Question

(The Gibbs phenomenon). Consider the function f(x) = { −1, −π≤x<0

1, 0its 2π periodic extension. (a) Compute its Fourier coefficients (because of the parity of the function you get a Fourier sine series);

Answer 1

The Fourier coefficients of the given function can be computed by using the formula: a_n = (1/π) ∫_0^2π f(x) sin(nπ x) dx. Since the function f(x) is odd, its Fourier series coefficients will be given by the Fourier sine series coefficients. The Fourier coefficients of the given function are (4/π) ((-1)^(n+1))/n.

Step-by-step explanation:

The Fourier series coefficients of the given function can be computed by using the formula:

an = \(\frac{1}{\pi}\) \(\int_{0}^{2\pi}\) f(x) sin(n\pi x) dx

Here, since the function f(x) is odd, its Fourier series coefficients will be given by the Fourier sine series coefficients.

Since f(x) = -1 for -\(\pi\) \u2264 x < 0 and f(x) = 1 for 0 < x < \(\pi\), and f(x) is periodic with period 2\(\pi\), we can write its Fourier series as:

f(x) = \(\frac{4}{\pi}\) \(\sum_{n=1}^{\infty}\) (\(\frac{(-1)^{n+1}}{n}\)) \(\sin(nx)\)

Therefore, the Fourier coefficients of the given function are \(\frac{4}{\pi}\) \(\frac{(-1)^{n+1}}{n}\).