2
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While solving quartic equation

isinh = t /.
Solve[-1 + (2 t)/x^2 - (-1 + 1/x^4 + e^2)\t^2 - 
(2\t^3)/x^2 + t^4/x^4 == 0, \t][[2]];
x = 10;
Plot[N[Im[isinh]], {e, 0, 2}, PlotRange -> Full,
Exclusions -> None, WorkingPrecision -> Infinity]

I tried playing with imaginary increments under square root. Do you think there is a way to avoid this singularity? If it requires hard hand work with Piecewise, do you think I can just omit it on the plot? (give me a hint how to plot it without singularity)

enter image description here

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1 Answer 1

4
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If you mean the vertical line, this is created by only one point.

p = Plot[N[Im[isinh]], {e, 0, 2}, PlotRange -> Full, 
   Exclusions -> None, WorkingPrecision -> Infinity];
list = Cases[p, Line[x_] :> x, -1];

ListPlot[list, PlotRange -> All]

enter image description here

If you delete this point then:

point = Cases[p, {x_ /; Abs[x - 1] < 0.01, y_ /; y < 0.5}, -1]

(*{{1., 0.0903386}}*)

DeleteCases[p, point[[1]], -1]

enter image description here

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1
  • $\begingroup$ Hi, the code failed in v13.3. $\endgroup$
    – lotus2019
    Aug 13, 2023 at 10:18

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