Is equality of terms for "real" numbers with roots, logarithm, exponential, sin, cos, and other trigonometric operations decidable with a Turing-machine? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T15:58:50Z http://mathoverflow.net/feeds/question/118972 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118972/is-equality-of-terms-for-real-numbers-with-roots-logarithm-exponential-sin Is equality of terms for "real" numbers with roots, logarithm, exponential, sin, cos, and other trigonometric operations decidable with a Turing-machine? tailcalled 2013-01-15T13:59:39Z 2013-01-15T18:55:06Z <p>If yes, how? Also, I know you can't do it for arbitrary statements about real numbers, but that's not what I'm asking, and by "real" numbers, I mean the numbers constructible from 1, -, /, and the operations mentioned in the title.</p> <p>Also, I don't care about numbers that can't be constructed from said operations and constant.</p> http://mathoverflow.net/questions/118972/is-equality-of-terms-for-real-numbers-with-roots-logarithm-exponential-sin/119013#119013 Answer by Gro-Tsen for Is equality of terms for "real" numbers with roots, logarithm, exponential, sin, cos, and other trigonometric operations decidable with a Turing-machine? Gro-Tsen 2013-01-15T18:50:33Z 2013-01-15T18:50:33Z <p>Assuming <a href="http://en.wikipedia.org/wiki/Schanuel%27s_conjecture" rel="nofollow">Schanuel's conjecture</a>, the answer seems to be yes, according to Daniel Richardson, "How to recognize zero" <em>J. Symbolic Comput</em> <strong>24</strong> (1997), 627–645 (<a href="http://dx.doi.org/10.1006/jsco.1997.0157" rel="nofollow">doi:10.1006/jsco.1997.0157</a>, available <a href="http://people.bath.ac.uk/masdr/rec.ps" rel="nofollow">here online</a>), in which the author defines a set of numbers he calls "elementary", which is algebraically closed and closed under exponential, logarithm and trigonometric functions, and for which equality is decidable (again, assuming Schanuel's conjecture).</p>