Probably theta, phi, psi cannot be retrived easily. But you could insert your own sliders for theta, phi, psi and rotate the plot using them (via orientation=...). This way you will know their values.

@mahmood180 

F:=(m,t) -> piecewise(m=0,1,m<=r,cos(m*t),m<=2*r,sin((m-r)*t));
b:=(n,m,t) -> piecewise(n-1<=N*t and N*t<n,F(m,2*Pi*N*t),0);

Yes, you are right. I have recommended to plot the hyperbola, but I did not :-)

Edit.

Actually, the corect solve call for this case would be:

solve(sys, useassumptions) assuming x*y+2*x+4*y+6<0;

and now the result is NULL.

@Markiyan Hirnyk 

You can't ask Maple such proofs. But it is a high-school problem!

P.S. Can you prove using Maple for example that if a number is a multiple of 9 then the sum of its digits is also a multiple of 9?

@Markiyan Hirnyk 

1. The post was not edited at all.

2. You are right here but it is not difficult to see that for any n, h(n) = h(k) for some k<100.

@Markiyan Hirnyk 

Did you read the second version?

Here is the Maple version of a the solution solution presented in the book.
It is a "one-liner"!

h:= n -> h(`+`(op(`^`~(convert(n,base,10),2)))): h(1):=1: h(4):=4;

So, the happy ages are:

select(t->h(t)=1, [$1..100]);
   [1, 7, 10, 13, 19, 23, 28, 31, 32, 44, 49, 68, 70, 79, 82, 86, 91, 94, 97, 100]

The reason why the recursive h works (and does not generate an "infinite recursion" message)
is because it can be proved that the only values h takes are 1, 2 and 4 (this is not mentioned
in the book, but see the wiki article also mentioned in Rouben Rostamian's worksheet), and f(2) =4.

A simple solution is possible in Maple even if this fact is not used/known.
It is obvious that the iterates f(n), f(f(n)), ... repeat. The next version of h
also computes this sequence if it is invoked with a 2nd argument.

h:=proc(n, dummy:=NULL)
local S:=NULL, a:=`+`(op(`^`~(convert(n,base,10),2)));
while not(a in {S}) do
  S := S,a;
  a:= `+`(op(`^`~(convert(a,base,10),2)))
od;
if dummy=NULL then min(S) else S fi
end;


This version can be used to verify that h(N) = {1,2,4}.

`union`( seq({h(n)},n=1..100) );
   {1,2,4}

@Annonymouse 

As I understand your system is polynomial (or ca be reduced to such).
Solving a system of polynomial equations seems to be an easy task but it is not. The most powerful tool is Groebner bases (in the packages Groebner and PolynomialTools).
If Groebner bases are used, the answer if Eq[n+1] is "dependant" of the previous ones could be easy.

The problem is that the complexity for Groebner bases could be huge, and this depends heavily on the form of the polynomials.
So, without testing is is impossible to tell if such a tool can be used.

Here I supposed that you want a symbolic approach. For a numeric solution, you could try/experiment with fsolve and DirectSearch.

@Markiyan Hirnyk 

My last post in this thread
(I declare the question solved).


c:=1:
p:=x->a*x^11+b*x^4+c: q:=x->b*x^11+c*x^4+a: r:=x->c*x^11+a*x^4+b:
# s:=solve([p(x),q(x),p(y),r(y)]):
ss:=solve([p(x),q(x),p(y),r(y),p(z),r(z)]):
nops([ss]);

==> 41

But ss is not what you have said in words. You mean:
sss:=solve([p(x),q(x),p(y),r(y),q(z),r(z)]):

Now, this (inutile as I have mentioned) system seems to reach Maple's patience too
and I had to interrupt it after several minutes.
You may try to wait and see if it ends successfully (as it should).

PS. Unfortunately nobody else entered for this question. It would have been interesting to see other opinions.

That's all.

@Markiyan Hirnyk 

Oh my god, it seems that you are incorrigible! So, any dialog with you is very difficult and exhausting!

Have you tested the system? Of course it is compatible. Practically it's the same: I have said (it's easy to see) that for this system if (p,q) and (p,r) have common roots then (q,r) also has!

@Markiyan Hirnyk 

solve(eval([p(z),q(z),r(z)],s[14])); # NULL

or

pqr:=eval([p(z),q(z),r(z)],s[14]):
Groebner[IsProper](pqr);   #false

The situations are not similar, here we have a symbolic solution + numerical confirmation.
This should be an acceptable solution by any standard!

Or, ... by hand.

 (Edited)

@Markiyan Hirnyk 

Then any result of a Maple solve command is "unbased".

We have found symbolically the example, solve agrees, numerical verification ok.
The last step would be to verify by hand. It will take a few weeks, but let's go! 

Edit.

PS. You see, it's another example of your multi-"unjustified objections"

@Markiyan Hirnyk 

The code was implicit.

c:=1:
p:=x->a*x^11+b*x^4+c: q:=x->b*x^11+c*x^4+a: r:=x->c*x^11+a*x^4+b:
s:=solve([p(x),q(x),p(y),r(y)]):
evalf(s[14]);  #the provided counterexample

@Markiyan Hirnyk 

Actually the real case results from the provided hint.
According to the Maple solution, there are only a finite number of polynomials
(one of them was posted) for which the conclusion fails; all of them have of course complex a,b,c.

The problem can be simplified as in the first russian solution (with A=a^2-bc etc)
and the exceptions have a smaller length for A,B,C.
[That was my first approach but I had no patience for all the cases when I gave the hint].

Note also that Maple verifies that a common root for p,q,r does not exist for a,b,c
in the provided example:
solve gives () and this can be also seen looking at the solutions (x,y,a,b) given by solve for
the system p(x)=q(x)=0=p(y)=r(y).

@Markiyan Hirnyk 

As I have mentioned, a,b,c are very large. If you relly want to see them, here they are:

c =1

a =

b =

P.S. What is the source of the problem? It was interesting.