@Markiyan Hirnyk Warning: This numeric integration takes an extreme amount of memory for me. I had to kill it from Windows Task Manager.

@Andriy "Just a name" means that k is a variable to which no numeric value has been assigned.

@mapleq2013 You wrote:

Could you please briefly explain why this is correct, if there is a reasoning at all?

Because you are extracting a procedure rather than an expression from dsolve's output.

I asked because I have never defined functions this way, I always do as the following example:

source(t):=sin(0.1* t)

I recommend that you never define a function that way. Maple will let you get away with that sometimes (especially if you use 2D input). The correct way to define a procedure (function) is

source:= t-> sin(0.1*t);

In the case of the dsolve output, you are not defining a procedure; rather, you are extracting a procedure that has already been defined, hence the t-> part is not needed.

There are two main syntaxes for plotting a procedure. The way that always works is

plot(f, a..b);

A syntax that works only if f is prepared to take symbolic input (as is source above) is

plot(f(t), t= a..b);

The procedures returned by dsolve are also prepared to take symbolic input.

I recommend that you never use the syntax

plot(f(t), a..b);

although, as I said, Maple will let you get away with it sometimes.

I've looked over the partial fractions for k= 0 through 7, and I am curious what pattern you saw (you said that you guessed the general decomposition). I can see that one fraction will always be 1/(k+1)/(x-1)^2. The other denominators are (x-1) and 2+sum(x^j, j= 1..k-1), but I can't guess their numerators.

@Kitonum You can make a factor-of-9 time improvement by including option remember in your procedure P. Without it plot3d calls dsolve repeatedly with the same value of k.

Please post your code.

If I correctly understand your situation, you should store the value computed at each iteration in a Vector. Then subtract from the true value outside the loop to get a value for the error at each iteration. I'll post an example in an Answer below.

@acer Thanks, Acer. I updated the Answer accordingly.

But I don't understand how your equality tester works. Specifically, I don't understand how member works. Groups are implemented as modules. According to ?member when M is a module, member(x, M) checks whether x is an export of M. But an element of a group is not an export of the module that represents that group! So how does it work?

@rodrigog So, if you don't want the purple shades, does that mean that you do want to go through orange, yellow, and green? To do that, and in reverse order, do

plots:-pointplot(Pts, color= [seq(COLOR(HUE, (N-k)*.65/N), k= 1..N)]);

If you don't want to go through orange, yellow, and green, then you need to tell me what you expect the color in the middle to be.

Please post your points as plaintext so that I can cut and paste them.

@Kurt Ewald:

It is an important issue that I don't know the answer to. I am hoping that someone who is expert in the GroupTheory package will see this and respond. In this case, the group F returned by FactorGroup is reported to be isomorphic to G!

You have 16 equations and 12 unknowns. Do you expect some of the equations to be redundant?

@Markiyan Hirnyk 

The code can be corrected simply by adding with(stats) and removing the extraneous {}* from before the plot command. The is command and the change of the form of the inequalities are unnecessary.

Kurt Ewald wrote:

Many thanks for the procedure. But some actions, for instance IsNilPotent(GH), fail.

Permutation groups are the most efficient representation, and all commands work on them AFAIK. The group returned by FactorGroup should be converted to a permutation representation:

GH:= FactorGroup(G,H):
GH:= PermutationGroup(GH);

IsNilpotent(GH);

   true

Kurt Ewald wrote:

I had found another way, to create the FactorGroup.
R:=map(Representative,LeftCosets(H,G)) and then
M:=Group(R).
The results were the same as with your procedure. But I was not sure, weather my way is allright.

That was my first idea. I had posted it, and then I deleted it. That method can only work if G has a subgroup isomorphic to G/H. That is often true, but not always.

What do you want each frame of the animation to represent? For example, each frame could represent the intersection of the surface with a horizontal plane (also known as the level curves).

@Preben Alsholm You're right. I got stuck by a bug in Maple 16.0 (only) where the whole curve is not plotted if there is a single undefined point. Using other Maple, everything is fine.