Fourier series in the interval (0,2π) Problem 4 | Engineering Mathematics
Expand f(x) = x(2π-x) as Fourier series in the interval (0,2π). Hence deduce that the sum of 1/1^2 +1/2^2 +1/3^2 +1/4^2 …= π^2/6
Expand f(x) = x(2π-x) as Fourier series in the interval (0,2π). Hence deduce that the sum of 1/1^2 +1/2^2 +1/3^2 +1/4^2 …= π^2/6
