Construct the Fourier series of the following periodic function.
f(x)={█(cosx,-π≤x<0,1,0≤x<π)┤
To find the Fourier series of the given periodic function, we will first determine the coefficients of the series. The Fourier coefficients can be calculated using the following formula:
a0 = (1/2π) ∫[0,π] f(x) dx
an = (1/π) ∫[0,π] f(x) cos(nx) dx
bn = (1/π) ∫[0,π] f(x) sin(nx) dx
Let's start by finding the coefficient a0:
a0 = (1/2π) ∫[0,π] f(x) dx
= (1/2π) ∫[0,π] 1 dx
= (1/2π) [x] from 0 to π
= (1/2π) (π - 0)
= 1/2
Next, calculate the coefficients an by integrating the given function multiplied by cos(nx):
an = (1/π) ∫[0,π] f(x) cos(nx) dx
For n = 1:
a1 = (1/π) ∫[0,π] (1)(cos(x)) dx
= (1/π) ∫[0,π] cos(x) dx
= (1/π) [sin(x)] from 0 to π
= (1/π) (sin(π) - sin(0))
= 0
For n > 1, the integral of the cos(nx) term over [0,π] will result in zero. Hence, all an (for n > 0) will be zero.
Now, let's calculate the coefficients bn by integrating the given function multiplied by sin(nx):
bn = (1/π) ∫[0,π] f(x) sin(nx) dx
For n = 1:
b1 = (1/π) ∫[0,π] (1)(sin(x)) dx
= (1/π) ∫[0,π] sin(x) dx
= (1/π) [-cos(x)] from 0 to π
= (1/π) (-cos(π) + cos(0))
= (1/π) (-(-1) + 1)
= 2/π
For n > 1, the integral of the sin(nx) term over [0,π] will result in zero. Hence, all bn (for n > 1) will be zero.
Therefore, the Fourier series of the given periodic function is:
f(x) = 1/2 + (2/π)sin(x)
This is the representation of the given function as an infinite sum of sine functions, weighted by the coefficients calculated.