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Whiffee Bollenbach

Full-range plotting with ParametricPlot?

Whiffee Bollenbach, retired

Posted 4 months ago

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A puzzle with ParametricPlot. The first plot masks part of the range, but nevertheless plots test points in the masked area. Can I get rid of the mask? The second plot shows no mask(s), but I don't know why it works. Can I get an informed comment? Thanks for your time, Whiffee Bollenbach d2 = ImplicitRegion[ 0 <= x < \[Infinity] \[And] -\[Infinity] < y < \[Infinity], {x, y}]; p2 = ParametricPlot[ Through[{Re, Im}[(x + I y)^0.5]], {x, y} \[Element] d2, PlotRange -> {{-1, 3.5}, {-3, 3}}, Frame -> True, ImageSize -> 200, AspectRatio -> Automatic, Epilog -> {{Blue, PointSize[0.025], Point[{Re[(x + I y)^0.5] /. {x -> 0, y -> -10}, Im[(x + I y)^0.5] /. {x -> 0, y -> -10}}]}, {Red, PointSize[0.025], Point[{Re[(x + I y)^0.5] /. {x -> 0, y -> -1}, Im[(x + I y)^0.5] /. {x -> 0, y -> -1}}]}, {Black, PointSize[0.025], Point[{Re[(x + I y)^0.5] /. {x -> 0, y -> 1}, Im[(x + I y)^0.5] /. {x -> 0, y -> 1}}]}}]; p3 = ParametricPlot[ Through[{Re, Im}[x + (I y)^3.5]], {x, y} \[Element] d2, PlotRange -> {{-1, 3.5}, {-3, 3}}, Frame -> True, ImageSize -> 200, AspectRatio -> Automatic, Epilog -> {{Blue, PointSize[0.025], Point[{Re[(x + I y)^0.5] /. {x -> 0, y -> -10}, Im[(x + I y)^0.5] /. {x -> 0, y -> -10}}]}, {Red, PointSize[0.025], Point[{Re[(x + I y)^0.5] /. {x -> 0, y -> -1}, Im[(x + I y)^0.5] /. {x -> 0, y -> -1}}]}, {Black, PointSize[0.025], Point[{Re[(x + I y)^0.5] /. {x -> 0, y -> 1}, Im[(x + I y)^0.5] /. {x -> 0, y -> 1}}]}}]; Row[{p2, p3}]

A puzzle with ParametricPlot. The first plot masks part of the range, but nevertheless plots test points in the masked area. Can I get rid of the mask? The second plot shows no mask(s), but I don't know why it works. Can I get an informed comment? Thanks for your time,

Whiffee Bollenbach

d2 = ImplicitRegion[
   0 <= x < \[Infinity] \[And] -\[Infinity] < y < \[Infinity], {x, y}];

p2 = ParametricPlot[
   Through[{Re, Im}[(x + I y)^0.5]], {x, y} \[Element] d2, 
   PlotRange -> {{-1, 3.5}, {-3, 3}}, Frame -> True, ImageSize -> 200,
    AspectRatio -> Automatic, 
   Epilog -> {{Blue, PointSize[0.025], 
      Point[{Re[(x + I y)^0.5] /. {x -> 0, y -> -10}, 
        Im[(x + I y)^0.5] /. {x -> 0, y -> -10}}]}, {Red, 
      PointSize[0.025], 
      Point[{Re[(x + I y)^0.5] /. {x -> 0, y -> -1}, 
        Im[(x + I y)^0.5] /. {x -> 0, y -> -1}}]}, {Black, 
      PointSize[0.025], 
      Point[{Re[(x + I y)^0.5] /. {x -> 0, y -> 1}, 
        Im[(x + I y)^0.5] /. {x -> 0, y -> 1}}]}}];

p3 = ParametricPlot[
   Through[{Re, Im}[x + (I y)^3.5]], {x, y} \[Element] d2, 
   PlotRange -> {{-1, 3.5}, {-3, 3}}, Frame -> True, ImageSize -> 200,
    AspectRatio -> Automatic, 
   Epilog -> {{Blue, PointSize[0.025], 
      Point[{Re[(x + I y)^0.5] /. {x -> 0, y -> -10}, 
        Im[(x + I y)^0.5] /. {x -> 0, y -> -10}}]}, {Red, 
      PointSize[0.025], 
      Point[{Re[(x + I y)^0.5] /. {x -> 0, y -> -1}, 
        Im[(x + I y)^0.5] /. {x -> 0, y -> -1}}]}, {Black, 
      PointSize[0.025], 
      Point[{Re[(x + I y)^0.5] /. {x -> 0, y -> 1}, 
        Im[(x + I y)^0.5] /. {x -> 0, y -> 1}}]}}];

