Hi!

I'm currently using Maple 17 and I'm trying to programatically export .eps 3D plots using the following code:

plotsetup(eps, plotoutput = `square_of_x`, plotoptions = `color=rgb , portrait, noborder, height=4in,width=4in, shrinkby=0.1`)

The problem that I'm facing is:

1- The z-axis label is outside of my figure (see uploaded figure). This happens even if I try to save the figure in .png and it seems that the 'shrinkby' option its not working properly;

Could you help me solving this?

Thanks!

Hello, 

I have just started using Maple, and it seems very powerful. I am trying to solve trigonometric equations and get all the solutions in a range, but when I use fsolve I only get one solution. 

Is this by design of the function or is there another way to do this? 

Tom

Hi!

In Maple, is it possible to define an element of a series like that:

(i actually tried to enter := instead of =, but html editor told me its not valid maple expression)

So at any moment i enter  and maple give me , or  ->  ?

How get answer of this integral

int(1/u.t.exp(-t/u), t = 0 .. infinity)

I want to plot the argument for a complex function. The input (x,y) represented in polar coordinates (r,phi) by default puts the cut at -I*Pi. Likewise the argument function:

argument(f(x)) plots the range -Pi..Pi.

However the function f(x)=x^2 could typically be plotted with 2 riemann surfaces on top of each other. When phi becomes 2Pi f(x) becomes 4Pi and only then I want to identify the 0 with 4Pi again since the points are equivalent in the preimage.

On the other hand the function f(x)=sqrt(x) never surpasses its own domain. The values always stay within the argument range of (0,2Pi) (in fact it only goes till Pi, or -Pi/2..Pi/2 in maple) when the preimage is taken to be (0,2Pi). Thus when plotting a preimage value of (x,y) with argument phi and 2Pi+phi they will have the same value since phi=2Pi+phi and I see a step in the plot. This step is actually there since the function has a cut at this point.

This step in the plotting image is also shown for f(x)=x^2 (e.g. at phi=+-Pi/2) but it is not of importance since it just comes from the argument function being constrained to -Pi..Pi.

So is it possible to change this behaviour?

I  encountered a non-integrable integral in the process of solving the following process, . How to achieve its numerical solution? Such as in a looping   code：

#######
pa[i] := pa[i-1]-(Int(subs(t = tau, Lpa[i-1]+Na1[i-1]-Na2[i-1]), tau = 0 .. t)); 

pw[i] := pw[i-1]-(Int(subs(t = tau, Lpw[i-1]+Nw1[i-1]-Nw2[i-1]), tau = 0 .. t)); u[i] := u[i-1]-(Int(subs(t = tau, Lu[i-1]+Nu1[i-1]+Nu2[i-1]), tau = 0 .. t))；

######
Detailed code see annexBC2.mw

Dear all,

I would like to ask you to help me with the following problem that I got error.

Thanks in advance

Is it possible to animate plot like this parametically:
animate(plot, [[cos(t), sin(t), t = 0 .. A]], A = 0 .. 2*Pi, scaling = constrained, frames = 50)

however given the x and y components as solutions of an implicit equation.

I know I could run RootOf. But it seems that there is a part missing due to a branch cut :-/

I had problems plotting the solutions of my PDE with the derivative as the axis. Anyone knows what went wrong?

p1 := pds:-plot(diff(f(x,y),y), y = 0 .. 10, x = 0, numpoints = 100)

I seem to be getting a plot of y=0. Any help would be greatly appreciated:))

fyp2.mw

I have a system of pde as follow,

PDE := [(x*y+1)*(diff(f(x, y), y, y, y))+(x+(3/4)*f(x, y))*(diff(f(x, y), y, y))-(1/2)*(diff(f(x, y), y))^2+T(x, y) = (1/4)*x*(diff(f(x, y), y))*(diff(diff(f(x, y), y), x))-(1/4)*x*(diff(f(x, y), x))*(diff(f(x, y), y, y)), (x*y+1)*(diff(T(x, y), y, y))/(.733)+(x/(.733)+(3/4)*f(x, y))*(diff(T(x, y), y)) = (1/4)*x*(diff(f(x, y), y))*(diff(T(x, y), x))-(1/4)*x*(diff(f(x, y), x))*(diff(T(x, y), y))]

sys_ode := diff(g(y), y, y, y)+(3/4)*g(y)*(diff(g(y), y, y))-(1/2)*(diff(g(y), y))^2+h(y) = 0, (diff(h(y), y, y))/(.733)+(3/4)*g(y)*(diff(h(y), y)) = 0

ics := g(0) = 0, h(0) = 1, (D(g))(10) = 0, g(10) = 0, h(10) = 0

sol2 := dsolve([sys_ode, ics], numeric)

BC := {T(0, y) = h(y), T(x, 0) = 1, T(x, 10) = 0, f(0, y) = g(y), f(x, 0) = 0, f(x, 10) = 0, (D[2](f))(x, 0) = 0}

pds := pdsolve(PDE, BC, numeric)

module() ... end module

pds:-plot(T, y = 0 .. 10, x = 0);

Error, (in pdsolve/numeric/plot) unable to compute solution for x<HFloat(0.0):
solution becomes undefined, problem may be ill posed or method may be ill suited to solution

When I try to use the solution of the Ode as the boundary condition for PDE, by subbing g(y) and h(y) into BC. The plot returns me the error. Anyone knows the reason behind this and how to solve? Any help would be really greatly appreciated. Thanks

hi.

i have problem with this eq.

i need your help.

N := 4;
print(`output redirected...`); # input placeholder
4
y := sum(A[2*n].cos(2.*n.x), n = 0 .. N);

eq1 := diff(y, `$`(x, 2))+(a+2*q*cos(2*x))*y

eq2 := map(combine, eq1, trig)

for i from 0 to 4 do eq4[i] := coeff(eq2, cos(2*n*x)) end do

From these I want to extract the co-ffficients of cos(0x),cos(2x),cos(4x)..

and form a simultaneous linear equation containg A0,A2,A4

The solution is 

aA0+qA2=0

2q*A0+(a-4)*A2+q*A4=0

Can anybody tell me how to do it

Am here again, pls help me check out this adm code, cant get a result.

below is the attached file

 adm_2_method.mw

Consider the following code snippet:

with(DifferentialGeometry):
DGsetup([x],M);
RemoveFrame(M);

By itself it is, of course, pretty meaningless, but that is not the point. My problem is that after Maple has executed it, it will annoyingly continue using M> for each new execution group that is inserted. Having removed the frame with the command RemoveFrame, I would have expected that not to be the case. How can the frame be truly removed?