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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.

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  • $\begingroup$ t_2 is interpreted as 2 t_, rather than t subscript 2 as a first comment. You also, have not defined E[x,y] to be what you want it to be. And x is just a variable in the above, not a function of y. $\endgroup$ – Konstantinos Oct 15 '18 at 11:35
  • $\begingroup$ I see, apologies for the subscript mistake, I wrote it this way because I thought it was more clear. From a mathematical point of view, writing E(x,y)=0 immediately means that x is a function of y and y is a function of x, as the constraint of setting the algebraic equation equal to zero means that x and y aren't independent. $\endgroup$ – Aran Oct 15 '18 at 11:54
  • $\begingroup$ please have a look at the answer I gave. It's more of an extended comment, but hopefully it will give your Series. Cheers!!! $\endgroup$ – Konstantinos Oct 15 '18 at 11:58
  • $\begingroup$ @Aran Your equationE[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? $\endgroup$ – Ulrich Neumann Oct 15 '18 at 12:59
  • $\begingroup$ It isn't a linear function in y, it's cubic in y and quartic in x. It is the four solutions for x as a function of y that I require. I apologise, I believe I entered the equation for F[x,y] over two lines so when copied into mathematica it reads as two separate equations, one which has only a linear term in y and one which has the rest. I will correct this. $\endgroup$ – Aran Oct 15 '18 at 13:21
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It is helpful to do a series expansion around $\sqrt[4]{y}$ since Series is now extra careful when dealing with fractional powers. Also, rather than using Series around the solution, it is much faster to replace $y$ with $y + O(y^{-n})$. That is, work the series from the inside out instead of the outside in. The downside of this approach is that you don't know the order that will be produced.

Let's see how this works. Take the first solution and replace $y$ with $s^4$:

xsol = x /. First @ Solve[F[x, Subscript[t, 3] x^2 + Subscript[t, 2]x - y] == 0, x] /. y -> s^4;

Next, replace $s$ with $s + O(s^{-n})$:

Assuming[Subscript[c, 1] > 0 && Subscript[c, 2] > 0 && Subscript[t, 2] > 0 && Subscript[t, 3] > 0 && s>0,
    Table[
        s[n_] = Series[xsol /. s -> Series[s, {s, Infinity, n}], {s, Infinity, 4}] //Simplify,
        {n, 2, 14}
    ]
] //Column //TeXForm

$\begin{array}{l} -\frac{s^2}{\sqrt{t_3}}+O\left(\frac{1}{\frac{1}{s}}\right) \\ -\frac{s^2}{\sqrt{t_3}}+O\left(\frac{1}{\left(\frac{1}{s}\right)^{3/4}}\right) \\ -\frac{s^2}{\sqrt{t_3}}+O\left(\frac{1}{\sqrt{\frac{1}{s}}}\right) \\ -\frac{s^2}{\sqrt{t_3}}+O\left(\frac{1}{\sqrt[4]{\frac{1}{s}}}\right) \\ -\frac{s^2}{\sqrt{t_3}}+O\left(\left(\frac{1}{s}\right)^0\right) \\ -\frac{s^2}{\sqrt{t_3}}-\frac{t_2-1}{2 t_3}+O\left(\sqrt[4]{\frac{1}{s}}\right) \\ -\frac{s^2}{\sqrt{t_3}}-\frac{t_2-1}{2 t_3}+O\left(\sqrt{\frac{1}{s}}\right) \\ -\frac{s^2}{\sqrt{t_3}}-\frac{t_2-1}{2 t_3}+O\left(\left(\frac{1}{s}\right)^{3/4}\right) \\ -\frac{s^2}{\sqrt{t_3}}-\frac{t_2-1}{2 t_3}+O\left(\left(\frac{1}{s}\right)^1\right) \\ -\frac{s^2}{\sqrt{t_3}}-\frac{t_2-1}{2 t_3}-\frac{i}{2 s t_3^{5/4}}+O\left(\left(\frac{1}{s}\right)^2\right) \\ -\frac{s^2}{\sqrt{t_3}}-\frac{t_2-1}{2 t_3}-\frac{i}{2 s t_3^{5/4}}+\frac{-t_2^2+4 t_2-3}{8 s^2 t_3^{3/2}}+O\left(\left(\frac{1}{s}\right)^3\right) \\ -\frac{s^2}{\sqrt{t_3}}-\frac{t_2-1}{2 t_3}-\frac{i}{2 s t_3^{5/4}}+\frac{-t_2^2+4 t_2-3}{8 s^2 t_3^{3/2}}+\frac{i \left(t_2^2-4 t_2+3\right)}{16 s^3 t_3^{7/4}}+O\left(\left(\frac{1}{s}\right)^4\right) \\ -\frac{s^2}{\sqrt{t_3}}-\frac{t_2-1}{2 t_3}-\frac{i}{2 s t_3^{5/4}}+\frac{-t_2^2+4 t_2-3}{8 s^2 t_3^{3/2}}+\frac{i \left(t_2^2-4 t_2+3\right)}{16 s^3 t_3^{7/4}}+\frac{1}{4 s^4}+O\left(\left(\frac{1}{s}\right)^5\right) \\ \end{array}$

