1
$\begingroup$

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 $4$.

EDIT

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?

$\endgroup$
4
$\begingroup$

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}
}
$\endgroup$
1
  • $\begingroup$ Hmmm... that seems as a new feature of mathematica. I have version 11. Anyway, I accept your answer. $\endgroup$ Dec 31 '19 at 21:12

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.