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How to plot a function specified in bipolar coordinates if straightforward substitution of expression of Cartesian ones via them is impossible?

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    $\begingroup$ "a function specified in bipolar coordinates" - do you have an example on hand? $\endgroup$ May 13 '16 at 9:11
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    $\begingroup$ Analogous solution should work: graph in the Polar Plane $\endgroup$
    – Kuba
    May 13 '16 at 9:13
  • $\begingroup$ Yes, the function I wish to see is Sqrt[Cosh[u] - Cos[v]]/Sinh[u] \!( (*SubsuperscriptBox[(f), ((-1)/2), (,)])[(Cosh[u])]) where Subscript[f, -1/2] is Legendre' s either P or Q of order - 1/2 of imaginary argument. Too complcated for a beginner. $\endgroup$ May 14 '16 at 16:20
  • $\begingroup$ you should edit the question with your function. $\endgroup$
    – george2079
    May 14 '16 at 17:40
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We can do this by implementing the transformation formula directly: (ref https://en.wikipedia.org/wiki/Bipolar_coordinates )

bipolar[a_] =  a {Sinh[#[[2]]], Sin[#[[1]]]}/
                     (-Cos[#[[1]]] + Cosh[#[[2]]]) &

Show[{ParametricPlot[ 
   Table[bipolar[1]@{s, t}, {t, 
     Cases[Range[-3, 3], Except[0]]}] , {s, -Pi, Pi}, 
   AspectRatio -> Automatic, PlotRange -> All],
  ParametricPlot[ 
   Table[bipolar[1]@{s, t}, {s, 
     Cases[Range[-3, 3], Except[0]]}] , {t, -Pi, Pi}, 
   AspectRatio -> Automatic, PlotRange -> All]}]

enter image description here

or use the built in CoordinateTransform

ParametricPlot[ 
   Evaluate[ CoordinateTransform[
         {{"Bipolar", {1}} -> "Cartesian"}, {s, 1}]] ,
            {s, -Pi, Pi}, AspectRatio -> Automatic, PlotRange -> All]]

(note Evaluate is essential here or it will be extremely slow)

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  • $\begingroup$ Thank you very much. I try to understand what is written. $\endgroup$ May 14 '16 at 16:21

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