Exercise: Find the Fourier series for f(x) = x on-2≤x≤ 2.
To find the Fourier series for f(x) = x on the interval -2 ≤ x ≤ 2, we can use the formula for the Fourier coefficients:
cn = (1/T) * ∫[T/2,-T/2] f(x) * e^(-i * 2π * n * x / T) dx
where T is the period of the function. In this case, the period is T = 4.
The integral becomes:
cn = (1/4) * ∫[-2,2] x * e^(-i * 2π * n * x / 4) dx
Evaluate the integral:
cn = (1/4) * [(x * (-4iπn) / (2πn))^(-2,2) - ∫[-2,2] (-4iπn) / (2πn) * e^(-i * 2π * n * x / 4) dx]
Simplify:
cn = (1/4) * [(x * (-4iπn) / (2πn))^(-2,2) + (4iπn)^(-2,2) * ∫[-2,2] e^(-i * 2π * n * x / 4) dx]
Using properties of complex exponentials, we know that ∫ e^(kx) dx = (1/k) * e^(kx). Plugging that into the integral:
cn = (1/4) * [(x * (-4iπn) / (2πn))^(-2,2) + (4iπn)^(-2,2) * (4 * (1/(2πn)) * [e^(-i * 2π * n * x / 4)])]
Further simplifying:
cn = (1/4) * [(x * (-4iπn) / (2πn))^(-2,2) + (4iπn)^(-2,2) * (1/(2πn)) * [e^(-i * 2π * n * x / 4)](-2,2)]
Simplify the exponentials and evaluate the integral bounds:
cn = (1/4) * [(x * (-4iπn) / (2πn))^(-2,2) + (4iπn)^(-2,2) * (1/(2πn)) * [(e^(-i * π * n) - e^(i * π * n))(1 - e^(i * π * n))]]
Simplify further using Euler's formula (e^(iθ) = cos(θ) + i * sin(θ)):
cn = (1/4) * [(x * (-4iπn) / (2πn))^(-2,2) + (4iπn)^(-2,2) * (1/(2πn)) * [(cos(πn) - i * sin(πn)) - (cos(πn) + i * sin(πn)))) * (1 - (cos(πn) + i * sin(πn)))]]
Simplify the trigonometric terms:
cn = (1/4) * [(x * (-4iπn) / (2πn))^(-2,2) + (4iπn)^(-2,2) * (1/(2πn)) * [(-2i * sin(πn)) * (2i * sin(πn))]]
Simplify further:
cn = (1/4) * [(x * (-4iπn) / (2πn))^(-2,2) + (4iπn)^(-2,2) * (1/(2πn)) * (-4 * sin^2(πn))]
We can simplify the expression inside the brackets and calculate the values of cn for different values of n to determine the Fourier series for f(x) = x.