of positive integers. More generally, the probability that k positive integers have GCD 1 is $1/\zeta(k)$ by a similar argument.</p> http://mathoverflow.net/questions/16991/what-are-the-connections-between-pi-and-prime-numbers/124656#124656 Answer by Geoff Robinson for What are the connections between pi and prime numbers? Geoff Robinson 2013-03-15T21:27:07Z 2013-03-15T21:27:07Z <p>Here is an example of a way to use $\pi$ to prove the infinitude of primes without calculating its value, or using the relatively deep fact that $\pi$ is irrational, but starting from the knowledge of $\zeta(2)$ and $\zeta(4).$ Suppose that there were only finitely many prime numbers $2= p_{1}, 3= p_{2}, \ldots, p_{k-1},p_{k}.$ From the formulae $\sum_{n=1}^{\infty} \frac{1}{n^{2}} = \frac{\pi^{2}}{6}$ and $\sum_{n=1}^{\infty} \frac{1}{n^{4}} = \frac{\pi^{4}}{90}$, we may conclude after the fashion of Euler that (respectively) we have: `$\prod_{j=1}^{k} \frac{p_{j}^{2}}{p_{j}^{2}-1} = \frac{\pi^{2}}{6}$' and $\prod_{j=1}^{k} \frac{p_{j}^{4}}{p_{j}^{4}-1} = \frac{\pi^{4}}{90}.$ Squaring the first equation and dividing by the second leads quickly to $\prod_{j=1}^{k} \frac{p_{j}^{2}+1}{p_{j}^{2}-1} = \frac{5}{2}$, so $5\prod_{j=1}^{k} (p_{j}^{2}-1) = 2 \prod_{j=1}^{k}(p_{j}^{2}+1).$ This is a contradiction, since the product on the left is certainly divisible by $3$, whereas every term in the rightmost product except that for $j = 2$ is congruent to $-1$ (mod 3), so we obtain $0 \equiv (-1)^{k}$ (mod 3), which is absurd. (I would be grateful if anyone knows a reference for a proof like this. I can't believe that I am the first person to think of it).</p>