Answer:
The given function f(x) is:
f(x) = { 1, -π < x < 0
{ -1, 0 < x < π
Since the function is odd and has period 2π, the Fourier series will only have sine terms. The Fourier series of f(x) can be written as:
f(x) = Σ[b_n sin(n x)], where n >= 1
where b_n is the n-th Fourier coefficient, given by:
b_n = (1/π) ∫[0,π] f(x) sin(n x) dx
We can evaluate the integral to find the Fourier coefficients:
b_n = (1/π) ∫[0,π] f(x) sin(n x) dx
= (1/π) [ ∫[0,π/2] sin(n x) dx - ∫[π/2,π] sin(n x) dx ]
= (1/π) [ -cos(n x)/n |[0,π/2] + cos(n x)/n |[π/2,π] ]
= (1/π) [ (-cos(n π/2)/n + 1/n) - (cos(n π) - cos(n π/2))/n ]
= (1/π) [ (1 - (-1)^n) / n ]
Therefore, the Fourier series of f(x) is:
f(x) = Σ[(1 - (-1)^n)/(n π) sin(n x)]
Where the sum is taken over all odd integers n.