I was in doubt if this question should be asked in Mathematics.SE or Mathematica.SE, but I've decided this site would be more appropriate, because I think those who read my question here will know about both mathematics and Mathematica.

I tried to plot $n(-1)^n$ with Plot, but it showed nothing. When I tried to plot it with DiscretePlot, it worked. Why did that happen?

  • 2
    $\begingroup$ $n (-1)^n$ is real only for integer $n$ and complex for everything else. Plot[] skips complex values, so... $\endgroup$ Apr 17, 2013 at 0:41
  • 1
    $\begingroup$ Use this instead Plot[Re[x Exp[i x \[Pi] ]], {x, -1, 1}] $\endgroup$
    – Spawn1701D
    Apr 17, 2013 at 0:42
  • $\begingroup$ ...or Plot[n (-1)^Floor[n], {n, -5, 5}]. $\endgroup$ Apr 17, 2013 at 0:45
  • $\begingroup$ Actually $(-1)^n$ becomes a multivariable function. It may take a real value apart from when $n$ is integer (but of course not when $n$ is irrational). $\endgroup$
    – Spawn1701D
    Apr 17, 2013 at 0:49

3 Answers 3


Take what you need:

  {Re, Im, Arg, Abs}[n (-1)^n] // Through,
  {n, 1, 10},
  Evaluated -> True

enter image description here

  • 4
    $\begingroup$ I can't help but to read Mr. Wizards posts in Professor Hubert J. Farnsworth's voice. $\endgroup$
    – jmlopez
    Apr 17, 2013 at 3:01

Perhaps @Gustavo Bandeira intended just discrete values of n. Then:

ListPlot[Table[n (-1)^n, {n, 1, 20}], PlotStyle -> PointSize[Large]]
  • 2
    $\begingroup$ OP did say that he has used DiscretePlot[]. $\endgroup$ Apr 17, 2013 at 3:31

See the mage of Araby
Get your functions, old and new
With a two-year guarantee
And a choice of colors, too

Map[Function[f, Plot[n (-1)^f[n], {n, -10, 10}, PlotLabel -> (n (-1)^f[n])]],
    {{Floor, Ceiling}, {Round, IntegerPart}}, {2}] // GraphicsGrid

Come on, pick one, sweetheart!

  • $\begingroup$ And what about the point $\frac{2}{3}$ for example? It is a real function there and to many other like this points. $\endgroup$
    – Spawn1701D
    Apr 17, 2013 at 9:31
  • $\begingroup$ @Spawn, Ah, for that you don't want the principal value of Power[]. I understand that Mathematica these days has the Surd[] function; that would be useful in the situation you think of. $\endgroup$ Apr 17, 2013 at 9:32
  • $\begingroup$ Actually 1 is the principal value for the odd roots the rest are complex of course. $\endgroup$
    – Spawn1701D
    Apr 17, 2013 at 9:36
  • $\begingroup$ @Spawn, I don't quite think so. To take $(-1)^{2/3}$ as an example, the real value is not the principal value, at least if we want our power function to have a branch cut that agrees with our branch cut choice for the logarithm. $\endgroup$ Apr 17, 2013 at 9:45
  • $\begingroup$ Just run it on wolfram alpha it will show you that 1 is the principal value. I suppose is a different definition for different purposes. $\endgroup$
    – Spawn1701D
    Apr 17, 2013 at 9:53

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