ASKSAGE: Sage Q&A Forum - Latest question feedhttp://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Tue, 31 Jan 2012 04:03:17 -0600simplify_full developmenthttp://ask.sagemath.org/question/8684/simplify_full-development/Hi! Could someone elaborate on the status of the simplification routines in SAGE? I've noticed that the simplify_full can't exactly be said to compete with e.g. mathematica's corresponding FullSimplify...
I don't know if the simplify functions operate as replacement rules or such, but if they do, I might be able to contribute in the development (I'm not much of a programmer though).
So could someone please explain about how simplify_full and other simplification functions work, and how I and others can participate? This post could work as an info for all who want to take part!H. ArponenTue, 31 Jan 2012 04:03:17 -0600http://ask.sagemath.org/question/8684/Round trip through Mathematica's FullSimplifyhttp://ask.sagemath.org/question/7845/round-trip-through-mathematicas-fullsimplify/The following command returns 1:
<code>
mathematica(-2/(1 + sqrt(3)*I)).FullSimplify().sage()
</code>
The issue is that Mathematica simplifies this expression to (-1)/(2/3), which it considers to be defined in terms of the primitive root. Sage on the other hand converts (-1)/(2/3) to 1, with the idea that any root will do. My question: is it a bug that putting this equation into Mathematica and bringing it back to Sage changes it from a complex number to a real number?
dstahlkeThu, 06 Jan 2011 16:54:26 -0600http://ask.sagemath.org/question/7845/