The Fourier series of the function, f(x) = 0, -π < x ≤ 0 = π - x, 0 < x < π in the interval [= π, π] is f (x) = \frac{\pi }{4} + \frac{2}{\pi } [\frac{\cos ⁡x}{1^{2}} + \frac{\cos ⁡3x}{3^{2}}\ldots\ldots\ldots] + [\frac{ \sin ⁡x}{1} + \frac{ \sin ⁡2x}{2} + \frac{ \sin ⁡3x}{3} + \ldots\ldots\ldots] The convergence of the above Fourier series at 𝑥 = 0 gives

Correct Answer is : ∑n=1∞⁡1(2n−1)2=π28

\sum ↙\stackrel{\infty}{n = 1}⁡\frac{1}{n^{2}} = \frac{\pi ^{2}}{6}

\sum ↙\stackrel{\infty}{n = 1}⁡ { (−1) ^{n + 1}}/{n^{2}} = \frac{\pi ^{2}}{12}

\sum ↙\stackrel{\infty}{n = 1}⁡\frac{1}{ (2n−1) ^{2}} = \frac{\pi ^{2}}{8}

\sum ↙\stackrel{\infty}{n = 1}⁡\frac{ (−1) ^{n + 1}}{2n−1} = \frac{\pi }{4}

GATE Civil Engineering (CE) 2016 Shift 2 Solved Paper

Question : 13 of 65

Marks: +1, -0

The Fourier series of the function,
f(x) = 0, -π < x ≤ 0
= π - x, 0 < x < π
in the interval [= π, π] is

f (x) = \;{π\;}/{4} + \;{2}/{π\;} [\;{\cos\;⁡x}/{1^{2}} + \;{\cos\;⁡3x}/{3^{2}}………] + [\;{\ sin\;⁡x}/{1} + \;{\ sin\;⁡2x}/{2} + \;{\ sin\;⁡3x}/{3} + ………]

The convergence of the above Fourier series at 𝑥 = 0 gives

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