Thanks again Aaron,
here is what I did :
import sympy as sp
sp.var('s t')
Hstab = 1/(s**3+s**2+5*s+4)
Hstab = sp.apart(Hstab,full=True).doit().evalf()
display(Hstab)
And I get :
[image: Capture d’écran 2020-08-10 à 23.00.45.png]
Which is great. However I would have liked the conjugate denominator
fractions combined. Is this possible ou should I write some code for this ?
Moreover : I am not sure I understood how is appart_list() interesting ?
Thanks again,
Mikhaël
Le lundi 10 août 2020 21:58:23 UTC+2, Aaron Meurer a écrit :
>
> If you replace the coefficients with floating point numbers, you can use
> full=True and get an answer with floats (although they aren't fully
> simplified for some reason, so you may want to use apart_list() instead).
> I'm not sure if it is possible to get an answer with just the real roots,
> so that there are no complex numbers added.
>
> Aaron Meurer
>
> On Mon, Aug 10, 2020 at 1:53 PM Mikhael Myara <[email protected]
> <javascript:>> wrote:
>
>> Thanks a lot Aaron.
>> However : for my usage, an approximate root would be sufficient. Is there
>> a way to allow this in Sympy ?
>> Thanks again,
>> Mikhaël
>>
>> Le lundi 10 août 2020 20:48:04 UTC+2, Aaron Meurer a écrit :
>>>
>>> The roots are the key difference. The second polynomial has a rational
>>> root, -1, meaning it can be split out in a partial fraction decomposition.
>>> If you don't call evalf() on the roots you can also see that the roots of
>>> the first polynomial are more complicated, because they are coming from the
>>> general cubic formula. You can also see the difference in the two if you
>>> call factor().
>>>
>>> >>> factor(s**3+s**2+5*s+4)
>>> s**3 + s**2 + 5*s + 4
>>> >>> factor(s**3+2*s**2+5*s+4)
>>> (s + 1)*(s**2 + s + 4)
>>>
>>> The first polynomial is irreducible over rationals, but the second
>>> factors. So if you do a partial fraction decomposition, it will not split
>>> because apart() only decomposes over rational numbers by default. If you
>>> want a full decomposition over all the roots, you can use something like
>>> apart(1/(s**3+s**2+5*s+4), full=True).doit().
>>>
>>> Note that it's not uncommon for polynomials to work like this, where if
>>> you change a coefficient it changes the behavior of it. That's because it's
>>> easy for a polynomial to have a rational root with one coefficient but not
>>> with another close coefficient, like in this case.
>>>
>>> Aaron Meurer
>>>
>>> On Mon, Aug 10, 2020 at 2:29 AM Mikhael Myara <[email protected]>
>>> wrote:
>>>
>>>> Dear all,
>>>>
>>>> I don't know if my problem is with the knowledge of fractionnal
>>>> decomposition itself or with symPy's implementation.
>>>>
>>>> I start with the following code :
>>>>
>>>> import sympy as sp
>>>> sp.var('s')
>>>>
>>>>
>>>> Hstab = 1/(s**3+s**2+5*s+4)
>>>> Hstab = sp.apart(Hstab)
>>>> display(Hstab)
>>>>
>>>>
>>>> The result is :
>>>>
>>>> [image: Capture d’écran 2020-08-10 à 10.18.18.png]
>>>> Now I check another similar expression :
>>>>
>>>> Hstab2 = 1/(s**3 + 2*s**2 + 5*s + 4)
>>>> Hstab2 = sp.apart(Hstab2,s)
>>>> display(Hstab2)
>>>>
>>>> and I get :
>>>>
>>>> [image: Capture d’écran 2020-08-10 à 10.21.14.png]
>>>>
>>>>
>>>>
>>>>
>>>> If now I check for the roots of the denominator for each case :
>>>>
>>>>
>>>> print("\n\nFirst case :")
>>>> P1=(1/Hstab).simplify()
>>>> display(P1)
>>>> for sol in sp.solve(P1,s):
>>>> display(sol.evalf())
>>>>
>>>>
>>>> print("\n\nSecond case :")
>>>> P2=(1/Hstab2).simplify()
>>>> display(P2)
>>>> for sol in sp.solve(P2,s):
>>>> display(sol.evalf())
>>>>
>>>> I get :
>>>>
>>>> [image: Capture d’écran 2020-08-10 à 10.26.18.png]
>>>> Which is very similar except that the real root of the first case is
>>>> not ass simple as the one of the second case. So I do not understand as I
>>>> can't see any mathematical impossibility, the two cases are very close.
>>>> Is there a way to reach the fractional decomposition for the first case
>>>> ?
>>>>
>>>> Thanks a lot,
>>>> Mike
>>>>
>>>>
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