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
>>>
>>
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