apart_list() does not seem to support the « full=True » mode.
So I guess I have to loop over the summed terms. Do you agree ?
thanks again,
  Mike

Le lundi 10 août 2020 23:05:42 UTC+2, Aaron Meurer a écrit :
>
> 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] 
> <javascript:>> 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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