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2

Using the option MaxRecursion reduces the number of PlotPoints needed. Clear["Global*"] f[x_, y_] := 2 (-4 + x^2) Sinh[(π x)/3] + 1/16 (((4 + x^2)^2 + 64 (-4 + x^2) Cos[y] Cosh[(2 π x)/3] + 256 x Sin[y] Sinh[(2 π x)/3]) Sinh[π x] - 2 (4 + x^2)^2 Sinh[(5 π x)/3] + (-12 + x^2)^2 Sinh[(7 π x)/3]); ContourPlot[f[x, y], {x, 3....

3

You can solve for $y$ to get the two level sets. Note that the plot in the question is inaccurate, even for a high number of plot points, and it seems to think they are left/right, but actually they are top/bottom: {f1, f2} = y /. Solve[f[x, y] == 0, y] /. C[1] -> 0; Plot[{f1, f2}, {x, 3.45, 3.5}] These expressions are both ArcTan of a large inner ...

1

Not a complete answer but one direction to go in: cp = ContourPlot[f[x, y] == 0, {x, 3.465728, 3.465729}, {y, 1.046786, 1.046795}, PlotPoints -> 500] {l1, l2} = Cases[Normal[cp], _Line, Infinity]; {{x1, y1}} = MinimalBy[First@l1, RegionDistance[l2]]; {{x2, y2}} = MinimalBy[First@l2, RegionDistance[l1]]; Show[ cp, ListPlot[{{x1, y1}, {x2, y2},}, ...

0

You're feeding it with the same precision so it's hard to expect different results, I think. In Mathematica itself: SphericalBesselY[1, 15.] // Accuracy will give you 17.35 [decimal] digits of accuracy. I'm not sure how WSTP treats 128bit floating numbers though By contrast: SphericalBesselY[1, 15.] (* -0.0399761*) SphericalBesselY[1,15100] (* telling it ...

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