most recent 30 from http://mathoverflow.net 2013-05-18T13:11:14Z http://mathoverflow.net/feeds/question/33817 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/33817/work-on-independence-of-pi-and-e muad 2010-07-29T17:56:46Z 2010-09-27T15:07:08Z

<p>It is an open problem to prove that $\pi$ and $e$ are algebraically independent (over $\mathbb{Q}$).</p> <ul> <li>What are some of the important results leading toward proving this?</li> <li>What are the most promising theories and approaches for this problem?</li> </ul>

http://mathoverflow.net/questions/33817/work-on-independence-of-pi-and-e/33820#33820 Evan Jenkins 2010-07-29T18:02:59Z 2010-07-29T18:02:59Z

<p><a href="http://en.wikipedia.org/wiki/Schanuel%27s_conjecture" rel="nofollow">Schanuel's conjecture</a> would imply this result. It states that if $z_1, \ldots, z_n$ are linearly independent over $\mathbb{Q}$, then $\mathbb{Q}(z_1, \ldots, z_n, e^{z_1}, \ldots, e^{z_n})$ has transcendence degree at least $n$ over $\mathbb{Q}$. In particular, if we take $z_1 = 1$, $z_2 = \pi i$, then Schanuel's conjecture would imply that $\mathbb{Q}(1, \pi i, e, -1) = \mathbb{Q}(e, \pi i)$ has transcendence degree 2 over $\mathbb{Q}$.</p>

http://mathoverflow.net/questions/33817/work-on-independence-of-pi-and-e/33837#33837 Stefan Geschke 2010-07-29T19:49:31Z 2010-07-29T19:49:31Z

<p>People in model theory are currently studying the complex numbers with exponentiation. Z'ilber has an axiomatisation of an exponential field (field with exponential function) that looks like the complex numbers with exp. but satisfies Schanuel's conjecture. He proved that there is exactly one such field of the size of $\mathbb C$. I would find it odd if Z'ilber's field turned out to be different from the complex numbers.</p> <p>By results of Wilkie, the reals with exponentiation are well understood, and the complex numbers with exponentiation is in some way the next step up. The model theoretic frame work (o-minimality) that works for the reals with exp. fails for the complex numbers, but there might be a similar theory that works for the complex field with exponentiation.</p>

http://mathoverflow.net/questions/33817/work-on-independence-of-pi-and-e/33838#33838 muad 2010-07-29T19:50:12Z 2010-07-29T19:50:12Z

<p>There is a proof of the algebraic independence of $\pi$ and $e^\pi$ in <a href="http://www.springer.com/mathematics/numbers/book/978-3-540-41496-4" rel="nofollow">Introduction to Algebraic Independence Theory</a> and <em>a detailed exposition of methods created in last the 25 years</em> although I have not read it.</p>