Question

Find the sum of \(\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} \) +……… using Fourier series expansion if f(x) = a when [0,π] and 2 π – x when [ π, 2 π].

(a) \(\frac{π^2}{8} \)

(b) \(\frac{π^2}{4} \)

(c) \(\frac{π^2}{16} \)

(d) \(\frac{π^2}{2} \)

The question was posed to me in unit test.

My question is from Fourier Series Expansions in portion Fourier Series of Engineering Mathematics

Answer 1

Correct option is (a) \(\frac{π^2}{8} \)

Explanation: \(a_0 = \frac{1}{π} (\int_0^π x dx)(\int_π^{2π} (2π-x)  dx) \)

\( = \frac{1}{π} [π^2+4π^2-4 \frac{π^2}{2}-2π^2+\frac{π^2}{2}] \)

\( = π.\)

\(a_n = \frac{1}{π} (\int_0^π (xcos(nx)dx) )(\int_π^{2π}((2π-x)cos(nx))dx) \)

\( = \frac{-4}{πn^2} \) when n is odd and 0 when n is even

\( b_n = 0 \)

now, substituting x=0 in the given function and the Fourier series expansion, we get,

\(0 = \frac{π}{2}-\frac{4}{π} (\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} +….) \)

Therefore, \(\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} +………= \frac{π^2}{8} \).