You had to have a formula to draw the graph, right? It would be easier to answer your questions using that formula than getting it directly from the graph.

Yes, you can just remove it. Then you'll just be left with the radial lines.

The command quo(a,b,x,'r') divides polynomial a by polynomial b with respect to variable x. It returns the quotient and stores the remainder in r. Each row of the matrix corresponds to one of our polynomials (that's the outer seq). To get the entry in the leftmost column, we divide by a[1]. To get the entry in the next column, we take the remainder of the division and divide it by a[2]. We proceed in this manner until we have divided by every variable.

The command quo(a,b,x,'r') divides polynomial a by polynomial b with respect to variable x. It returns the quotient and stores the remainder in r. Each row of the matrix corresponds to one of our polynomials (that's the outer seq). To get the entry in the leftmost column, we divide by a[1]. To get the entry in the next column, we take the remainder of the division and divide it by a[2]. We proceed in this manner until we have divided by every variable.

@Andriy 

I have decided that it is a typo as I can't understand this construction:

N:= (i,j,sigma,f)-> sum(sum(sum(nn(i,j,sigma), sigma= 1..2), j= 1..3), i= 1..f):

You're right: It should be N:= f-> .... What I wrote is nonsense! Sorry about that. Try this:

nn:= ()-> `if`([args]::list(integer), ap.am(args), 'procname'(args)):

N:= proc(f)
local i,j,sigma;
     Sum(Sum(Sum(nn(i,j,sigma), sigma= 1..2), j= 1..3), i= 1..f)
end proc:

I can't help with the simplification of the nn operators.

@Andriy 

I have decided that it is a typo as I can't understand this construction:

N:= (i,j,sigma,f)-> sum(sum(sum(nn(i,j,sigma), sigma= 1..2), j= 1..3), i= 1..f):

You're right: It should be N:= f-> .... What I wrote is nonsense! Sorry about that. Try this:

nn:= ()-> `if`([args]::list(integer), ap.am(args), 'procname'(args)):

N:= proc(f)
local i,j,sigma;
     Sum(Sum(Sum(nn(i,j,sigma), sigma= 1..2), j= 1..3), i= 1..f)
end proc:

I can't help with the simplification of the nn operators.

@Markiyan Hirnyk asked: Could you explain the occurrence of or in your code?

I also wondered at it at first. But pay close attention to the direction of the inequalities that are connected by or. Preben had to do it this way because two-argument arctan always returns a value between -Pi and Pi. So the shaded region is considered as two regions by inequal.

@Markiyan Hirnyk asked: Could you explain the occurrence of or in your code?

I also wondered at it at first. But pay close attention to the direction of the inequalities that are connected by or. Preben had to do it this way because two-argument arctan always returns a value between -Pi and Pi. So the shaded region is considered as two regions by inequal.

@Jimmy I think (just guessing here) that when the fit is very good, as it is in your situation, that it can work without the parameter ranges.

@Jimmy I think (just guessing here) that when the fit is very good, as it is in your situation, that it can work without the parameter ranges.

@Andriy But you never tested what I actually wrote! You have N:= f-> ..., but I wrote N:= (i,j,sigma,f)-> ....

@Andriy But you never tested what I actually wrote! You have N:= f-> ..., but I wrote N:= (i,j,sigma,f)-> ....

Your procedure can be much simpler:

P:= proc(N1,N2)
local q,r;
     if N1 > N2 then error end if;
     q:= iquo(floor(N2)-ceil(N1)+1, 12, 'r');
     2*q+`if`(r>3, `if`(r>8, 2, 1), 0)
end proc;

Your procedure can be much simpler:

P:= proc(N1,N2)
local q,r;
     if N1 > N2 then error end if;
     q:= iquo(floor(N2)-ceil(N1)+1, 12, 'r');
     2*q+`if`(r>3, `if`(r>8, 2, 1), 0)
end proc;

@mehdi jafari I think that one problem is that in your eq1 (in problem 3.mws) there occurs both a[1](t) and a[1] without the (t). Is that what you really intended? I can't think of any reason why someone would want to use the same variable as both a function and a non-function in one expression.

For your sample probem with eq1, eq2, and eq3, can you write down "by hand" what the Matrix A should be? That is, does there exist A such that A . < a1(t),a2(t),a3(t) > = < eq1,eq2,eq3 >. It is not clear to me that such A exists.