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Official PYQ · 1993Q01.5Calculus: Limits, Continuity, DifferentiabilityMCQ1 mark

GATE CS 1993 Q01.5 — Calculus: Limits, Continuity, Differentiability

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Fourier series of the periodic function (period 2π) defined by

f(x)={0,−p<x<0x,0<x<p is π4+∑[1πn2(cos⁡nπ−1)cos⁡nx−1ncos⁡nπsin⁡nx]f(x) = \begin{cases} 0, -p < x < 0\\x, 0 < x < p \end{cases} \text{ is }\\ \frac{\pi}{4} + \sum \left [ \frac{1}{\pi n^2} \left(\cos n\pi - 1 \right) \cos nx - \frac{1}{n} \cos n\pi \sin nx \right ]

But putting x=πx = \pi, we get the sum of the series

1+132+152+172+⋯ is 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \cdots \text{ is } 

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Answer: C

By Dirichlet's theorem, at a point of discontinuity x=πx = \pi, the Fourier series of a periodic function f(x)f(x) with period 2π2\pi converges to the average of its left-hand and right-hand limits:

Sum of the series at x=π is f(π−)+f(−π+)2=π+02=π2\text{Sum of the series at } x = \pi \text{ is } \frac{f(\pi^-) + f(-\pi^+)}{2} = \frac{\pi + 0}{2} = \frac{\pi}{2}

Now, we substitute x=πx = \pi into the given Fourier series:

π4+∑n=1∞[1πn2(cos⁡nπ−1)cos⁡nπ−1ncos⁡nπsin⁡nπ]=π2\frac{\pi}{4} + \sum_{n=1}^{\infty} \left[ \frac{1}{\pi n^2} (\cos n\pi - 1) \cos n\pi - \frac{1}{n} \cos n\pi \sin n\pi \right] = \frac{\pi}{2}

Since sin⁡nπ=0\sin n\pi = 0 and cos⁡nπ=(−1)n\cos n\pi = (-1)^n for all integers nn:

π4+∑n=1∞1πn2((−1)n−1)(−1)n=π2\frac{\pi}{4} + \sum_{n=1}^{\infty} \frac{1}{\pi n^2} ((-1)^n - 1)(-1)^n = \frac{\pi}{2} π4+∑n=1∞1πn2(1−(−1)n)=π2\frac{\pi}{4} + \sum_{n=1}^{\infty} \frac{1}{\pi n^2} (1 - (-1)^n) = \frac{\pi}{2}

The term 1−(−1)n1 - (-1)^n is 00 when nn is even, and 22 when nn is odd. Thus, only odd terms n=1,3,5,…n = 1, 3, 5, \dots survive:

π4+2π(1+132+152+172+⋯ )=π2\frac{\pi}{4} + \frac{2}{\pi} \left( 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \cdots \right) = \frac{\pi}{2}

Solving for the required sum SS:

2πS=π2−π4=π4  ⟹  S=π28\frac{2}{\pi} S = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \implies S = \frac{\pi^2}{8}

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