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I came across this bizarre behaviour whilst noodling around - a change in 'Plot' results...

testPlot1 = Plot[Log[HarmonicNumber[n]], 
   {n, 1, 5}]; 
testPlot2 = Plot[Log[HarmonicNumber[n]], 
   {n, 1, 43}]; 
Show[testPlot1, PlotRange -> {{0, 5}, {0, 1}}]
Show[testPlot2, PlotRange -> {{0, 5}, {0, 1}}]

The line moves away from the origin, crossing the n line increasingly far to the right, and giving no negative output - it just vanishes at the n line.

At first I thought it happened exclusively for n>=43, but actually, it seems to happen randomly for different domain ranges. Using Table gives the correct result.

What's going wrong, and how do I fix it?

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  • 1
    $\begingroup$ In the first case, you are not setting PlotRange at all, but in Show you are using specific PlotRange. So why do you expect the results to look the same? $\endgroup$
    – Nasser
    Commented Jun 17, 2019 at 12:54

2 Answers 2

Reset to default
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Use the plotoption AxesOrigin->{0,0}:

testPlot1 =Plot[Log[HarmonicNumber[n]], {n, 1, 5}, AxesOrigin -> {0, 0}] ;
testPlot2 =Plot[Log[HarmonicNumber[n]], {n, 1, 43}, AxesOrigin -> {0, 0}] ;

Show[testPlot1, PlotRange -> {{0, 5}, {0, 1}}]

enter image description here

Show[testPlot2, PlotRange -> {{0, 5}, {0, 1}}]

enter image description here

same plot in both cases!

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  • $\begingroup$ Hi @Ulirch. Do you mean AxesOrigin? (With an 'e') $\endgroup$
    – Richard Burke-Ward
    Commented Jun 17, 2019 at 12:49
  • $\begingroup$ Sorry: Yes I meant AxesOrigin $\endgroup$
    – Ulrich Neumann
    Commented Jun 17, 2019 at 12:54
  • $\begingroup$ I'm afraid that doesn't seem to work.If I specify AxesOrigin -> {0, 0} for testPlot2, the curve still hits the n line at n=1 - which is mathematcally incorrect. If I use it for testPlot1 and testPlot2 the curve still intersects the n line in different places for each plot. Mathematica is changing the actual values of the plot, not where the axes are. $\endgroup$
    – Richard Burke-Ward
    Commented Jun 17, 2019 at 12:55
  • $\begingroup$ I would expect Log[HarmonicNumber[1]]==0 $\endgroup$
    – Ulrich Neumann
    Commented Jun 17, 2019 at 13:02
  • $\begingroup$ You're right @Ulrich (and that was what I expected too). I must have entered something wrongly. Much appreciated. $\endgroup$
    – Richard Burke-Ward
    Commented Jun 17, 2019 at 13:03
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Another solution: increase the PlotPoints in testPlot2:

testPlot1 = Plot[Log[HarmonicNumber[n]], {n, 1, 5}];
testPlot2 = 
  Plot[Log[HarmonicNumber[n]], {n, 1, 43}, PlotPoints -> 100];
Show[testPlot1, PlotRange -> {{0, 5}, {0, 1}}]
Show[testPlot2, PlotRange -> {{0, 5}, {0, 1}}]

Mathematica graphics Mathematica graphics

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  • $\begingroup$ Hi @Chris. Also good. I have ticked Ulrich's because he got there first, but I appreciate the suggestion. $\endgroup$
    – Richard Burke-Ward
    Commented Jun 17, 2019 at 13:15
  • $\begingroup$ No prob -- nothing stops you from upvoting multiple answers BTW :) $\endgroup$
    – Chris K
    Commented Jun 17, 2019 at 13:24
  • $\begingroup$ The magic number of PlotPoints must be >5, but the option PlotPoints -> {Automatic, {1}}] works to#o! $\endgroup$
    – Ulrich Neumann
    Commented Jun 17, 2019 at 13:37

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