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I wanted to plot a function involving Floor, here is my code:

epsilon = 0.1;
f[t_] = Floor[t/epsilon];
Plot[{f[t]}, {t, 0, 5}]

it gives me the following, which has smaller bars as t increases: enter image description here

I know that I can plot Floor with DiscretePlot, as suggested here, but I would like to understand, why Plot cannot handle it correctly. Thanks!

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  • $\begingroup$ Related: (29346) $\endgroup$
    – Mr.Wizard
    Commented Apr 15, 2015 at 16:07

1 Answer 1

Reset to default
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You need to specify more PlotPoints

epsilon = 0.1;
f[t_] = Floor[t/epsilon];

Plot[{f[t]}, {t, 0, 5},
 PlotPoints -> 250,
 ImageSize -> 500]

enter image description here

epsilon = 0.5;
f[t_] = Floor[t/epsilon];

Plot[{f[t]}, {t, 0, 5},
 ColorFunction -> "Rainbow",
 PlotPoints -> 250,
 PlotStyle -> Thickness[0.01],
 PlotTheme -> "Detailed",
 ImageSize -> 500]

enter image description here

epsilon = 0.5;
f[t_] = Floor[t/epsilon];

Plot[{f[t]}, {t, 0, 5},
 Filling -> Axis,
 Frame -> True,
 FrameTicks -> {{All, None}, {Range[0, 5, 0.5], None}},
 GridLines -> {Range[0, 5, 0.5], Range[0, 10, 1]},
 PlotPoints -> 250,
 PlotStyle -> Thickness[0.001],
 ImageSize -> 500]

enter image description here

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  • $\begingroup$ thanks, that solves the problem. but do you have an intuitive understanding why mathematica decides to make the bars smaller? So, why does it work better for small t as for large t? $\endgroup$
    – Mario Krenn
    Commented Oct 5, 2014 at 16:57
  • $\begingroup$ @NicoDean I don't understand it either. With too few PlotPoints you just get unpredictable results. I don't even see a pattern "within" this unpredictability. $\endgroup$
    – eldo
    Commented Oct 5, 2014 at 17:13
  • 2
    $\begingroup$ The bar length is determined by where the adaptive sampling for Plot selects the plot points. This is affected by the initial sampling points set by the option PlotPoints. $\endgroup$
    – Bob Hanlon
    Commented Oct 5, 2014 at 19:26

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