[sage-support] Re: Inequalities

2017-04-06 Thread Emmanuel Charpentier
That works because your inequation has exactly one variable. Currentkly, Maxima (and therefore Sage) can't solve an inequation with more than one variable (i. e. express the solution for one variable using otrher variables as parameters). HTH, -- Emmanuel Charpentier Le mardi 4 avril 2017

[sage-support] Re: Inequalities

2017-04-03 Thread kcrisman
On Sunday, April 2, 2017 at 5:41:48 PM UTC-4, Emmanuel Charpentier wrote: > > Sorry for the late answer. > > Symbolic inequations solving seems currently broken in Maxima, and, hence, > in Sage (at least if my interpretation of the error messages is correct...). > For other readers, I should

[sage-support] Re: Inequalities

2017-04-02 Thread Emmanuel Charpentier
Sorry for the late answer. Symbolic inequations solving seems currently broken in Maxima, and, hence, in Sage (at least if my interpretation of the error messages is correct...). Furthermore, our current system of conversion to other systems (fricas, sympy, maple, mathematica, usw...)

[sage-support] Re: Inequalities

2017-02-28 Thread Andrey Novoseltsev
No idea, but perhaps sage-support will help! On Tuesday, 28 February 2017 03:39:22 UTC-7, Ingo Dahn wrote: > > solve provides the solution of inequalities as a list of equations and > inequalities. Is there a way to transform this into a simplified union of > open-closed intervals? > For

[sage-support] Re: Inequalities solving

2009-05-11 Thread Jason Grout
Leonardo Passos wrote: Is there a way to find the domain of x given a simple inequality like 5*x + 2 x - 6 in Sage? Mapple does support this...but it ain't free software. I was wondering if could use Sarge in a high school math class, so this is of big relevance. One way is to use the

[sage-support] Re: Inequalities solving

2009-05-10 Thread Leonardo
I found a message on sage-devel list (http://www.mail-archive.com/sage- de...@googlegroups.com/msg10351.html) about an inequality solver. However, each time x is a divisor of something (therefore it cannot be zero), the isolver function fails, as it tries to find the zeros of x as a solving