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.