We see that in order to get a result good to order $O(s^{-5})$ we needed to replace $s$ with $s + O(s^{-15})$.

Finally, recall that $s$ is actually $\sqrt[4]{y}$. So, the final result is:

Series[Normal[s[14]] /. s->y^(1/4), {y, Infinity, 1}] //TeXForm

$-\frac{\sqrt{y}}{\sqrt{t_3}}-\frac{t_2-1}{2 t_3}-\frac{i \sqrt[4]{\frac{1}{y}}}{2 t_3^{5/4}}+\frac{\left(-t_2^2+4 t_2-3\right) \sqrt{\frac{1}{y}}}{8 t_3^{3/2}}+\frac{i \left(t_2^2-4 t_2+3\right) \left(\frac{1}{y}\right)^{3/4}}{16 t_3^{7/4}}+\frac{1}{4 y}+O\left(\left(\frac{1}{y}\right)^{5/4}\right)$

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This will solve for x(y); I think I have not made any mistakes

Ftest[y_] := 
 x[y]^4 - 2 y*
   x[y]^3 - (1/Subscript[t, 3]) y^3 + ((1 - Subscript[t, 2])/
     Subscript[t, 3]^2) y^2 + (x[y]^2)*(y^2) - ((Subscript[t, 2] + 2)/
     Subscript[t, 3])*y*
   x[y]^2 + ((Subscript[t, 3]^2 - Subscript[t, 2]^2)/
     Subscript[t, 3]^2)*
   x[y]^2 + ((Subscript[t, 2] + 2)/Subscript[t, 3])*x[y]*
   y^2 + ((Subscript[t, 2]^2 - Subscript[t, 3]^2)/
     Subscript[t, 3]^2) x[y]*y + Subscript[c, 1]*x[y] + Subscript[c, 2]

Solve[Ftest[y] == 0, x[y]]

And then hopefully you will be able to get the Taylor series, by defining each solution as a function of y. Not the most automated method, but it should do the trick.

If you want to make the Taylor faster, then I suggest that you do some Factor and Simplify but not in the full thing. Rather take smaller bits, simplify them and then reconstruct the result out of the simplified versions.

Hopefully helpful.

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  • $\begingroup$ Thank you for your answer! Unfortunately, I believe the code for extracting the series seems to give the same as mine. The procedure of factor and simplify has helped a lot though. Thanks again. $\endgroup$ – Aran Oct 15 '18 at 12:01
  • $\begingroup$ It kind of does need some time. Also, you might want to consider the command PowerExpand when you take each term individually. This might make things a bit simpler. Glad I could help, at least partially. $\endgroup$ – Konstantinos Oct 15 '18 at 12:02
  • 1
    $\begingroup$ @Konstantinos In the definition of Ftest I see a part x^4 which obviously should be ` x[y]^4`? $\endgroup$ – Ulrich Neumann Oct 15 '18 at 12:36
  • $\begingroup$ @UlrichNeumann yes, that was a mistake. I fixed it now. Thanks for pointing it out. $\endgroup$ – Konstantinos Oct 15 '18 at 12:38

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