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let's say that I have a sphere in $N$-dimensions

$$x_1^2+x_2^2+...+x_N^2=R$$

If I want to know the projection on the $x_1-x_2$ plane in this case I can figure out myself that is a circle, but how can I get Mathematica to do the projection and plot it for me?

Of course I need to apply this to a more complicated case where the N-dimensional surface is not as simple and can only be defined implicitly by $f(x_1,...,x_N)=0$.

I hope I have not forgot too much math from school that I am asking mathematica to do this for me.

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  • $\begingroup$ Actually it's a disk in this case. (You probably realize that but, if not, the difference is important.) $\endgroup$
    – Daniel Lichtblau
    Commented Feb 7, 2014 at 15:25
  • $\begingroup$ Can you give an example of an actual function you plan to work with? The $n$-sphere is trivial, but your actual problem might still have some structure to it that will make it easier to solve than the completely general case. $\endgroup$
    – Szabolcs
    Commented Feb 7, 2014 at 20:28

1 Answer 1

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Here's an approach using FindInstance and RegionPlot. Unfortunately it is incredibly slow.

A funky function for us to test on:

f[x_, y_, z_, w_] := Sin[x] Cos[y] + Sin[y] Cos[z] + Sin[z] Cos[w] + Sin[w] Cos[x] - 1

Check whether a point $(x,y)$ corresponds to a solution $f(x,y,z,w)=0$ for some $z$ and $w$:

g[x_?NumericQ, y_?NumericQ] := Length@FindInstance[f[x, y, z, w] == 0, {z, w}, Reals] > 0

Plot the region where $g(x,y)$ is True:

RegionPlot[g[x, y], {x, -Pi, Pi}, {y, -Pi, Pi}, PlotPoints -> 5, MaxRecursion -> 1]

enter image description here

Even with PlotPoints and MaxRecursion turned way down it takes like ten minutes to create the plot.

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  • $\begingroup$ Why not ContourPlot[f[x, y, 0, 0] == 0, {x, -Pi, Pi}, {y, -Pi, Pi}] ? $\endgroup$
    – b.gates.you.know.what
    Commented Feb 7, 2014 at 10:18
  • $\begingroup$ My understanding of the question is we want the projection of the entire set $f(x,y,z,w)=0$ onto the $xy$ plane, not a slice through the $xy$ plane. That is, if $f(1,1,1,1)=0$ then $(1,1)$ should be included in the plot even if $f(1,1,0,0)\ne0$. $\endgroup$
    – user484
    Commented Feb 7, 2014 at 10:23
  • $\begingroup$ As said by Rahul I need not to just plot a slice, but the entire projection. For the N-sphere it would be a filled circle, not the perimeter of a circumference of some radius. The FindInstance is a good function to think about for this issue and the proposed example is at least food for thoughts, thanks. $\endgroup$
    – Rho Phi
    Commented Feb 7, 2014 at 14:08
  • $\begingroup$ I see that the function gets incredibly slower as the number of dimensions increases. I am wondering if putting some more information in the game could help. For instance my function is Likelyhood, so I know that there is a maximum at a given point that is contained in the area that will result from the projection. I was thinking to use this information by drawing the projection with a random shot of points starting from a seed based of the location of the maximum point. This, I think, should scale much more friendly when one increases the number of dimensions that are projected out. $\endgroup$
    – Rho Phi
    Commented Feb 7, 2014 at 14:49
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    $\begingroup$ Show@Table[ ContourPlot[f[x, y, z, w] == 0, {x, -3, 3}, {y, -3, 3}], {z, -3, 3, .3}, {w, -3, 3, .3}] gives some hints about the shape and it's much faster. Of course it's far from precise ... $\endgroup$
    – Szabolcs
    Commented Feb 7, 2014 at 20:48

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