3
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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
6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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