So for a rational function one could do this to get the series approximation?
def rseries(p, x, o): """Return truncated series of univariate rational p in variables x up to order o about the point x = 0. """ n, d = p.as_numer_denom() if not all(i.is_polynomial() for i in (n, d)): return R, r = ring(x.name, QQ) def rp(p): rv = 0 for a in Add.make_args(p.expand()): c, b = a.as_coeff_Mul() if b == 1: e = 0 else: e = b.as_base_exp()[1] rv += c*r**e return rv n, d = map(rp, (n, d)) return rs_mul(n, rs_series_inversion(d, r, o), r, o) >>> x = var('x') >>> rseries(x/(1-x+x**2+3*x**10),x,12) -4*x**11 - x**10 + x**8 + x**7 - x**5 - x**4 + x**2 + x >>> _.as_expr().coeff(x**11) -4 Given that this is so much faster, I wonder why this is not implemented for series.Perhaps that is part of the work that can yet be done on series: using input type to tailor that method used to give the output. BTW, this runs for Anane's expression without timing out on live.sympy.org. /c On Friday, February 16, 2018 at 10:51:29 PM UTC-6, Leonid Kovalev wrote: > > There are much more efficient tools in sympy/polys/ring_series.py that > could in principle replace series but are more limited in the expressions > they support. > For example, this function is the reciprocal of a polynomial. Introducing > this polynomial (called g) and calling rs_series_inversion yields the > answer at once (683772*t**783): > > from sympy import * > from sympy.polys.ring_series import rs_series_inversion > R, t = ring('t', QQ) > g = (1-t) * (1-t**5) * (1-t**10) * (1-t**25) * (1-t**50) * (1-t**100) > rs_series_inversion(g, t, 784) > > Well, it took a bit of time, measured with timeit at 654ms. But that's > nothing compared to series which took 28.8 seconds. > > To get the specific coefficient, one can use > > rs_series_inversion(g, t, 784).coeff(t**783) > > > > On Friday, February 16, 2018 at 8:50:46 PM UTC-5, Chris Smith wrote: >> >> Just out of curiosity, do we have anything to generate the coefficients >> of terms of a rational function's Taylors series? I read the wiki ( >> https://en.wikipedia.org/wiki/Rational_function) on the "method of >> generating functions" but that seems to be pretty significant computation >> for this particular problem. I was able to compute the term with the series >> (as Belkiss indicated) but is there a better way? >> >> On Monday, February 12, 2018 at 2:09:05 AM UTC-6, Kalevi Suominen wrote: >>> >>> >>> >>> On Monday, February 12, 2018 at 9:56:16 AM UTC+2, Belkiss Anane wrote: >>>> >>>> Hello! I have SymPy on my computer but it crashed and I'm using >>>> someone else's to do some math problems. Unfortunately, the online SymPy >>>> times out really fast. Can someone run these commands for me please? >>>> >>>> from sympy import * >>>> t = Symbol('t') >>>> f = 1 / ((1-t) * (1-t**5) * (1-t**10) * (1-t**25) * (1-t**50) * (1-t**100)) >>>> f.series(t,0,784) >>>> >>>> >>>> I am just looking for the coefficient of t**783! >>>> >>>> >>>> Thank you very much, you'll be saving my night! >>>> >>>> >>> I get 683772⋅t^783, but I have not checked that for correctness. >>> >>> Kalevi Suominen >>> >> -- You received this message because you are subscribed to the Google Groups "sympy" group. To unsubscribe from this group and stop receiving emails from it, send an email to sympy+unsubscr...@googlegroups.com. To post to this group, send email to sympy@googlegroups.com. Visit this group at https://groups.google.com/group/sympy. To view this discussion on the web visit https://groups.google.com/d/msgid/sympy/c85f092d-4ff3-467c-996b-4fec20b6b4aa%40googlegroups.com. For more options, visit https://groups.google.com/d/optout.