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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?

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2  
$n (-1)^n$ is real only for integer $n$ and complex for everything else. Plot[] skips complex values, so... –  J. M. Apr 17 '13 at 0:41
    
Use this instead Plot[Re[x Exp[i x \[Pi] ]], {x, -1, 1}] –  Spawn1701D Apr 17 '13 at 0:42
    
...or Plot[n (-1)^Floor[n], {n, -5, 5}]. –  J. M. Apr 17 '13 at 0:45
    
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). –  Spawn1701D Apr 17 '13 at 0:49

3 Answers 3

up vote 10 down vote accepted

Take what you need:

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

enter image description here

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2  
I can't help but to read Mr. Wizards posts in Professor Hubert J. Farnsworth's voice. –  jmlopez Apr 17 '13 at 3:01

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!

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And what about the point $\frac{2}{3}$ for example? It is a real function there and to many other like this points. –  Spawn1701D Apr 17 '13 at 9:31
    
@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. –  J. M. Apr 17 '13 at 9:32
    
Actually 1 is the principal value for the odd roots the rest are complex of course. –  Spawn1701D Apr 17 '13 at 9:36
    
@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. –  J. M. Apr 17 '13 at 9:45
    
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. –  Spawn1701D Apr 17 '13 at 9:53

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

ListPlot[Table[n (-1)^n, {n, 1, 20}], PlotStyle -> PointSize[Large]]
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2  
OP did say that he has used DiscretePlot[]. –  J. M. Apr 17 '13 at 3:31

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