Ah, I see. I didn't know that A would be a constant Matrix. Still, if you are simplifying something containing things like sin(3), that will still require some symbolic simplification.
Aaron Meurer On Mar 26, 2011, at 8:53 AM, Yuri Karadzhov wrote: > On Mar 24, 6:22 pm, "Aaron S. Meurer" <[email protected]> wrote: >> Things look good so far. I would include, for example, a description of how >> you can reduce it to Ax=0. >> >> Also, you might find that you have to improve some simplification algorithms >> to solve Ax=0, because the present ones will not always be able to reduce >> large expressions to 0 (especially ones containing trigonometric functions). >> >> >> Aaron Meurer >> >> On Mar 24, 2011, at 6:01 AM, Yuri Karadzhov wrote: > > A - is constant matrix, x - is constant vector. I there a problem in > linear algebra module??? > > the idea is simple yet realization is much harder. > > by default f1(t),...,fn(t) - linearly dependent if there exist > coefficients x1,...,xn : x1^2+...+xn^2!=0 > and x1*f1(t)+...+xn*fn(t)=0 > > if you choose n good points t0,...,tn you will have linear system Ax=0 > > the only problem is - to choose "good" points, but it is solvable. > > -- > You received this message because you are subscribed to the Google Groups > "sympy" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/sympy?hl=en. > -- You received this message because you are subscribed to the Google Groups "sympy" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/sympy?hl=en.
