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I have an algebraic curve given by

F1[x_,y_] = x^4-2 x^3 y+Subscript[c, {5,1}]+y Subscript[c, {5,2}]+y^2(1/Subsuperscript[t, 3, 2]-Subscript[t, 2]/Subsuperscript[t, 3, 2])+x^2 (1+y^2-Subsuperscript[t, 2, 2]/Subsuperscript[t, 3, 2]+(y (-2-Subscript[t,2]))/Subscript[t, 3])+x (y (-1+Subsuperscript[t, 2, 2]/Subsuperscript[t, 3, 2])+(y^2 (2+Subscript[t, 2]))/Subscript[t, 3])-y^3/Subscript[t, 3]

where $t_3, t_2, c_{(5,1)}$ and $c_{(5,2)}$ are constants, fixed to make the curve genus $0$. It is possible to find the three solutions to this curve $y_i(x)$ for $i=1,2,3$ asymptotically.

F1sol = Solve[F1[x,y]==0,y];
Collect[PowerExpand[Normal[Series[y /. F1sol[[1]], {x, \[Infinity], 1},Assumptions -> x > 0]]],x]

and the same for the other two solutions.

Now I have a modified curve, given by

Fh[x_,y_]=x^4-4 x^3 y \[Alpha]+1/16 y^4 (1-4 \[Alpha]^2)^2+Subscript[c, 
{5,1}]+y Subscript[c, {5,2}]+x^2 (1+y^2 (-(1/2)+6 \[Alpha]^2)-
Subsuperscript[t, 2, 2]/Subsuperscript[t, 3, 2]+(y (-2-Subscript[t, 
2]))/Subscript[t, 3])+x (-y^3 \[Alpha] (-1+4 \[Alpha]^2)+y (-2 \[Alpha]+(2 \
[Alpha] Subsuperscript[t, 2, 2])/Subsuperscript[t, 3, 2])+(2 y^2 \[Alpha] 
(2+Subscript[t, 2]))/Subscript[t, 3])+(y^3 (-2-8 \[Alpha]^2-(-1+4 \
[Alpha]^2) Subscript[t, 2]))/(4 Subscript[t, 3])+(y^2 (4-4 Subscript[t, 2]+(1-4 \[Alpha]^2) Subsuperscript[t, 2, 2]+(-1+4 \[Alpha]^2) Subsuperscript[t, 3, 2]))/(4 Subsuperscript[t, 3, 2])

which reduces to the previous curve for $\alpha \rightarrow \frac{1}{2}$.

Now when I find the solutions to this curve I have four solutions for $y_i (x)$.

However, I can't seem to find the correct way to take the limit of $\alpha \rightarrow \frac{1}{2}$ such that I recover the solutions to the previous curve (one of the solutions will blow up in this limit since it is a singular perturbation).

I have attempted doing a series expansion about $\alpha = \frac{1}{2}$ and taking only the bits that don't disappear as in the limit, and then doing a series expansion about $x=\infty$ and this recovers some of the terms in the original solution, but not the non-analytic terms (such as $\sqrt{x}$ and $1/\sqrt{x}$ terms).

Normal[Series[Normal[Series[y /. F4sol[[4]], {\[Alpha], 1/2, 0},  
Assumptions -> \[Alpha] > 0]], {x, \[Infinity], 1}, Assumptions -> x > 0]]

Is there any way of implementing the limit in mathematica that correctly reproduces the solutions of the original equation?

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  • $\begingroup$ Could you include the code for the definition of your functions? Some people might not be too keen to copy it out by hand. $\endgroup$ – aardvark2012 Sep 21 '17 at 11:20

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