@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 

@Yee Voon 

OK then, I just wanted to prevent a typo.

@acer 

Yes, heatmap displays the image as a background.
I don't understand why some of the PLOT objects are not documented
(even if they are not supposed to be used by the user).

You should consider the following issues about your posts:

1. Nobody wants to see a same problem posted again and again.
If it has not been answered, probably it cannot be answered
(for various reasons).

2. You use Maple 12. Very few users are able to run this version.

3. You use the document mode. This mode should be used only for presentations,
when the code works.
For testing and debugging, this would imply a conversion
and probably most users prefer to do something else with their time.

4. If you discovered a strange result produced by Maple,
it would be useful to exhibit the simplest version where
this abnormal result is still present.

@Rouben Rostamian  

Yes, because evalf/evalhf introduces some 0.*I
plot(Re(ans), ...)  solves the problem.

@alimahmood 

n is the dimension of the square matrix Z. Maple cannot use a symbolic n, so, n must be specified (any positive integer).

Probably y(...) means  y*(...)  but  sigma[s] = ?

@AmirHosein Sadeghimanesh 

You should be aware that it is not correct to use Riemann sums for improper integrals. E.g.

Int(sin(1/t^6)/t^6, t = 0 .. 1);

Here the Riemann sums diverge but the integral converges (not absolutely).

@AmirHosein Sadeghimanesh 

I assume that you have the absolute value in the integral. Then, by Fubini-Tonelli one may swap the integrals and the inner one is +oo.
If you remove the absolute value, some computations are needed and it seems that the iterated integral still diverges (but I had not the time for complete analysis in this case).