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These are answers submitted by BrianB

Hi Doug

Thank you for your reply.

It might be that I do not fully understand what you are writting but I cannot see that the precision of the numbers in the expression is the problem.

The following two equivalent expressions gives two different results.

(exp(-I*Pi))^(1/2-(1/100)*(I*8)); evalf(%);
exp(-I*Pi*(1/2))*exp(-8*Pi*(1/100)); evalf(%);

1.285730980*I and -.7777676792*I, respectivly.


By the way I'm using maple 13

Hi Robert

This also saves the image  to something that looks like a very hard compressed jpeg picture.

Could someone please try this exampl:

> restart;
> with(plots);
> CEQ1 := .9875000000*(exp(I*Pi*(lambda+I*epsilon)))^2+1.012500000;
> P := plot3d([subs(abs(CEQ1))], lambda = -1/2 .. 10*(1/2), epsilon = -.3 .. .7, axes = boxed, grid = [70, 70]);
> plotsetup(eps, plotoutput = "c:/myplot.eps", plotoptions = "portrait,noborder,height=3in,width=3in,leftmargin=0in,bottommargin=0in");
> P;
> plotsetup(default);

The two above mentioned methods for exporting eps pictures works only for simpler graphs.

Regards Brian


I agree that this works with some graphs. I worked with a simple 2 plot. When I do the same with a 3d plot with a lot "data" it does not export it as a vectorized picture. The following example cannot be exported:

plot3d([subs(abs(CEQ1))], lambda = -1/2 .. 10*(1/2), epsilon = -.3 .. .7, axes = boxed, grid = [70, 70]);

Hi Joe

Thank you for your answer.

Regards Brian


Thank you for your answer. I can now do the following which was what I wanted.

> unassign(phi[0])

> diff(phi[0](z), z)
> phi[0] := z -> a*z

> diff(phi[0](z), z)

Hi Robert

Thank for your answer.

Is it correctly understood then that it is only unknown functions that figures in an expression that you write as phi[0](z) so that diff(phi[0](z), z) is understood and not just 0? When a funtion on the other hand is declared you do not need this because it follows from the expression what it is a funtion of, right?.

Hi Thomas

Thank you that was what I was looking for.

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