Is $\zeta(3)/pi^3$ rational? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T22:13:43Z http://mathoverflow.net/feeds/question/60595 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/60595/is-zeta3-pi3-rational Is $\zeta(3)/pi^3$ rational? Thomas Bloom 2011-04-04T19:27:43Z 2011-05-09T21:40:46Z <p>Apery proved in 1976 that $\zeta(3)$ is irrational, and we know that for all integers $n$, <code>$\zeta(2n)=\alpha \pi^{2n}$</code> for some $\alpha\in \mathbb{Q}$. Given these facts, it seems natural to ask whether we can have <code>$\zeta(n)=\alpha \pi^n$</code></p> <p>for all $n$ (I'm mainly interested in the case $n=3$). The proofs I've seen of the irrationality of $\zeta(3)$ don't seem to give this information.</p> <p>My gut feeling is that the answer is no, but I can't find any reference proving this fact. I know that the answer hasn't been proven to be yes ($\zeta(3)$ isn't even known to be transcendental), but ruling out this possibility seems an easier problem.</p> http://mathoverflow.net/questions/60595/is-zeta3-pi3-rational/60610#60610 Answer by Gerald Edgar for Is $\zeta(3)/pi^3$ rational? Gerald Edgar 2011-04-04T21:11:26Z 2011-04-04T21:11:26Z <p>The same method that gives you those even cases also gives an answer in the odd cases. But (for both) it is the sum of $1/k^n$ over all nonzero integers $k$ ... so of course in the odd case you get $0$, and in the even case you divide by $2$ go get $\zeta(n)$.</p> http://mathoverflow.net/questions/60595/is-zeta3-pi3-rational/60624#60624 Answer by Gerry Myerson for Is $\zeta(3)/pi^3$ rational? Gerry Myerson 2011-04-05T00:20:48Z 2011-04-05T01:08:35Z <p>I can't find references but I know it has been shown that if $\zeta(3)/\pi^3=a/b$ is rational then $a$ and $b$ are enormous. </p> <p>EDIT: I found a reference, but not in a formal publication. At <a href="http://tech.groups.yahoo.com/group/primenumbers/message/22659?threaded=1&amp;p=2" rel="nofollow">http://tech.groups.yahoo.com/group/primenumbers/message/22659?threaded=1&amp;p=2</a> it says, </p> <p>"Re: Numerology about the Apery Constant $\zeta(3)$</p> <p>"I also attempted to use PSLQ to figure out whether $\zeta(3)/\pi^3$ was a low-degree low-height algebraic number. Result: If it has degree $\le10$ then its height is at least $10^{91}$."</p> <p>This was posted by someone identifying himself as Warren Smith. </p> http://mathoverflow.net/questions/60595/is-zeta3-pi3-rational/64397#64397 Answer by user02138 for Is $\zeta(3)/pi^3$ rational? user02138 2011-05-09T14:35:22Z 2011-05-09T21:24:42Z <p>If $\alpha, \beta > 0$ such that $\alpha \beta = \pi^{2}$, then for each non-negative integer $n$, \begin{align} \alpha^{-n} \left( \frac{\zeta(2n+1)}{2} + \sum_{k \geq 1} \frac{k^{-2n-1}}{e^{2 k \alpha} - 1} \right) &amp; = (- \beta)^{-n} \left( \frac{\zeta(2n+1)}{2} + \sum_{k \geq 1} \frac{k^{-2n-1}}{e^{2 k \beta} - 1} \right) - \end{align} \begin{align} \qquad 2^{2n} \sum_{k = 0}^{n+1} (-1)^{k} \frac{B_{2k} \ B_{2n- 2k + 2}}{(2k)! \ (2n - 2k + 2)!} \alpha^{n - k + 1} \beta^{k}. \end{align} where $B_n$ is the $n^{\text{th}}$-Bernoulli number.</p> <p>For odd positive integer $n$, \begin{align} \zeta(2n+1) = -2^{2n} \left( \sum_{k = 0}^{n+1} (-1)^{k} \frac{B_{2k} \ B_{2n- 2k + 2}}{(2k)! \ (2n - 2k + 2)!} \right) \pi^{2n+1} - 2 \sum_{k \geq 1} \frac{k^{-2n-1}}{e^{2 \pi k} - 1}. \end{align}</p> <p>In particular, for $n = 1$, \begin{align} \zeta(3) = -4 \left( \sum_{k = 0}^{2} (-1)^{k} \frac{B_{2k} \ B_{2- 2k + 2}}{(2k)! \ (2 - 2k + 2)!} \right) \pi^{3} - 2 \sum_{k \geq 1} \frac{k^{-3}}{e^{2 \pi k} - 1}. \end{align}</p> <p>Observe that the coefficient of $\pi^{3}$ is <em>rational</em>, however, it is my understanding that nothing is known about the algebraic nature of the infinite sum.</p> http://mathoverflow.net/questions/60595/is-zeta3-pi3-rational/64430#64430 Answer by DamienC for Is $\zeta(3)/pi^3$ rational? DamienC 2011-05-09T20:58:15Z 2011-05-09T21:10:51Z <p>It is definitely not known if $\zeta(3)/\pi^3$ is rational or not. By the way, there is <a href="http://arxiv.org/abs/0808.2762" rel="nofollow">a paper of Felder and Willwacher</a> where they prove that the weight of a certain graph appearing in <a href="http://arxiv.org/abs/q-alg/9709040" rel="nofollow">Kontsevich's formality quasi-isomorphism</a> is, up to a rational, $\zeta(3)/\pi^3$. The question whether Kontsevich's quasi-isomorphism is defined over $\mathbb{Q}$ or not, is still open. If the answer to this question would be "yes", then the <a href="http://arxiv.org/abs/0906.0187" rel="nofollow">associator defined by Alekseev and Torossian</a> would have rational coefficients... and that would definitely be a great result!</p> <p>Among the main recent advances concerning rationality of zeta values, there are the works of <a href="http://math.univ-lyon1.fr/homes-www/rivoal/articles.html" rel="nofollow">Tanguy Rivoal</a> and <a href="http://wain.mi.ras.ru/publications.html" rel="nofollow">Wadim Zudilin</a>. One of the most advanced results is that there is at least one irrational in $\zeta(5)$, $\zeta(7)$, $\zeta(9)$, $\zeta(11)$. </p> http://mathoverflow.net/questions/60595/is-zeta3-pi3-rational/64432#64432 Answer by YBL for Is $\zeta(3)/pi^3$ rational? YBL 2011-05-09T21:40:46Z 2011-05-09T21:40:46Z <p>In this <a href="http://www.zentralblatt-math.org/zmath/en/advanced/?q=an%3Apre05770900&amp;format=complete" rel="nofollow">article</a>, Takaaki Musha proves that $\zeta(2n+1) \notin (2\pi )^{2n+1} \mathbb{Q}$. I haven't read it so I can't say more.</p> <p>See this <a href="http://mathoverflow.net/questions/38190/values-of-zeta-at-odd-positive-integers-and-borels-computations" rel="nofollow">question</a>.</p>