On Jan 15, 2008 10:44 AM, Martin Albrecht <[EMAIL PROTECTED]> wrote: > > > It might be a lot easier to help if you gave the rational function. > > Depending on how complicated the denominator is, you basically just have to > > compute the Taylor series of the rational function, by differentiation and > > evaluation (using Taylor's formula), i.e., kind of like this is doing, but > > over GF(p): > > > > sage: f = (x^3 + x +1)/((x^4 + x^2 + 2)*x^3*(x^3-5)) > > sage: f.taylor(x, 0, 4) > > -1/(10*x^3) - 1/(10*x^2) + 1/(20*x) - 7/100 + x/200 + 17*x^2/200 - > > 103*x^3/2000 - 23*x^4/2000 > > Hi, > > sorry for not being specific enough earlier. In my particular application > > f(t) = p(t)/(1-t)^n > > where p is a polynomial with integer coefficients. So I am not actually > working over GF(p) and in that case the Taylor expansion seems to give me > what I want. However as I am looking into this now, I try to come up with > something more general. I am wondering what Magma is doing (maybe just Taylor > as well?) and if we want this too, e.g. that > > sage: L.<t> = LaurentSeriesRing(IntegerRing()) > sage: L(f) > > returns the expansion? Would that make sense? Is it feasible?
Yes that makes sense and would be a great idea to do. In your particular case above, you should write p(t)/(1-t)^n = p(t) * (1/(1-t))^n then expand 1/(1-t) out as a geometric series, raise it to the power of n, and multiply it by p(t). You actually will get a power series rather than a Laurent series, (1-t) has no pole at 0. William --~--~---------~--~----~------------~-------~--~----~ To post to this group, send email to sage-support@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-support URLs: http://sage.math.washington.edu/sage/ and http://sage.scipy.org/sage/ -~----------~----~----~----~------~----~------~--~---
