Question

(1) Obtain the fourier series of a function,
f(x)=0 ; -π<x<0
=1 ;0<x<π​

Answer 1

Answer:

Find the Fourier series expansion of the following periodic function f(x) of period p = 2pi. f(x) {x + pi - pi < x less than ore qualto 0 pi 0 < x less than ore qualto pi By letting x = 0 in the result obtained in prove that pi^2/8 = sigma^infinity_n = 1 1/(2n - 1)^2.