Yes, that is what I mean by only getting the real roots. I don't think it's
possible now, though it shouldn't be hard to do it manually with
apart_list().

Aaron Meurer

On Mon, Aug 10, 2020 at 3:03 PM Mikhael Myara <[email protected]>
wrote:

> 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]>
>> 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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