Separation of a rational fraction

Questions
1

Hi
i do not understand why :

Apart[1/(1+x^4)]

does not calculate something like

(1+2x^2+x^4)-2x^2

etc…
Is there no means to apart really all rational fractions ?

2

Hi again Adrien,

you probably mean:

expr = 1/(1 + x)^4;
ExpandDenominator[expr]

Regards – Henrik

3

(1) I do not see how your proposed result is related to your input.

(2) Apart gives a partial fraction decomposition over the rationals, that is, the denominator is factored over Q.

4

Hi Henrik my aim is not expanding but separating into rational fraction hi Daniel, thanks for the information that Apart works in Q ; i had not seen that in the help menu of Apart now, how can i separate this ? :

Apart[1/(x^4 + 1)]
Apart[1/(x^2 - 2)]
5

Is this what you need?

Apart[Factor[1/(1 + x^4), Extension -> {Sqrt[I]}]]
Apart[Factor[1/(x^2 - 2), Extension -> {Sqrt[2]}]]
6

Hi Gianluca

yes for the second :

Apart[Factor[1/(x^2 - 2), Extension -> {Sqrt[2]}]]

But the question is : do i have to calculate by hand the roots of the denominator each time i want Mathematica to separate my f(X) in simple elements ?

For the first one :

  • how do you think i could avoid these unusual (for me) (-1)^{1/4}, in order to have for example an exponential form exp(I Pi/4) instead ?

  • how could i get the separation in simple elements in R, i.e. how make Mathematica guess without my help that :

      1/((1 + x^2 + x*Sqrt[2]) (1 + x^2 - x*Sqrt[2])) == 1/(x^4 + 1)

and then give the decomposition ?

7

I am not an expert, however you can make Mathematica find the appropriate extension, and you can use ExpToTrig to convert the unusual powers:

Apart[Factor[1/(x^2 - 2), Extension -> (x /. Solve[x^2 - 2 == 0])]]
Apart[Factor[1/(x^4 + 1), 
  Extension -> (x /. Solve[x^4 + 1 == 0] // ExpToTrig)]]
8

fine, thank you very much