Fourier expansions:

Consider a square pulse on [-π,π] (see Note at the end): 
	 f(x) = 1 for |x|≤1 , = 0 for 1<|x|≤ π  

It is an even function on [-π,π], so its Fourier expansion is a
   Fourier cosine series:  F(x) = a0/2 + ∑ancos(nx)   (sum from n=1 to ∞) 
   with Fourier coefficients:  an = 1/π -π∫π f(x)cosnx dx , n=0,1,2,... 
(the Fourier sine coefficients are zero since f(x)sin(nx) is odd function).
Note that a0/2 is the average of f(x) over [−π,π] (called the DC coefficient).

According to the theory of Fourier expansions:
 Since f∈L2(-π,π), the series converges to it in L2-sense, so F=f in L2-sense.
 Since f and f′ are piecewise continuous on [-π,π], the series also converges 
  pointwise to the 2π-periodic extension of f(x) ∀x∈(-∞,∞) and F(x)=(f(x-)+f(x+))/2.
  Thus, F(x)=f(x) when f(x) is continuous at x, and F(x)=average of values when 
  f(x) has a jump at x.  The F.S. simply does the best it can!

1. Sketch the graph of f(x) and its 2π-periodic extension F(x).
   To what value will the F.S. converge at x=0 ?  at x=1 ?  at x=2π-1 ?
2. Find the Fourier coefficients an explicitly (evaluate integral by hand), for any n=0,1,2,...
	 
3. To see how the Fourier Series approximates the function as the number of terms increases, 
   write a code that evaluates FN(x) = N-th partial sum of the Fourier series (sum from 0 to N), 
   and plot f(x) and FN(x) on [-π,π] on the same plot
   Compute the 2-norm of f-FN .
   Debug with N=4 terms using M=10, 50, 100 evaluation points.

4. Try N=10, 50, 100, 200.  Must use many evaluation points x for plotting, 
   at least 800 for high N, to capture the oscillations.
   How well does FN(x) approximate f(x) as N increases ?
   Is the error decreasing at all x ?  what happens at the jumps?

Note: The choice [-π,π] is for convenience, it simplifies expressions.
On an interval [-L,L], the basis functions would be cos(nπx/L) and
      a_n = 1/L _-L∫^L f(x)cos(nπx/L) dx , n=0,1,2,...