So even though rs_series can produce the answer quickly, is this still the best way to do this? Following the wiki I was able to cobble together a recurrence relationship generator but the problem posed here at the outset would be "cruel and unusual" for a homework problem, wouldn't it? :-) By that I mean that the recurrence relationship has many relationships that need to be solved before even generating the coefficient of such a high ordered term. But the expression factors nicely and perhaps the individual recurrence relationships can be found fairly easily and then combined to find the answer.
factored expression: ``` (t - 1)**6* (t + 1)**3* (t**2 + 1)* (t**4 - t**3 + t**2 - t + 1)**3* (t**4+ t**3 + t**2 + t + 1)**5* (t**8 - t**6 + t**4 - t**2 + 1)* (t**20 - t**15 + t**10 - t**5 + 1)**2* (t**20 + t**15 + t**10 + t**5 + 1)**3* (t**40 - t**30 + t**20 - t**10 + 1) ``` On Saturday, February 17, 2018 at 12:22:54 PM UTC-6, Leonid Kovalev wrote: > > > I wonder why this is not implemented for series > > It seems the idea (as stated on GSoC page > <https://github.com/sympy/sympy/wiki/GSoC-2018-Ideas#series-expansions>) > was to replace series with rs_series after rs_series is expanded to handle > all functions. There was not much recent progress on that. Like with > assumptions or solveset, complete replacement is hard. One can make gradual > process by making series call rs_series first, and if NotImplementedError > is raised, proceed with the usual series expansion. In this case, I think > that the logic in your rseries function could be a part of rs_series > function. Or the existing logic of rs_series function could be improved to > handle quotients of two functions where each can be handled by rs_series > already. I was surprised that rs_series could not handle a rational > function directly, it looks like the primary use case for ring-based series > manipulations. > > > On Saturday, February 17, 2018 at 12:36:10 PM UTC-5, Chris Smith wrote: >> >> 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. 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