Re: How to set y always greater than or equal to x for a function?

*To*: mathgroup at smc.vnet.net*Subject*: [mg46630] Re: How to set y always greater than or equal to x for a function?*From*: drbob at bigfoot.com (Bobby R. Treat)*Date*: Thu, 26 Feb 2004 17:53:19 -0500 (EST)*References*: <c1h0s4$9sc$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

That works great, this is just a quibble, but... Turning a Rule into a function and using MapAt seemed distasteful to me, so I came up with an equivalent solution, learning a thing or two in the process: f[x_, y_] := If[y > x, Sqrt[y - x], 0] plot1 = Graphics3D@Plot3D[f[x, y], {x, -100, 100}, {y, -100, 100}, DisplayFunction -> Identity]; rule = {x_?NumberQ, y_?NumberQ, z_?NumberQ} /; y ? x -> Sequence[]; plot2 = Head[plot1][First@plot1 /. rule, Sequence @@ Rest@plot1]; Show[plot2, DisplayFunction -> $DisplayFunction]; Bobby "David Park" <djmp at earthlink.net> wrote in message news:<c1h0s4$9sc$1 at smc.vnet.net>... > Albert, > > I think this is an interesting plotting problem whose solution is not > totally obvious. If you are just beginning to learn Mathematica, then it is > probably not the first problem to tackle. > > Plot3D[Sqrt[y - x], {x, -100, 100}, {y, -100, 100}]; > > gives warning messages and a jagged edge to the surface. > > We could define a function > > f[x_, y_] := If[y > x, Sqrt[y - x], 0] > > Then > > Plot3D[f[x, y], {x, -100, 100}, {y, -100, 100}]; > > works more or less. But it has the extra xy-plane surface, which is not part > of the surface you are specifying. > > A few years ago I learned the following trick from MathGroup. (I can't > remember exactly who posted it.) Generate the plot above and then throw out > all points that have y <= x. > > plot1 = Graphics3D@ > Plot3D[f[x, y], {x, -100, 100}, {y, -100, 100}, > DisplayFunction -> Identity]; > > plot2 = MapAt[# /. {x_?NumberQ, y_?NumberQ, z_?NumberQ} /; y <= x -> > Sequence[] &, plot1, 1]; > > We had to convert the plot from SurfaceGraphics to Graphics3D. We had to map > our function onto the first part of plot1. The first part contains all the > primitive graphics for the surface. Any point in a Polygon for which y <=x > gets replaced by Sequence[], which effectly gets rid of it. We then Show the > new plot. > > Show[plot2, DisplayFunction -> $DisplayFunction]; > > That gives a pretty good representation of the surface. > > The DrawGraphics package at my web site gives some other techniques for > handling plotting problems of this type. One of its advantages is that it > automatically extracts the primitive graphics from any plot and it is then > easy to manipulate or combine the various elements. > > Needs["DrawGraphics`DrawingMaster`"] > > The following implements the same trick as above. > > Module[{g}, > g = Draw3D[f[x, y], {x, -100, 100}, {y, -100, 100}]; > Draw3DItems[ > {g /. {x_?NumberQ, y_?NumberQ, z_?NumberQ} /; y <= x -> Sequence[]}, > BoxRatios -> {1, 1, 1/2}, > Axes -> True]]; > > Another method is to reparametrize the surface. If we were going to > integrate a two variable function we could use > Integrate[f[x,y],{x,-100,100},{y,x,100}]. The second iterator depends on x. > But we can't directly do that when plotting surfaces. DrawGraphics has the > command > > IteratorSubstitution[{y, Sqrt[y - x]}, {y, x, 100}] > {{(-w)*(-100 + x) + x, Sqrt[(-w)*(-100 + x)]}, {w, 0, 1}} > > It reparametrizes y and Sqrt[y-x] for {y,x,100} in terms of x and w. w has > the fixed iterator {w,0,1}. We can then plot this new function and then use > DrawingTransform3D to get back from the xw-plane to the xy-plane. > > Draw3DItems[ > {Draw3D[Sqrt[(-w)*(-100 + x)], {x, -100, 100}, {w, 0, 1}] /. > DrawingTransform3D[#1 & , (-#2)*(-100 + #1) + #1 & , #3 & ]}, > > BoxRatios -> {1, 1, 1/2}, > Axes -> True, > Background -> Linen, > ImageSize -> 450]; > > But this isn't too nice for this particular case because the polygons are > crimed together at one corner. > > Another idea is to plot the function Sqrt[y], where we can use fixed ranges > and then rotate it by 45 degrees. I plot it in two sections to get a better > representation of the rising section. > > Draw3DItems[ > {{Draw3D[Sqrt[y], {x, -150, 150}, {y, 0, 2}, PlotPoints -> {25, 2}], > Draw3D[Sqrt[y], {x, -150, 150}, {y, 2, 150}, PlotPoints -> 25]} // > UseRotateShape[-45°, 0, 0]}, > BoxRatios -> {1, 1, 1/2}, > PlotRange -> {{-100, 100}1.01, {-100, 100}1.01, {-0.1, 15.1}}, > Axes -> True, > Background -> Linen, > ImageSize -> 450]; > > This is probably a little nicer than the non DrawGraphics plot because the > "mesh" lines are either level, or the sqrt curves. > > As an extra, here is the same surface combined with -Sqrt[y-x] and then spun > around for better viewing. g is just the primitive graphics for the surface > we plotted above. We draw it and then we draw it again with the z component > reversed. > > plot3 = > Module[{g}, > g = {Draw3D[Sqrt[y], {x, -150, 150}, {y, 0, 2}, PlotPoints -> {25, > 2}], > Draw3D[Sqrt[y], {x, -150, 150}, {y, 2, 150}, PlotPoints -> 25]} > // > UseRotateShape[-45°, 0, 0]; > Draw3DItems[ > {g, > g /. DrawingTransform3D[#1 &, #2 &, -#3 &]}, > BoxRatios -> {1, 1, 1}, > Boxed -> False, > PlotRange -> {{-100, 100}1.01, {-100, 100}1.01, {-12.1, 12.1}}, > Background -> Linen, > ViewPoint -> {1.779, -2.844, 0.447}, > ImageSize -> 450]]; > > SpinShow[plot3] > SelectionMove[EvaluationNotebook[], All, GeneratedCell] > FrontEndTokenExecute["OpenCloseGroup"]; Pause[0.5]; > FrontEndExecute[{FrontEnd`SelectionAnimate[200, AnimationDisplayTime -> 0.1, > AnimationDirection -> Forward]}] > > David Park > djmp at earthlink.net > http://home.earthlink.net/~djmp/ > > > > > > > From: Albert Franco [mailto:francoatnsuok at earthlink.net] To: mathgroup at smc.vnet.net > > I recently purchased Mathematica 5.0, and I am in the process of > learning how to take advantage of it's many features. > > This is an example of one line I would like to calculate in Mathematica: > > Plot3D[Sqrt[y-x],{x,-100,100},{y,-100,100}]; > > > How can I best edit this line to make y always greater than or equal to x > during > calculation so the program doesn't try to take the square root of a > negative number? > > Any help will be appreciated. > > Albert Franco

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