Why does Apart not give full solutions? E.g.:
ApartSquareFree[x/(x^1 + x^2 + x^3 + x^4 + x^5), x]
1/(1 + x + x^2 + x^3 + x^4)
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$\begingroup$ What would you like it to give? $\endgroup$– mikadoCommented Sep 15, 2016 at 18:58
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$\begingroup$ a sum of terms with $a_{ij}/(x-x_i)^j$ (or $(b_{ij}x+c_{ij})/(x-z_i)^j(x-\bar{z_i})^j$) $\endgroup$– divBCommented Sep 15, 2016 at 19:00
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1 Answer
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This works for your specific problem:
Clear[y]
y = x/(x^1 + x^2 + x^3 + x^4 + x^5);
ApartSquareFree[
Numerator[y]/Factor[Denominator[y], Extension -> GoldenRatio]]
$\frac{2 x+\sqrt{5}+1}{\sqrt{5} \left(2 x^2+\sqrt{5} x+x+2\right)}+\frac{-2 x+\sqrt{5}-1}{\sqrt{5} \left(2 x^2-\sqrt{5} x+x+2\right)}$
