0
$\begingroup$

I want to conduct partial fraction decomposition to rational function on $\mathbb C$ and tried this

PFD[R_, x_] := 
 Apart[Numerator[Simplify@R]/
   Factor[Denominator[Simplify@R], Extension -> All], x]

Normally it works fine like this

In[1]:=PFD[(2 (-1 + x^2) (6 x - 3))/((1 + x^2)^2 (3 + x^2)^3), x]

Out[1]:=-((3/2 - 3 I)/(-I + x)^2) + (6 + (3 I)/2)/(-I + x) - (
 3/2 + 3 I)/(I + x)^2 + (6 - (3 I)/2)/(I + x) - (
 Sqrt[3] (I + 2 Sqrt[3]))/(-I Sqrt[3] + x) - (
 Sqrt[3] (-I + 2 Sqrt[3]))/(I Sqrt[3] + x)

However, when the denominator has multiple complicated complex roots, Apart will automatically simplify the result

In[2]:=PFD[( x^4 (1 + x^2 + x^4 + x^6))/((-1 + x^2)^4 (1 + x^16)), x]

Out[2]:=2/(-1 + x^2)^4 - 1/(-1 + x^2)^3 - 14/(-1 + x^2)^2 + 43/(
 2 (-1 + x^2)) + (
 65 + 77 x^2 + 77 x^4 + 65 x^6 + 43 x^8 + 15 x^10 - 15 x^12 - 
  43 x^14)/(2 (1 + x^16))

rather than leaving the parts of different poles alone. In other cases, it will output a tremendously long expression

In[3]:=PFD[ (x^4) /((-1 + x^2)^4 (1 + x^5)), x]

Out[3]:=(*too long for here*)

but it turns out to be not that long after simplification

In[4]:=FullSimplify /@ % // ToRadicals

Out[4]:=1/(32 (-1 + x)^4) - 1/(64 (-1 + x)^3) - 13/(128 (-1 + x)^2) + 41/(
 256 (-1 + x)) + 1/(80 (1 + x)^5) - 3/(160 (1 + x)^3) - 3/(
 160 (1 + x)^2) - 1/(6400 (1 + x)) + (
 2 (5 I + Sqrt[5 (5 + 2 Sqrt[5])]))/(
 25 (Sqrt[50 - 10 Sqrt[5]] - 5 I (-3 + Sqrt[5]) - 
    4 Sqrt[5 (5 - 2 Sqrt[5])] x)) - (-2 + Sqrt[5])/(
 5 (1 + (-1)^(1/5)) (2 - 4 (-1)^(1/5) + (-1)^(2/5) + 2 (-1)^(3/5) + 
    I Sqrt[5 (5 - 2 Sqrt[5])] x)) + (
 2 (-I (-2 + Sqrt[5]) + 1/Sqrt[85 + 38 Sqrt[5]]))/(
 5 (Sqrt[250 - 110 Sqrt[5]] - 5 I (-1 + Sqrt[5]) + 
    4 Sqrt[5 (5 - 2 Sqrt[5])] x)) + (5 + Sqrt[5])/(
 25 (5 - Sqrt[5] + (-Sqrt[5] - I Sqrt[5 (5 - 2 Sqrt[5])]) x))

Question Is there away to stop Apart from simplifying the complex expression and output a nice result efficiently?

In addition, I would be grateful if anyone could that tell me there exist a readily made PFD package.

Thanks in advance.

Improve this question
$\endgroup$
4
  • $\begingroup$ is there a reason why you can't just called Apart? as in expr = (2 (-1 + x^2) (6 x - 3))/((1 + x^2)^2 (3 + x^2)^3); Apart[expr, x] gives !Mathematica graphics What is the purpose of your PFD function? $\endgroup$
    – Nasser
    Mar 17 at 6:09
  • $\begingroup$ Note the first line. I am trying to factor on complex numbers, that is, the denominator should be of the form $(x-z)^n$ where $z$ is one of its complex root (no quadratic terms in the power). $\endgroup$
    – Po1ynomial
    Mar 17 at 9:55
  • $\begingroup$ I originally supposed Apart also has the Extension option but found none. $\endgroup$
    – Po1ynomial
    Mar 17 at 9:56
  • $\begingroup$ OK, so why can't you always just use FullSimplify /@ % // ToRadicals afterwords like you showed if this fixes it and gives you the desired simplified output now? I also do not know of method to make it work automatically over complex. Maple does have one such method, but not Mathematica as builtin. $\endgroup$
    – Nasser
    Mar 17 at 14:11

1 Answer 1

Reset to default
3
$\begingroup$

Maybe something like this?

ResourceFunction["ExtendedApart"][(x^4)/((-1 + x^2)^4 (1 + x^5)), x]

(* Out[480]= (1 + (-1)^(1/5) + (-1)^(2/5))/(25 (-1)^(1/5) - 25 x) + (
 1 - (-1)^(1/5) + (-1)^(3/5))/(25 (-1)^(3/5) - 25 x) + 1/(
 32 (-1 + x)^4) - 1/(64 (-1 + x)^3) - 13/(128 (-1 + x)^2) + 41/(
 256 (-1 + x)) + 1/(80 (1 + x)^5) - 3/(160 (1 + x)^3) - 3/(
 160 (1 + x)^2) - 1/(6400 (1 + x)) + (-1 + (-1)^(2/5) - (-1)^(4/5))/(
 25 ((-1)^(2/5) + x)) + (-1 + (-1)^(3/5) + (-1)^(4/5))/(
 25 ((-1)^(4/5) + x)) *)
Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.