Comment #5 on issue 2548 by asmeurer: Partial fraction decomposition with linear denominators
http://code.google.com/p/sympy/issues/detail?id=2548
I still think we should do what I suggested in comment 1 (up to a possible renaming of suggested options).
Using extension=I is basically a non-solution. How would this work if you have apart(1/(x**3 + x + 1))? Or how about apart(1/(x**5 + x + 1))? Bronstein's algorithm creates the information needed to return an expression with linear denominators. The problem is that RootSum destroys this information. There needs to be a way to tell RootSum to either return true linear denominators (e.g., compute the splitting extension), or if this is not possible (or is too expensive), to stay in an unevaluated form that has linear denominators, or, alternately, to use RootOf objects.
By the way, the way things are now, is there any difference between the output of apart() and apart(full=True)? As far as I can tell, there isn't, because of the way RootSum puts things back together, but I wasn't very extensive in my testing.
-- You received this message because you are subscribed to the Google Groups "sympy-issues" group. To post to this group, send email to sympy-issues@googlegroups.com. To unsubscribe from this group, send email to sympy-issues+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/sympy-issues?hl=en.