Row[{p2, p3}]

POSTED BY: Whiffee Bollenbach

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Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 4 months ago

You may have just misplaced a parenthesis in `x + (I y)^3.5`: ParametricPlot[ReIm[(x + I y)^3.5], {x, y} \[Element] d2, PlotRange -> {{-1, 3.5}, {-3, 3}}, Frame -> True, AspectRatio -> Automatic, Epilog -> {{Blue, PointSize[0.025], Point[ReIm[(x + I y)^0.5] /. {{x -> 0, y -> -10}, {x -> 0, y -> -1}, {x -> 0, y -> 1}}]}}]

You may have just misplaced a parenthesis in x + (I y)^3.5:

ParametricPlot[ReIm[(x + I y)^3.5], {x, y} \[Element] d2, 
 PlotRange -> {{-1, 3.5}, {-3, 3}}, Frame -> True, 
 AspectRatio -> Automatic, 
 Epilog -> {{Blue, PointSize[0.025], 
    Point[ReIm[(x + I y)^0.5] /. {{x -> 0, y -> -10}, {x -> 0, 
        y -> -1}, {x -> 0, y -> 1}}]}}]

POSTED BY: Gianluca Gorni

Answer

Whiffee Bollenbach

Whiffee Bollenbach, retired

Posted 4 months ago

Thanks for reply. The parenthesis was not misplaced. The question is how to get the square root function to show full range in ParametricPlot. The 3.5 exponent on imaginary was just a kludge, but if made general, it saturates the plot.

POSTED BY: Whiffee Bollenbach

Answer

Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 4 months ago

You are using an unbounded region. It seems that Mathematica will automatically truncate it: Show[Region@ ImplicitRegion[ 0 <= x < \[Infinity] \[And] -\[Infinity] < y < \[Infinity], {x, y}], Frame -> True, PlotRange -> 5] Show[DiscretizeRegion@ ImplicitRegion[ 0 <= x < \[Infinity] \[And] -\[Infinity] < y < \[Infinity], {x, y}], Frame -> True, PlotRange -> 5] Give a suitably bounded and discretized (why?) region to `ParametricPlot`: d2 = DiscretizeRegion@ ImplicitRegion[0 <= x < 50 \[And] -30 < y < 30, {x, y}]; ParametricPlot[ReIm[(x + I y)^0.5], {x, y} \[Element] d2, PlotRange -> {{-1, 3.5}, {-3, 3}}, Frame -> True, ImageSize -> 200, AspectRatio -> Automatic]

You are using an unbounded region. It seems that Mathematica will automatically truncate it:

Show[Region@
  ImplicitRegion[
   0 <= x < \[Infinity] \[And] -\[Infinity] < y < \[Infinity], {x, 
    y}], Frame -> True, PlotRange -> 5]
Show[DiscretizeRegion@
  ImplicitRegion[
   0 <= x < \[Infinity] \[And] -\[Infinity] < y < \[Infinity], {x, 
    y}], Frame -> True, PlotRange -> 5]

Give a suitably bounded and discretized (why?) region to ParametricPlot:

d2 = DiscretizeRegion@
   ImplicitRegion[0 <= x < 50 \[And] -30 < y < 30, {x, y}];
ParametricPlot[ReIm[(x + I y)^0.5], {x, y} \[Element] d2, 
 PlotRange -> {{-1, 3.5}, {-3, 3}}, Frame -> True, ImageSize -> 200, 
 AspectRatio -> Automatic]

POSTED BY: Gianluca Gorni

Answer