Lets say I would like to construct a procedure which 

a) Allowed to call the view option from plots:-display by a different name? Lets say zoom. 

So the argument would PlotGraph(x^2,zoom = [-5..5,-5..5]) 

Is that by any means theoretically possible ? 

PlotGraph := proc(func::anything, zoom := {(x::range := a .. b), (y::range := c .. d)}) plots:-display(plot(func), view = zoom); end proc;

PlotGraph(x^2 , zoom = [0 .. 1, -5 .. 5]);

Error, (in plots:-display) expecting option view to be of type {"default", list({"default", range(realcons)}), range(realcons)} but received zoom = [0 .. 1, -5 .. 5]

So question to forum what am I doing wrong? 
 

Hello, I have a question about converting an output of a expression into another.

solve({-1 < x, 0 < 2*x/(x^2 - 1), x < 1}, {x})

This is my expression which evaluates into: . Which is correct but I would like it to be a classic range, more precisely (-1,0).

Is there a command to do that?

Thanks,

David.

I am looking to solve the following ode numerically for different values of w, 

diff(y(x),x)^2=y(x)^(-3w-1), along with the intial condition y(1)=1. 

i have tried to solve this for w=-1/3 and w=0 but i am getting the error:

Error, (in DEtools/convertsys) unable to convert to an explicit first-order system

I believe it probably has something to do with the fact that the first derivative is squared but i dont know how to procede with fixing that. 

Any help would be greatly appreciated. 

I  can't seem to get a solution to the following problem.  Can anyone see where I am going wrong I thought I had correct IBC s but they may be wrong/ill-posed

Melvin
 

Download BurgersEqns.mw

Here it is:

#hBar := 'hBar': m := 'm':Fu := 'Fu': Fv := 'Fv': # define constants
hBar := 1:m := 1:Fu := 0.2:Fv := 0.1: # set constant values - same as above ...consider reducing
At O( 
`ℏ`^2;
) the real quantum potential term is zero, leaving the classical expression:
pdeu := diff(u(x,t),t)+u(x,t)/m*(diff(u(x,t),x)) = Fu;
              / d         \           / d         \      
      pdeu := |--- u(x, t)| + u(x, t) |--- u(x, t)| = 0.2
              \ dt        /           \ dx        /      
On the otherhand, the imaginary quantum potential equation for v(x,t) has only O( 
`ℏ`;
) terms and so is retained as semiclassical
pdev := diff(v(x,t),t)+u(x,t)/m*(diff(v(x,t),x))-hBar*(diff(u(x,t),x$2))/(2*m)+v(x,t)*(diff(u(x,t),x))/m = Fv;
                                                     2         
         / d         \           / d         \   1  d          
 pdev := |--- v(x, t)| + u(x, t) |--- v(x, t)| - - ---- u(x, t)
         \ dt        /           \ dx        /   2    2        
                                                    dx         

              / d         \      
    + v(x, t) |--- u(x, t)| = 0.1
              \ dx        /      
By inspection of the derivatives in above equations we now set up the ICs and BCs for 
u(x, t);
 and 
v(x,t).;
The quantum initial and boundary conditions are similar to the classical case, but also comprise additional boundary condition terms for 
v;
 and for 
u;
, notably a 1st derivative BC term for 
u;
.
IBCu := {u(x,0) = 0.1*sin(2*Pi*x),u(0,t) = 0.5-0.5*cos(2*Pi*t),D[1](u)(0,t) = 2*Pi*0.1*cos(2*Pi*t)};# IBC for u
          IBCu := {u(0, t) = 0.5 - 0.5 cos(2 Pi t), 

            u(x, 0) = 0.1 sin(2 Pi x), 

            D[1](u)(0, t) = 0.6283185308 cos(2 Pi t)}
IBCv := {v(x,0) = 0.2*sin((1/2)*Pi*x),v(0,t)=0.2-0.2*cos(2*Pi*t)};# IBC for v
                   /                                 
          IBCv := { v(0, t) = 0.2 - 0.2 cos(2 Pi t), 
                   \                                 

