Consider following expression: $$ \sqrt{-\left(2 \sqrt{x^{-8/3}-x^{-4/3}+1}\, x^{4/3}+5\right) x^{4/3}+4 \sqrt{x^{-8/3}-x^{-4/3}+1}\, x^{4/3}+2 x^{8/3}+5} $$ In code:
f[x_] := Sqrt[
-(2 Sqrt[x^(-8/3) - x^(-4/3) + 1] x^(4/3) + 5) x^(4/3)
+ 4 Sqrt[x^(-8/3) - x^(-4/3) + 1] x^(4/3) + 2 x^(8/3) + 5
]
Why in Mathematica one can see non smoothness at $x\sim10^4$?
Plot[f[x], {x, 0, 10^4}]
I guess, it is a numerical issue, but then how one can deal with it?
In addition to using WorkingPrecision as pointed out in the comments, it is often more efficient and reduces numerical noise to Simplify (or in some cases FullSimplify) the function's definition.
Original definition
f1[x_] := Sqrt[-(2 Sqrt[x^(-8/3) - x^(-4/3) + 1] x^(4/3) + 5) x^(4/3) +
4 Sqrt[x^(-8/3) - x^(-4/3) + 1] x^(4/3) + 2 x^(8/3) + 5]
Simplify once by using Set rather than SetDelayed.
f2[x_] = Sqrt[-(2 Sqrt[x^(-8/3) - x^(-4/3) + 1] x^(4/3) + 5) x^(4/3) +
4 Sqrt[x^(-8/3) - x^(-4/3) + 1] x^(4/3) + 2 x^(8/3) + 5] // Simplify;
Or use Evaluate with SetDelayed.
f3[x_] := Evaluate[
Sqrt[-(2 Sqrt[x^(-8/3) - x^(-4/3) + 1] x^(4/3) + 5) x^(4/3) +
4 Sqrt[x^(-8/3) - x^(-4/3) + 1] x^(4/3) + 2 x^(8/3) + 5] // Simplify]
Timing[Plot[#[x], {x, 0, 10^4}, WorkingPrecision -> 15];][[1]] & /@ {f1, f2,
f3}
{1.472757, 0.066547, 0.066083}
WorkingPrecision:Plot[f[x], {x, 0, 10^4}, WorkingPrecision -> 20]gives this. – Öskå Aug 08 '14 at 10:04WorkingPrecision -> Infinityfrom an answer to [7019] does not work on this example; butPPlot[N[f[x], 8], {x, 0, 10^4}, Exclusions -> None, WorkingPrecision -> Infinity]does. I'll propose [3152] as a duplicate. – Michael E2 Aug 08 '14 at 12:59