@Katatonia 

SYS1:=u>=0, r^2 - 2*r*c -r +2*c = u^2, -r+c+1+u>0,  0<=c, c<=1 , 0<r, r<1, r<2*c:
SolveTools:-SemiAlgebraic([SYS1], [r,u], parameters=[c]);

(Here, of course, the auxiliary variable u can be ignored when reading the answer.)

@s265344 I see now that you want real solutions. My answer is for complex ones and for your first version.
For the second one use Sys:=(lhs-rhs)~(sys);  but the conclusion is the same.
For real solutions it's more difficult; don't use a1>0 because the algorithms used by Maple are different and the system is too large for them.

For real solutions the DirectSearch approach could be OK but we must be sure not to obtain just an approximation of the null solution.

@Kitonum Unfortunately we should admit that this is just a speculation, Maple does not manage such problems. It does not work if we write {Limit(u(x, t), x=infinity)=0, Limit(u(x, t), x=-infinity)=0} or  {limit(u(_X, t), _X=infinity)=0, limit(u(_X, t), _X=-infinity)=0}.

Also, it does not work for A := int(diff(sin(x)*u(x,t),x), x=-infinity..infinity, continuous);

@mmcdara Once we know the cause, a workaround is simple. In your code, instead of instance := sol(...), just use:

instance := convert(sol(parameters=data), `global`);

@Earl The use of complex numbers is not essential, it is just a convenient way to obtain orthogonal coordinates (and conformal maps).

@Kitonum I think that the OP obviously knew the answer, and just wanted an explanation.

The question is not very clear. What means that f is continuous over an interval J? Must f be continuous at each point of J, or the restriction f|J must be continuous? How e removable discontinuity is it treated?

Note. Suppose that discont returns the empty set. We cannot conclude that f is continuous on R, because probably discont ignores the branches which define f (even if usually they are chosen such that f be continuous on R when possible).

@Assinat 

A^+ is the transpose of the matrix A. 

column-red.mw

@Carl Love Not quite. E.g. for p=1 your result is complex.
 

@Earl  The proc uses practically "homogeneous coordinates". See Homogeneous coordinates - Wikipedia and the references therein. You may want to check directly that the image under f of three collinear points are also collinear. 

@mmcdara You can use this version of BruteForce and then 5/121 works. It also includes your first proc and obtains other representations too, e.g. BF(13/12, 2) = 1/2+1/3+1/4

BF := proc(a, mindenom:=1)
  local A:=a, b:=mindenom-1, s:=NULL:
  while A > 0 do
    b := max(ceil(1/A),b+1):
    A := A-1/b:
    s := s, 1/b:
  end do:
  return s;
end proc:

@student_md 

It's not difficult to see yourself that  int(Psi(s), s=0..t) = P(1)*Psi(t)

implies Psi(t)=0. [it reduces to a simple linear ODE].

@student_md 

Let's take for simplicity the dimension N=1.
Then the first line in Eq (1) ==> Psi(t) = 0. So, your method does not seem to be useful.

@pjv70 

It is hard without sources. If you just want the correct series, simply add 1, or use:

JT3:=convert(JacobiTheta3(z,q), Sum, dummy = n) assuming abs(q)<1:
series(JT3, z=0, 4);