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

  • $\begingroup$ See this question: 68824. $\endgroup$ – Mahdi Jun 22 '15 at 17:47
  • $\begingroup$ @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 $\endgroup$ – grdgfgr Jun 22 '15 at 17:56
  • $\begingroup$ What is expected result for the first one? For the second one: Apart@Factor[1/(1 + x^2), Extension -> {(-1)^(1/2), I}]? $\endgroup$ – Mahdi Jun 22 '15 at 18:05

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.

  • $\begingroup$ would there be any way to do something like: ExpToTrig[ Apart[Factor[1/(1 + x^4), Extension -> Roots[1 + x^4 == 0, x]]]] $\endgroup$ – grdgfgr Jun 22 '15 at 18:27
  • $\begingroup$ @grdgfgr How about ExpToTrig[Apart[Factor[1/(1 + x^4), Extension -> (x /. Solve[1 + x^4 == 0, x])]]]? $\endgroup$ – kirma Jun 22 '15 at 18:36
  • $\begingroup$ @ 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. $\endgroup$ – Dr. Wolfgang Hintze Jun 22 '15 at 19:43

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