New answers tagged precision-and-accuracy
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[1,15`100] (* telling it ...
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