RE: Re: Plot3D question

*To*: mathgroup at smc.vnet.net*Subject*: [mg70677] RE: [mg70641] Re: Plot3D question*From*: "Erickson Paul-CPTP18" <Paul.Erickson at Motorola.com>*Date*: Mon, 23 Oct 2006 02:49:45 -0400 (EDT)

I think it was just a matter of chance. The numbers you selected (i.e., +/-10 and with 100 points) didn't happen to hit on an exact 0 in the denominator. You could get closer with an odd number of plotpoints and plotrange all, but still might not hit it depending on the platform. However, I did hit it and you might as well with: Plot3D[Sin[x]/y, {x, -8, 8}, {y, -8, 8}, PlotPoints -> 129] This selection of range of y being of the form 2^n and the number of points being 1+2^m would likely get some complaints starting out with either divide by 0 or result of infinity. Hope that helps ? Paul -----Original Message----- From: dimitris [mailto:dimmechan at yahoo.com] To: mathgroup at smc.vnet.net Subject: [mg70677] [mg70641] Re: Plot3D question I would like to apologise for the mistakes which obviously contained my previous post. Here comes the correct one: --------------------------------------------------------------- I would like to know why the Plot3D function does not complain about the following command (note there is a singularity at y=0) Plot3D[Sin[x]/y, {x, -10, 10}, {y, -10, 10}] whereas it does complain (a lot!) about the following Plot3D[Sin[x]/y, {x, -10, 10}, {y, 0 10}] I believe it has to do with the non-adaptive algorithm used by Plot3D, but even the command Plot3D[Sin[x]/y, {x, -10, 10}, {y, -10, 10},PlotPoints->100] does not change the behavior. Note that I don't have any problem why the Plot function does not complain about the following command Plot[Sin[x]/x, {x, 0, 10}]; since I understand that the adaptive algorithm actually does not sample points at x=0 Cases[%, {(x_)?NumberQ, (y_)?NumberQ} /; x >= 0 && y > 0.9, Infinity] I really appreciate any comments/guidance/insight. Thanks Dimitris