It is possible to create a polynomial over the symbolic ring, i.e. SR ['x,y']. I don't know if this'd be faster though.
On May 22, 2008, at 1:08 PM, Andrey Novoseltsev wrote: > > Maybe, but the coefficients are symbolic non-polynomial (and non- > rational) expressions. Can it be done anyway? I also had problems with > subtracting 1 from an expression, which I got by substituting rational > functions into a polynomial, so I switched to symbolic representation. > > On May 22, 12:54 pm, Robert Bradshaw <[EMAIL PROTECTED]> > wrote: >> It almost sounds to me like you'd rather be working in a multivariate >> polynomial ring (which will be much, much faster). >> >> - Robert >> >> On May 22, 2008, at 12:48 PM, Andrey Novoseltsev wrote: >> >> >> >>> Actually, yesterday was the first time I really tried to use Sage >>> for >>> symbolic computations and it was quite frustrating for me. >> >>> 1) It would be really nice if it was faster. >> >>> 2) It seems to me that >>> ((x+y)*y).coeff(y^2) >>> and >>> ((x+y)*y).expand().coeff(y^2) >>> should return the same coefficient 1 (while the first one returns >>> zero). >> >>> 3) I think that >>> ((x+x*y)/x).simplify() >>> should return 1+y, instead of the original expression, otherwise, >>> what >>> is simplify simplifying? >> >>> 4) It would be nice to be able to get the part of given degree in >>> the >>> given list of variables, and write >>> f3 = fp.homogeneous_part([up, vp], 3) >>> instead of something like >>> f3 = (fp.coeff(up,3).coeff(vp,0)*up^3*vp^0 >>> + fp.coeff(up,2).coeff(vp,1)*up^2*vp^1 >>> + fp.coeff(up,1).coeff(vp,2)*up^1*vp^2 >>> + fp.coeff(up,0).coeff(vp,3)*up^0*vp^3)- Hide quoted text - >> >> - Show quoted text - > > --~--~---------~--~----~------------~-------~--~----~ To post to this group, send email to sage-devel@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/sage-devel URLs: http://www.sagemath.org -~----------~----~----~----~------~----~------~--~---