                             /1     \\ 
            v(x, 0) = 0.2 sin|- Pi x| }
                             \2     // 
IBC := IBCu union IBCv;
         /                                 
 IBC := { u(0, t) = 0.5 - 0.5 cos(2 Pi t), 
         \                                 

   u(x, 0) = 0.1 sin(2 Pi x), v(0, t) = 0.2 - 0.2 cos(2 Pi t), 

                    /1     \  
   v(x, 0) = 0.2 sin|- Pi x|, 
                    \2     /  

                                           \ 
   D[1](u)(0, t) = 0.6283185308 cos(2 Pi t) }
                                           / 
pds:=pdsolve({pdeu,pdev},IBC, time = t, range = 0..0.2,numeric);# 'numeric' solution
                     pds := _m2606922675232
The following quantum animation is in contrast with the classical case, and illustrates the delocalisation of the wave form caused by the quantum diffusion and advection terms:
T:=2; p1:=pds:-animate({[u, color = green, linestyle = dash], [v, color = red, linestyle = dash]},t = 0..T, gridlines = true, numpoints = 2000,x = 0..0.2):p1;
                             T := 2
Error, (in pdsolve/numeric/animate) unable to compute solution for t>HFloat(0.0):
matrix is singular
                               p1
 

The help page for dsolve,numeric,events says that "A call to a procedure-form dsolve numeric procedure" can return the time that one or more events fired by coding the eventfire option.

I do not know how to code for this.

Please provide a simple example of a procedure-form dsolve numeric with an event that exhibits this feature.

sys.pdf

Can  anyone help me to proof that this solution is the right solution to this system ?

Good day to you all.

I would like to construct a plot of several data sources and would like to understand how I can animate each event.

For instance - consider a plot of 2 data sets (a simple example is in the attached worksheet), one as a point and another as a column. What is the most efficient way to construct this? Following that, how can both data sets be animated in synch?

Thanks for reading!

MaplePrimes_Example.mw

Hi everyone:

I'm going to write code to give me the following expression, how can I do it?
n is Natural and k=0..n. 

Just want labels but no axes. Also seems with axes=none sometimes I get wierd artifacts in the surfacewireframe where the wireframe is some wierd color. If I set the axes to normal it goes away and then usually it is ok. Sometimes half the graph is colored this way. It is with MatrixPlot.

How does one set the size of a 3D plot? I get tired of dragging it's size when it resets it to the default size. size=[w,h] does not work with plot3d.

matrixplot(A, style=surfacewireframe,heights=histogram);

when A has zero values they are graphed, is there any way to not have them show? Choose any matrix for A I just want to see non-zero elements.

Also when looking down on the 3d graph I'd like the tops of the histogram cells to be colored as if I'm looking at a 2d plot. When I do it now I can't tell any difference between any of the cells. This looks to be a lighting issue I suppose. There is some variation but it is very difficult to tell. Some light settings work better than others. Z-Hue seems to be the best but still a little washed out.

I've always found that the way maple handles rotating graphs to be very unintuitive. It is very difficult for me to get the exact view I want by rotating the graph.

Usually i want to view for 1 2d projection then rotate to view the other. In most programs one can do this easier and there usually is some modifier key. In maple this is not the case and how it handles rotation is bizzaire.

For example, if I set the orientation to be x axis and start to rotate by dragging the mouse horizontally the 3d graph starts to rotate all 3 axis. It seems to depend on where the mouse is exactly at in the graph. That is, if I drag only in the horizontal but the mouse starts at the top then it is different than when it starts at the bottom. This is not intuitive it it requires me to know exactly how maple does things and thing work backwards...

Many programs have a modifier key that will snap one direction/angle to not change. This is idea. rather than have all 3 angles changing arbitrarily(it is nice when one just wants to perview the graph in 3D but not good when one wants to get a specific angle).

Is there any easy way to get what I want? I know I can set the orientation manually. That is not what I'm after. I'm after changing the rotation algorithm that maple uses or to add a modifier key to restrict rotation to only 2 angles.

Draw the surface area of the paraboloid z=12-x^2-y^2 which is inside the cone z=sqrt(x^2+y^2)?

Hello,

I was working one of the problems for my course in structural dynamics and came across the following function that needs to be plotted. How can we do it in maple 2018.

i have a acceleration of record earthquake(in excel ) and i get fourier and power spectra from another software (seismosignal) .Now i want to get it in maple but i try a lot but i couldn't.
i very Eager to learn it

rr.xlsx