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.

@Markiyan Hirnyk 

Please be more explicit. Are you saying that the solution plotted is not correct?
The question was to plot the curve, not to prove the unicity (I know how to do it if needed).


 

@Jjjones98 

The Fourier series in cos is the same as the Fourier series (sin&cos), the function being even in [-Pi,Pi].

Anyway, in your case:

restart;
f:=cos(alpha*x):
A:=n->1/Pi*int(f*cos(n*x),x=-Pi..Pi); # Fourier coeffs
alpha:=3/4:
f6:=A(0)/2+add(A(n)*cos(n*x),n=1..6):
numapprox:-infnorm(f-f6,x=0..Pi);

     0.5206935226e-1

You may want to compute the L^2 norm.

You cannot have two distinct Fourier series; it is unique.

@gaurav_rs 

Just change 'complex' to 'real'.

@gaurav_rs 

Probably you use an old version of Maple. In recent versions the answer is very fast.

Please post a link to such a question, I did not find one.

Your second Identity is wrong (some +/- are inverted).

@Adam Ledger