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Re: Restriction on the domain of Plot3D


Hi,
One way that I could think of, is to multiply your function with an
appropriate UnitStep function. I'm not sure that this is efficient or
even what you had in mind but it will make the plot at zero level for
the undesired region. I renamed your function. Please look at the
following

eqt1[q_, N_, M_] := -(3*q/(8*Pi))*(Sum[(
  Cos[4*Pi*n*d]/(2*n*d)) - (Sin[4*Pi*n*
          d]/(8*Pi*(n*d)^2)) - (Cos[4*Pi*n*d]/(32*Pi^2*(n*
            d)^3)), {n, 1, N}] - Sum[(Cos[4*Pi*
                Abs[z0 - m*d]]/Abs[z0 - m*d]) - (
            Sin[4*Pi*Abs[
                  z0 - m*d]]/(4*Pi*Abs[z0 - m*d]^2)) - (Cos[4*Pi*Abs[z0 - m*
          d]]/(16*Pi^2*Abs[z0 - m*d]^3)), {m, -M, M}])UnitStep[d - 2z0]
Plot3D[Evaluate[eqt1[1, 10, 10]], {z0, 10^-6, 3}, {d, 1/10, 6}, PlotPoints -> 
      50];

I hope this is helpful.
By the way, did you used the ViewPoint option to rotate the graph

yehuda

On Mon, 21 Mar 2005 03:01:52 -0500 (EST), Steeve Brechmann (schumi)
<steevebrechmann at yahoo.ca> wrote:
> 
> Hi everyone,
> 
> I want to plot the following function with Plot3D :
> 
> eqt[q_,N_,M_]:= -
> (3*q/(8*Pi))*(Sum[(Cos[4*Pi*n*d]/(2*n*d))-(Sin[4*Pi*n*d]/(8*Pi*(n*d)^2))-(Cos[4*Pi*n*d]/(32*Pi^2*(n*d)^3)),{n,1,N}] -
> Sum[(Cos[4*Pi*Abs[z0-m*d]]/Abs[z0-m*d])-(Sin[4*Pi*Abs[z0-m*d]]/(4*Pi*Abs[z0-m*d]^2))-(Cos[4*Pi*Abs[z0-m*d]]/(16*Pi^2*Abs[z0-m*d]^3)),{m,-M,M}])
> 
>  Plot3D[Evaluate[eqt[1,10,10]],{z0,10^ - 6,3},{d,1/10,6},PlotPoints->50];
> 
> But, how can i specify to Mathematica, that i want just the plot for 2*zz0 < dd ?
> 
> Thanks for the help !
> 
> Steeve Brechmann
> Physics Department, Laval University
> steevebrechmann at yahoo.ca
> 
>


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