This two one-line codes should represent the same thing, i.e. the first root of polynomial in $y$:

(y/.Solve[y^4+x y^3-x^2+x y==0,y])[[1]]
Root[y^4+x y^3-x^2+x y/.y->#1&,1]

Then I computed identical series of the two expressions at $x=0$.

Input 1:

Series[(y/.Solve[y^4+x y^3-x^2+x y==0,y])[[1]],{x,0,3}]//Normal

Output 1:

$\frac{2 x^{5/3}}{9}-\frac{34 x^{7/3}}{81}+\frac{2 x^3}{3}-\sqrt[3]{x}-\frac{2 x}{3}$

Input 2:

Series[Root[y^4+x y^3-x^2+x y/.y->#1&,1],{x,0,3}]//Normal

Output 2:

-some error message-

But why I got right answer in the first case and error message in the second one?

Is this correct behavior?

What if I wanted to compute the same way series of $y^5+x y^3-x^2+x y$. Notice I changed the first term $y^4$ to $y^5$ and in this case output of "Solve" is in terms of "Root" because then radicals are impossible for 5th degree polynomial so I am unable to compute such series. How to do it also in this case? If I am not mistaken this series are called "Puiseux expansion". I can not find any other command in Matheamtica to compute such expansion for polynomials of degree bigger than $4$.


I just noticed I got the same answer also in the second case but also with an error message because of which I overlooked the answer.

Anyway, is there a better method in Mathematica to compute such series - Puiseux expansion?


1 Answer 1


You can use AsymptoticSolve:

AsymptoticSolve[y^4 + x y^3 - x^2 + x y == 0, {y, 0}, {x, 0, 3}, Reals]
  {y -> x - 2 x^3}, 
  {y -> -x^(1/3) - 2x/3 + 2x^(5/3)/9 - 34x^(7/3)/81 + 2x^3/3}
  • $\begingroup$ Hmmm... that seems as a new feature of mathematica. I have version 11. Anyway, I accept your answer. $\endgroup$ Commented Dec 31, 2019 at 21:12

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