Partial fraction expansion with complex numbers

Question

$$\text{Apart}\left[\frac{1}{x^4+1}\right]$$

Does nothing. How can I get it to expand it. Sometimes it is useful.

@Mahdi $$\text{Apart}\left[\text{Factor}\left[\frac{1}{x^4+1},\text{Extension}\to i\right]\right]$$ gives only $$\frac{i}{2 \left(x^2+i\right)}-\frac{i}{2 \left(x^2-i\right)}$$ and $$\text{Apart}\left[\text{Factor}\left[\frac{1}{x^4+2},\text{Extension}\to i\right]\right]$$ straight up does not work — Gappy Hilmore, Jun 22 '15 at 17:56
What is expected result for the first one? For the second one: Apart@Factor[1/(1 + x^2), Extension -> {(-1)^(1/2), I}]? — Mahdi, Jun 22 '15 at 18:05
Your question is a partial case of the Mittag-Leffler's theorem for the expansion of meromorphic functions. There is a nice demo here. — yarchik, Sep 27 '20 at 15:47

score 8 · Accepted Answer · answered Jun 22 '15 at 18:10

8

I found by trial and error that Extension-> Sqrt[I] does the job.

ExpToTrig[Apart[Factor[1/(1 + x^4), Extension -> Sqrt[I]]]]

$$\frac{\frac{1}{4}+\frac{i}{4}}{\sqrt{2} \left(-x+\frac{1+i}{\sqrt{2}}\right)}+\frac{\frac{1}{4}+\frac{i}{4}}{\sqrt{2} \left(x+\frac{1+i}{\sqrt{2}}\right)}-\frac{\frac{1}{4}-\frac{i}{4}}{\sqrt{2} \left(-x-\frac{1-i}{\sqrt{2}}\right)}-\frac{\frac{1}{4}-\frac{i}{4}}{\sqrt{2} \left(x-\frac{1-i}{\sqrt{2}}\right)}$$

Here ExpToTrig is not really required but it does the final beautifying.

answered Jun 22 '15 at 18:10

Dr. Wolfgang Hintze

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would there be any way to do something like: ExpToTrig[ Apart[Factor[1/(1 + x^4), Extension -> Roots[1 + x^4 == 0, x]]]] – Gappy Hilmore Jun 22 '15 at 18:27
@grdgfgr How about ExpToTrig[Apart[Factor[1/(1 + x^4), Extension -> (x /. Solve[1 + x^4 == 0, x])]]]? – kirma Jun 22 '15 at 18:36
@ kirma : that seems to be the general rule setting the Extension to the roots of the polynomial in question. From Help: Extension is an option for various polynomial and algebraic functions that specifies generators for the algebraic number field to be used. – Dr. Wolfgang Hintze Jun 22 '15 at 19:43
@GappyHilmore Apart[Factor[1/(1 + x^4), Extension -> All]] works and so does Apart[Factor[1/(2 + x^4), Extension -> All]] (in V13 but maybe it didn't back then). – Michael E2 May 05 '22 at 13:37

score 3 · Answer 2 · answered May 05 '22 at 06:04

3

There's an internal function:

Integrate`ComplexApart[1/(1 + x^4), x]
(*
-((-1)^(1/4)/(4 (-(-1)^(1/4) + x))) +
 (-1)^(1/4)/(4 ((-1)^(1/4) + x)) -
 (-1)^(3/4)/(4 (-(-1)^(3/4) + x)) +
 (-1)^(3/4)/(4 ((-1)^(3/4) + x))
*)

answered May 05 '22 at 06:04

Michael E2

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fyi, in Maple they have 2 functions. parfrac and fullparfrac for using complex domain. screen shot It looks like fullparfrac is like ComplexApart you show. – Nasser May 05 '22 at 08:13
@Nasser With so many top-level (System) functions in Mathematica, you would think ComplexApart would be among them. – Michael E2 May 05 '22 at 13:33

Partial fraction expansion with complex numbers

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