Re: [sympy] Re: Can someone help me with this command?

[Aaron Meurer]

Here is some background on this problem
https://www.johndcook.com/blog/2014/07/09/making-change/.

Aaron Meurer

On Sun, Feb 18, 2018 at 5:00 PM, Chris Smith  wrote:
> 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) 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:
>>
>>
>>
>

[sympy] Re: Can someone help me with this command?

[parag goyal]

This code is working fine for me .I am using sympy- 1.1.1


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).coeff(t**783)
...:
Out[50]: 683772

On Monday, February 12, 2018 at 1:26:16 PM UTC+5:30, 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! 
>
>

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[sympy] Re: Can someone help me with this command?

[Chris Smith]

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 
> ) 
> 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 prob

[sympy] Re: Can someone help me with this command?

[Leonid Kovalev]

> I wonder why this is not implemented for series

It seems the idea (as stated on GSoC page 
) 
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   

>>>

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[sympy] Re: Can someone help me with this command?

[Chris Smith]

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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[sympy] Re: Can someone help me with this command?

[Leonid Kovalev]

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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[sympy] Re: Can someone help me with this command?

[Chris Smith]

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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[sympy] Re: Can someone help me with this command?

[Kalevi Suominen]

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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