@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.

@did 

You have no choice. A perfect CAS does not exist and never will.

P.S. Fourier series? Trivial result?

@Carl Love 

gamma is known by the assume facility. So its tables contain the information.

about(gamma);
gamma:
  is assumed to be: gamma

@Markiyan Hirnyk 

The question was: why reject the 0 answer and accept the other one without some math investigations?
I think that Carl's suggestion (+ some evalf testss) is more reliable.

BTW. See this:

restart: 
int( exp((1+a+b*I)*exp(I*x)) , x=-Pi..Pi)  assuming real;  # a  and  b  are assumed to be real
simplify(%)  assuming real;

    0
    0

@Kitonum 

So, we have 2 symbolic results. How can we decide which one is true.  Is it always true? (e.g. a=0 in your answer).
Hard to answer without some maths.

 

@tomleslie 

can be found with a few computations by hand using the inequalities  u-1 < floor(u) <= u.
They can be obtained also like this:

Download bounds.mw