Answer:
The Fourier series of a periodic function f(x) with period 2L can be expressed as:
f(x) = a0/2 + Σ[n=1 to ∞] (ancos(nπx/L) + bnsin(nπx/L))
where
a0 = (1/L) ∫[-L,L] f(x) dx
an = (1/L) ∫[-L,L] f(x)*cos(nπx/L) dx
bn = (1/L) ∫[-L,L] f(x)*sin(nπx/L) dx
In this case, we have f(x) = 1 for 2k - 0.25 <= x <= 2k + 0.25, and f(x) = 0 otherwise. The period is 0.5, so L = 0.25.
First, we can find the value of a0:
a0 = (1/0.5) ∫[-0.25,0.25] 1 dx = 1
Next, we can find the values of an and bn:
an = (1/0.5) ∫[-0.25,0.25] 1*cos(nπx/0.25) dx = 0
bn = (1/0.5) ∫[-0.25,0.25] 1*sin(nπx/0.25) dx
Since the integrand is odd, we have:
bn = (2/0.5) ∫[0,0.25] 1*sin(nπx/0.25) dx
Using the substitution u = nπx/0.25, du/dx = nπ/0.25, dx = 0.25du/(nπ), we get:
bn = (4/nπ) ∫[0,nπ/4] sin(u) du = (4/nπ) (1 - cos(nπ/4))
Therefore, the Fourier series of f(x) can be written as:
f(x) = 1/2 + Σ[n=1 to ∞] [(4/nπ) (1 - cos(nπ/4))] * sin(nπx/0.25)
for 2k - 0.25 <= x <= 2k + 0.25, and f(x) = 0 otherwise.