So, you have a kind of "almost" space filling curve  (see wiki).
- What is the source for this curve?
- Do you know to prove mathematically the claimed property? Or you want some kind of "incomplete" proof using Maple?
Or maybe you want to use this curve for something else?

@tomleslie 

It is not difficult to explain.
When SolveEquations solves f(x)=0, it actually minimizes an associated function F(x)  (such as F(x) = f(x)^2).

A minimization with constraints g(x)<=0  is much more difficult (mainly for several variables). It is not possible in general just to minimize F and then select the solutions satisfying g(x)<=0  because in most cases the set of such solutions could be empty. It is faster for the user to select himself the desired solutions if he knows that such solutions exist and they are not too many.

@Earl 

You have mentioned the tangency. In this case of a not differentiable curve it should be interpreted as having at least a common point and disjoint interiors. This can be done but it's of course more complicated.

So, without shades of grey and without thickness? Then make a custom legend using textplot, line, pointplot.

@weidade37211 

@Earl 

If you really want exact tangency with a circle  you should solve the problem separately.
Since the curve is not differentiable for t=k*Pi (k in Z), these points need special atention, fsolve/DS only is not enough.

@tomleslie 

Combining yours with mine:

r1:=t->piecewise(t<Pi, sqrt(exp((t+(1/4)*Pi)*(1/2))),
                t<2*Pi, sqrt(exp((9*Pi/4-t)*(1/2)))
               ):
r:=t->r1(t - 2*Pi*floor(t/(2*Pi))):
plots:-animate(plot, 
[[[r(t-u)*cos(t),r(t-u)*sin(t),t=0..2*Pi] ,
[[0,r(Pi/2-u)*sin(Pi/2)],[0,2+r(Pi/2-u)*sin(Pi/2)]]   ],thickness=[1,6]],
u=0..2*Pi, scaling=constrained);

@Adam Ledger 

For this specific s()  it does not matter too much; it's only a shift of the sequence with one position. 
Of course for a recurrence  s(n) = f(n, s(n-1)) rather than  s(n) = f(s(n-1)),  the starting index could be important.

Just a remark: this way the climber will never reach the top of the mountain :-)   [it will take an ethernity!]

You should isolate the problem. You have a large execution group in 2D input which is difficult (time consumming) to split/manage.

@tsunamiBTP 

Your equation Q1=0  is polynomial of degree m in X = cos(2*Pi*x/T).
So, it cannot be solved analitically except possibly in some very special situations; you will have to compute the Galois group in order to find these situations. This is very complicated and useless for this problem.

P.S. Beying polynomial, you don't need RootFinding; you can simply find with solve all the roots of the polynomial (numerically).

Anyway the splash screen is essential because the Java interface is slow (even on decent computers) and the user has to see something until the worksheet appears.

@Markiyan Hirnyk 

You started the game asking for maths and proofs. I gave you a distilled version of the mathematical result, and you don't like it. I cannot do anything about it, I'm sorry too.

@Markiyan Hirnyk 

1. The addition of x>0 is just my method to get rid of the curve not defined at t=0.
2. These are standard notions, see wiki or any topology textbook. The theorem is elementary.
f,g are the roots (depending on t) of the quadratic equation in x (OP's f(x,t)=0).

@Markiyan Hirnyk 

OK, let's play.

1. There is only one example here.

2. The following theorem applies:
If X is a connected topological space, Y a Hausdorff space and f,g: X --> Y are continuous such that f(x)<>g(x) for x in X
then the multifunction F : X -> 2^Y, F(x) = {f(x), g(x)} has exactly two continuous selections.

Use this for X = (0,4].

@Markiyan Hirnyk 

For this example it is easy to see that there is a unique continuous curve defined in [0,4] and as shown, it can be obtained at once.
Of course, for a more general case, Preben's method should be used.