Download failed_plot_ac.mw

I don't understand what that Matlab colormap is doing, so I'm not sure how to reproduce that exact coloring.

But you could look at this old Post, and adjust the formula for generating the hue component.

Less fast and a bit more clunky are results from complexplot3d, or densityplot (of argument). Eg,

plots:-complexplot3d(ln((exp(1/z))), z=-0.5-0.5*I..0.5+0.5*I,
                     style=surface, orientation=[-90,0,0], grid=[501,501],
                     lightmodel=none, size=[600,600]);
plots:-densityplot(argument(exp(1/(x+I*y))), x=-0.5..0.5, y=-0.5..0.5,
                   axes=boxed, colorbar=false, grid=[301,301],
                   #colorstyle=HUE
                   colorscheme=["zgradient",["Yellow","Blue"],colorspace="HSV"]
                 );

You might compute the derivatives by using the D operator.

When plotted, evalf of such D calls may go through fdiff to perform numeric differentiation. Such approximation of a derivative -- by a finite difference scheme -- of a pde solution numerically approximated via finite difference methods, may need care. I would not expect high accuracy, even if you do try and adjust the step-size(s).

Double-check everything, and adjust as required.

Download badPDE_ac.mw

Try,

   X := Sample(roll,[n,n]):

For Maple 2023 you could use the new colorbar option for 2D contour plots, as well as colorscheme to get that spread of hue values.

For Maple 2022 you could use one of the following approaches:

1) Use DocumentTools:-Tabluate, like in this old response of mine. Adjust weights and dimensions and borders, to taste. You could use the colorscheme option with densityplot, to get the exact colorbar you want, eg. blue to red in hue, etc.

2) Use this old Post of mine, for getting a colored legend along with contourplot (and densityplot too if you want that kind of graded filled effect), or use a specialized form of the legend option with the contourplot command.

3) Do something like the following, using polygons for that filled legend effect. Naturally, you could also merge/display the following with added contour lines.

Download M2022_colorbar_fun.mw

I don't know the full details of your example (all ranges, etc), but here's another:

restart;
with(plots): with(plottools,polygon): with(ColorTools,Color):
f := cos(t)*sin(x)-sin(x)+(1/2)*sin(x)*t^2-(1/24)*sin(x)*t^4:
(a,b,c,d) := -Pi,0,-4,1:
Q := plot3d(f, x=a..b, t=c..d):
(zmin,zmax) := (min,max)(op([1,3],Q)):
display(densityplot(f, x=a..b, t=c..d, 'style'=':-surface',
                        'colorscheme'=["zgradient",["Blue","Red"],
                                       'colorspace'="HSV"]),
  seq(polygon([[0,0],[0,0]],'thickness'=0.3, 'legend'=sprintf("%5.3f",z),
              'color'=Color("HSV",[:-Hue(Color("Blue"))*(zmax-z)/(zmax-zmin),1,1])),
      z=zmax..zmin, 'numelems'=24),
  'legendstyle'=['location'=':-right'], 'size'=[580,500], labels=["",""],
  'axis[2]'=['location'=':-high','tickmarks'=[],'color'="White"],
  'axis[1]'=['location'=':-low','tickmarks'=[],'color'="White"]);

I found it by looking at the list of extra options on the ?dsolve,numeric,DAE Help page's Description section.

Near a mention of the differential option, that page had a cross-reference link to the ?dsolve,numeric,DAE_extension Help page, which contains the details of these extension methods for the numeric DAE solvers.

The default extent of the axis view comes from the data points themselves (or a fitted curve).

You can override it by using the usual view option for plotting commands.

For example,

restart; with(Statistics):
N:=200: U:=Sample(Normal(0,1),N):
X,Y:=<seq(1..N)>,<seq(sin(2*Pi*i/N)+U[i]/5,i=1..N)>:
ScatterPlot(X,Y,view=[-20..220,-2..2]);
ScatterPlot(X,Y,view=[-20..220,default]);
ScatterPlot(X,Y,view=[default,-2..2]);

The first sentence in the Options section of the ScatterPlot Help page (Maple 2018) states, with a cross-reference link to the Help page for plot options, "The options argument can contain one or more of the options shown below. All unrecognized options will be passed to the plots[display] command. See plot[options] for details."

What that means is that most of the 2D plotting command common options can be simply passed straight into calls to the ScatterPlot command.

Download OverflowError_ac.mw

An alternative to dharr's nice idea, for this example, is to selectively utilize limit.

Naturally, it's not as fast as dharr's switch. But I didn't have to consider the form.

Download Discontinuous_contours_ac.mw

ps. for implicitplot this use of gridrefine=3 above seems to produce a slightly nicer result that the original gridrefine=4 does there.

Your other set of contour values is also improved:

You can do this simply by indexing into C.

You shouldn't be using the deprecated linalg:-bloackmatrix to construct C. That also produces a deprecated lowercase matrix, by the way. You can use the <...> constructor instead, which here produces a Matrix.

I've edited my response to include the possibility that you'd want to make the V[i] into columns of a Matrix.

Download Vec_indx.mw

Sure, you could apply log10 to the x and (discarding negatives) the y values. Then force new tickmarks by powering 10, then set the axis locations to `low`, etc. But getting realy nice smooth curves by adaptive sampling becomes tricky. Why reinvent that wheel?

You can compare the loglogplot result with a plot call specifying mode=log for both axes.

I've added gridlines. Notices that the log mode does occur for the vertical axis, though the values are close enough to make the effect less apparent.

loglog_ac.mw

The expressions log[10](R) and log10(R) will both evaluate to ln(R)/ln(10). So the change occurs even before solve get's its arguments.

You may use the inert version instead. You can utilize the value command later, if you'd like.

For example (using %log[10] since you used log[10], though it's similar with %log10 ),

Download log10_ac.mw

Is this the kind of thing you're trying to accomplish?

note: I realize that I could have used mod (or iquo,irem), to handle your integer examples. But I wanted to give you something that you might also use for floating-point angles. I'm not sure how you plan to use any results, however...
 

Download QuestionMesPrincipale_ac.mw

You have the code,

for j from 1 to m-1 do
  for i from 0 to n-1 do
   u[i,j+1]:=(9*u[i-1,j]+14**u[i,j]+9*u[i+1,j])/16-u[i,j-1];
  end do;
end do:

That inner loop starts at i=0, but the lower bound of the first index of array u is 0, so you have an error when that loop tries to access u[i-1,j]. That index value becomes i-1=-1 when i=0, and is less than the lower bound 0. And so that access attempt is out-of-bounds. It's trying to access an element of u which does not exist. Perhaps you intended something like,
  for i from 1 to n-1 do
instead.

There is also an insteance of [i]/10 in the code. But [i] makes a list. Your code tries to add that to scalar values. Did you really intend for that? Perhaps you intended just i/10 instead?

There is also an instance of 4**u[i,j] in the code. The double-asterisk is syntax for powering, eg.  x**2 = x^2.  I suspect you intended just 4*u[i,j].

You could simply use mul to multiply together the finite (and concrete) number of terms, instead of stuffing in unevaluation quotes to try avoiding premature evaluation under product.

note. You could also use just Matrix(3,f) instead of Matrix(3,(i,j)->f(i, j)). And the entries are already floats, so you don't need to wrap it with evalf.

Download rzarouf_ac.mw