How to plot this function in Mathematica?
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dirichlet[x_] := If[IntegerQ[Numerator[Rationalize[x]]], 1, 0]
dirichlet[1.24898]
dirichlet[Pi]
1
0
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$\begingroup$ How about dirichlet[N[Sqrt[Pi/Exp[1]], 10]]? $\endgroup$ – user64494 Dec 19 '17 at 18:10
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$\begingroup$ @user64494 gives me 0 in MMA11.1 - is that wrong? $\endgroup$ – Sumit Dec 20 '17 at 7:16
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$\begingroup$ @Sumit: Up to the definition, it should result 1. $\endgroup$ – user64494 Jan 19 '18 at 8:19
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$\begingroup$ @Kuba, I don't have a rational answer for that. $\endgroup$ – Sumit Jan 19 '18 at 9:00
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Faked by ListLinePlot and improved code borrowed from bill s
d[x_] := Piecewise[{{1, Abs[Rationalize[x, 0.0003] - x] > 0.0002}, {0,True}}];
ListLinePlot[Table[{x, d[x]}, {x, -1/2, 1, 0.0001}],
MeshStyle -> PointSize[Small], Mesh -> {Range[-1/2, 1, 0.0001]},
MeshFunctions -> {#1 &}, MeshShading -> {White, White},
PlotRange -> {Automatic, {-1/2, 3/2}},
PlotLegends -> {"Dirichlet function"}]
answered Jan 19 '18 at 8:13
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Of course you can't really plot this function, since it is discontinuous everywhere. But you can fake it:
d[x_] := Piecewise[{{1, Abs[Rationalize[x, 0.01] - x] > 0.004}, {0, True}}];
DiscretePlot[d[x], {x, 0, 1, 0.001}]
By playing with the second argument of the Rationalize and the value in the inequality, you can change the detailed appearance of the function.
Rationalize$\endgroup$ – Sumit Dec 19 '17 at 14:41Plot[{0, 1}, {x, -1, 2}, PlotStyle -> Black, PlotRange -> {Automatic, {-1/2, 3/2}}]. Anyway, you can't really plot Dirichlet function... $\endgroup$ – anderstood Jan 18 '18 at 16:15Thomae's functionclick here: mathworld.wolfram.com/notebooks/NumberTheory/… $\endgroup$ – Mariusz Iwaniuk Jan 19 '18 at 16:47