@Markiyan Hirnyk 
OK, I'll include this time the mathematical part (even if I have non seen many math proofs in this forum!).

@Markiyan Hirnyk 
The fact that a product of powers of A and B has the provided form is not proved too. Both must be done by hand but are simple. (Maple was useful to guess the result and this is the essential step.)

@Carl Love 

"Functional form" is of course ok, my point was the consistency.
`!` is a protected symbol in Maple so it cannot be assigned. Even unprotected, it will not work (I've just checked). It can be overloaded, but this is not discussed here.

P.S. I want to congratulate you for the very clear and useful syntheses you post sometimes. They should appear on top when searching mapleprimes (but unfortunately the indexing mechanism is not able to do this).

@Carl Love 

Unfortunately the terminology is not very consistent in the documentation.
If f is a binary operator, the infix notation is x f y of course (x,y being the operands).
The prefix notation would be  f x y  but in Maple this notation is not used.
f(x,y) is the functional notation.
The prefix notation is used for unary operators, e.g.  not x. The associated functional notation used in Maple is `not`(x).
The unary operator ! is used with a postfix notation:  x!. The corresponding Maple procedure in Maple is factorial (`!` is not a procedure) so the functional notation is factorial(x).
 

@asa12 

Your aa0  is actually  []  i.e.  the identity (but in an incorrect Maple notation). In this case if you want to use orbit  construct the permgroup first:

pg:=permgroup(9, {[]});
                    pg := permgroup(9, {[]})
orbit(pg,2);
                      {2}

P.S. You can use your phone to create a “Personal Hotspot” for the laptop ("tethering"). It's simple.

@asa12 

Sorry, I don't understand.

You should state the complete problem (and preferably the desired answer).
Why don't you try to formulate it in your language and then use Google Translate?

You should also upload the worksheet, not an image.

@asa12 

a^6 was computed.  a^0 = []  by definition.

For orbit the first argument must be a set.

# FixedPoint
FP:=(a,n) -> {seq(i,i=1..n)} minus {op(map(op,a))}: # n is the order of the permutation a
p:=[[1,2],[3,4,5]];
                    p := [[1, 2], [3, 4, 5]]
FP([[1,2],[3,4,5]],9);
                          {6, 7, 8, 9}
orbit({p},3);
                           {3, 4, 5}

@danielpf 

Instead of Dirac(i-j)   it should be KroneckerDelta(i,j)  or (better) K(i-j),  where  K := i -> piecewise(i=0,1,0);

To obtain your result, it's possible to do something like:

ex:=sum(sin(x[i]),i=1..N):
diff(subs(x[i]=(x(i)(t)),ex),t):
subs(diff(x(i)(t),t)=K(i-j), x(i)(t)=x[i], %);

@tomleslie 

The OP probably wants a parametric plot such as  plot([y1(x), y2(x), x=a..b]);
But for polynomials of degree 2 this is a conic section, not his graph.

@taro 

You have right single quotes (' ')  used to delay evaluation. See ?quotes
Note that for left single quotes  `f`(x)  evaluates directly to  x^2 + 1.
 

@Carl Love 
Think about the frustrations of a student seeing this:

@Carl Love 

The method would work for systems of the form

sum( a[i,j]*floor(f[i,j]), j=1..n ) = b[i],  i = 1,...,n

where f[i,j] is affine in x[1],...,x[n]  and a[i,j], b[i] are integers.
But such systems may have infinitely many regions. In this case we could use the parametric form of SemiAlgebraic but complications will appear.
Note also that even in the finite case SemiAlgebraic (or solve) will most probably fail if n is > 3 say.
Not to mention that the number of regions could be huge.

@TechnicalSupport 

You should also change
x(t), y(t); # plain x(t) and y(t)
to
x(u), y(u); # plain x(u) and y(u)
(or similar).

1. A typo:

soln := dsolve( sys ); # { x(t) = cos(t), y(t) = -sin(t) }
assign( sol );

Should be assign(soln). So,  x(t), y(t) are assigned but not as the desired procedures [remember tables; return the procnames]. Hence  x(u), y(u)  will not work.

2. @Carl
The type  Type is indeed very clever! Isn't idtype a better name for it?

@did 

I know it is simple, I have mentioned it. But you must use Taylor series, recognize the Fourier series and its convergence, integrate it. Using residues is even simpler.
But a CAS does not work as a human. Among other things, for integration it uses special functions. They have branches and so a lot of complications as in our case.