What is wrong with this code?

Question

I am plotting polar curves with PolarPlot and I tried plotting a parabola with $\frac{2}{1-\cos x}$ and $\sec x$.

Obviously there are two singularities that have to be considered, $\cos x=1$ and $\cos x=0$. I dont know how to specify multiple singularities for the same function or variable. This is what I have tried so far:

PolarPlot[{2/(1 - Cos[x]), Sec[x]}, {x, 0, 2 \[Pi]}, 
 Exclusions -> {Cos[x] == 0, 1}]

PolarPlot[{2/(1 - Cos[x]), Sec[x]}, {x, 0, 2 \[Pi]}, 
 Exclusions -> {x == 0, 2 \[Pi], \[Pi]/2, (3 \[Pi])/2}]

PolarPlot[{2/(1 - Cos[x]), Sec[x]}, {x, 0, 2 \[Pi]}, 
 Exclusions -> {Cos[x] == 0, Cos[x]==1}]

What is the right way to do this?

Any help appreciated. Thank you.

Answer 1

I would have expected your last form to work. Avoid the issue by putting the exclusion points {-Pi/2, 3Pi/2} at the boundaries of the iterator.

PolarPlot[{2/(1 - Cos[x]), Sec[x]},
 {x, -Pi/2, 3 Pi/2},
 AspectRatio -> 1,
 PlotRange -> {{-2, 20}, {-10, 10}},
 PlotLegends -> "Expressions"]

EDIT: Supposition about behavior of Exclusions -> Automatic.

$PerformanceGoal

"Quality"

From the documentation for Exclusions: "Exclusions -> Automatic is effectively equivalent to Exclusions -> True if $PerformanceGoal is 'Quality', and to Exclusions -> None otherwise. Exclusions -> True excludes all subregions where discontinuities are found in functions being plotted." For the function 2/(1 - Cos[x]) the discontinuities are easy to detect; however, the discontinuities in Sec[x] are apparently too subtle for whatever algorithm is used and are overlooked unless they are at the iterator boundaries. Presumably, the functions are evaluated at the iterator boundaries and when the singularities of Sec[x] are at the boundaries, they are discovered and excluded.

Answer 2

Interestingly enough, there are no errors if you change the order of the two functions, and it doesn't seem to require any explicit exclusions:

PolarPlot[{Sec[x], 2/(1 - Cos[x])}, {x, 0, 2 \[Pi]}, 
 PlotRange -> {{-5, 5}, {-5, 5}}]

As Rahul Narain suggested, I also added a plot range.

Answer 3

my best effort at that plot, manually clipping the angle range:

 Show[{
      PolarPlot[ 2/(1 - Cos[x]), {x, .1, 2 Pi - .1} ,
         AspectRatio -> 1 , PlotStyle -> Red  ],
      PolarPlot[Sec[x], {x, 0, 2 Pi} , AspectRatio -> 1 ]},
         PlotRange -> {{-5, 20}, {-10, 10}}]

after playing with this a bit I've concluded PolarPlot is simply badly behaved regarding Exclusion and even PlotRange options.

If we make that first range {0,2 Pi} the curve disappears all together, unless we specify some large number for PlotPoints :

 Show[{
     PolarPlot[ 2/(1 - Cos[x]), {x, 0, 2 Pi} ,
      AspectRatio -> 1 , PlotStyle -> Red  , PlotPoints -> 2000],
     PolarPlot[Sec[x], {x, 0, 2 Pi} , AspectRatio -> 1 ]}, 
          PlotRange -> {{-5, 20}, {-10, 10}}]

The curve fills in if you make PlotPoints even larger..not a very satisfactory solution though.

Answer 4

The RegionFunction option allows you to parametrically define what portion of the function to plot

PolarPlot[2/(1 - Cos[x]), {x, 0, 2 \[Pi]}, AspectRatio -> 1, 
 PlotRange -> 30 {{-1, 1}, {-1, 1}}, 
 RegionFunction -> Function[{x, y, \[Theta], r}, r <= 25]]