I have the following algebraic equation:
F[x_, y_] := x^4 - 2*x^3*y + Subscript[c, 1] + y*Subscript[c, 2] +
y^2*(1/Subscript[t, 3]^2 - Subscript[t, 2]/Subscript[t, 3]^2) +
x^2*(1 + y^2 - Subscript[t, 2]^2/Subscript[t, 3]^2 +
(y*(-2 - Subscript[t, 2]))/Subscript[t, 3]
) +
x*(y*(-1 + Subscript[t, 2]^2/Subscript[t, 3]^2) +
(y^2*(2 + Subscript[t, 2]))/Subscript[t, 3]) -
y^3/Subscript[t, 3]
or:
$$F(\text{x$\_$},\text{y$\_$})\text{:=}x^4-2 x^3 y+c_1+y c_2+y^2 \left(\frac{1}{t_3^2}-\frac{t_2}{t_3^2}\right)+x^2 \left(1+y^2-\frac{t_2^2}{t_3^2}+\frac{y \left(-2-t_2\right)}{t_3}\right)+x \left(y \left(-1+\frac{t_2^2}{t_3^2}\right)+\frac{y^2 \left(2+t_2\right)}{t_3}\right)-\frac{y^3}{t_3}$$
where $c_1,c_2,t_3,t_2$ are all constants.
From this I can construct the following curve, $E(x,y)=F(x,t_3 x^2 + t_2 x -y)$, and I would like to find the $x(y)$ solutions, $E(x(y),y)=0$.
This part is easy, just solve the above equation for x. Since $E(x,y)$ is of degree 4 in $x$ you get 4 solutions for $x(y)$.
What I want to do is compute the asymptotic expansion of each of these solutions. I have tried
sol=Solve[E[x,y]==0,x]
Series[
x/.sol[[i]],
{y,Infinity,1},
Assumptions->{y>0,Subscript[t,3]>0,Subscript[t,2]>0,Subscript[c,1]>0,Subscript[c,2]>0}
]
where $i$ denotes one of the four solutions of $E$, but for some reason the computation is taking an exceedingly long time.
I suspect that either I have made a mistake at some point, not noticed some deficiency in the solution, or I am not using the most efficient method, but I can't seem to identify anything I have done wrong.
Any help with this computation would be greatly appreciated.
E[x,y]==0
is a linear function in y, so it's easy to evaluate the unique solution ` y=f[x]`. What's your reason to use the inverse of this function? $\endgroup$