GATE 2016 CE (CE2) – Question 13

Engineering Mathematics · Calculus: Integrals, partial and total derivatives, gradient, divergence, curl, line, surface and volume integrals · 1 mark · Multiple choice

On $[-\pi,\pi]$, $f(x)=0$ for $-\pi<x\le0$ and $f(x)=\pi-x$ for $0<x<\pi$. Its Fourier series is $f(x)=\pi/4+(2/\pi)\sum_{n=1}^{\infty}\cos((2n-1)x)/(2n-1)^2+\sum_{n=1}^{\infty}\sin(nx)/n$. Convergence at $x=0$ gives ______.

$\sum_{n=1}^{\infty}1/n^2=\pi^2/6$
$\sum_{n=1}^{\infty}(-1)^{n+1}/n^2=\pi^2/12$
$\sum_{n=1}^{\infty}1/(2n-1)^2=\pi^2/8$
$\sum_{n=1}^{\infty}(-1)^{n+1}/(2n-1)=\pi/4$

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Correct answer: (C) $\sum_{n=1}^{\infty}1/(2n-1)^2=\pi^2/8$

Explanation

At x=0 the one-sided limits are0 andπ, so the Fourier series converges to their averageπ/2. Substitution in the series eliminates the sine terms and sets all cosine terms to1. Thus π/2=π/4+(2/π)sum[1/(2n−1)²], giving the odd-reciprocal-square sum **π²/8, C**.