PYQs / GATE CE / 2016 / Set 2 / Q13 GATE CE 2016 Set 2 — Question 13 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. MCQ +1 / -0.33 Medium Fourier Series Partial Differential Equations Engineering Mathematics
Engineering Mathematics → Partial Differential Equations → Fourier Series
Last updated 5 September 2026
Question The Fourier series of the function,
f ( x ) = 0 , − π < x ≤ 0 f(x) = 0, -\pi < x \leq 0 f ( x ) = 0 , − π < x ≤ 0 = π − x , 0 < x < π = \pi - x, 0 < x < \pi = π − x , 0 < x < π in the interval
[ − π , π ] [-\pi, \pi] [ − π , π ] is
f ( x ) = π 4 + 2 π [ cos x 1 2 + cos 3 x 3 2 + … ] + [ sin x 1 + sin 2 x 2 + sin 3 x 3 + … ] . f(x) = \frac{\pi}{4} + \frac{2}{\pi} \left[ \frac{\cos x}{1^2} + \frac{\cos 3x}{3^2} + \dots \right] + \left[ \frac{\sin x}{1} + \frac{\sin 2x}{2} + \frac{\sin 3x}{3} + \dots \right]. f ( x ) = 4 π + π 2 [ 1 2 cos x + 3 2 cos 3 x + … ] + [ 1 sin x + 2 sin 2 x + 3 sin 3 x + … ] . The convergence of the above Fourier series at
x = 0 x = 0 x = 0 gives
Correct answer (C) Σₙ₌₁^(∞) (1)/((2n-1)²) = (π²)/(8)
Solution 1. According to Dirichlet's theorem, at a point of discontinuity x = x 0 x = x_0 x = x 0 f ( x 0 − ) + f ( x 0 + ) 2 \frac{f(x_0^-) + f(x_0^+)}{2} 2 f ( x 0 − ) + f ( x 0 + ) 2. At x = 0 x = 0 x = 0 f ( 0 − ) = 0 f(0^-) = 0 f ( 0 − ) = 0 f ( 0 + ) = π − 0 = π f(0^+) = \pi - 0 = \pi f ( 0 + ) = π − 0 = π 3. The series converges to 0 + π 2 = π 2 \frac{0 + \pi}{2} = \frac{\pi}{2} 2 0 + π = 2 π 4. Substituting x = 0 x = 0 x = 0 f ( 0 ) = π 4 + 2 π [ cos 0 1 2 + cos 0 3 2 + … ] + [ sin 0 1 + sin 0 2 + … ] f(0) = \frac{\pi}{4} + \frac{2}{\pi} \left[ \frac{\cos 0}{1^2} + \frac{\cos 0}{3^2} + \dots \right] + \left[ \frac{\sin 0}{1} + \frac{\sin 0}{2} + \dots \right] f ( 0 ) = 4 π + π 2 [ 1 2 c o s 0 + 3 2 c o s 0 + … ] + [ 1 s i n 0 + 2 s i n 0 + … ] π 2 = π 4 + 2 π ∑ n = 1 ∞ 1 ( 2 n − 1 ) 2 + 0 \frac{\pi}{2} = \frac{\pi}{4} + \frac{2}{\pi} \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} + 0 2 π = 4 π + π 2 ∑ n = 1 ∞ ( 2 n − 1 ) 2 1 + 0 5. Rearranging the terms:
π 2 − π 4 = 2 π ∑ n = 1 ∞ 1 ( 2 n − 1 ) 2 \frac{\pi}{2} - \frac{\pi}{4} = \frac{2}{\pi} \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} 2 π − 4 π = π 2 ∑ n = 1 ∞ ( 2 n − 1 ) 2 1 π 4 = 2 π ∑ n = 1 ∞ 1 ( 2 n − 1 ) 2 \frac{\pi}{4} = \frac{2}{\pi} \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} 4 π = π 2 ∑ n = 1 ∞ ( 2 n − 1 ) 2 1 ∑ n = 1 ∞ 1 ( 2 n − 1 ) 2 = π 2 8 \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \frac{\pi^2}{8} ∑ n = 1 ∞ ( 2 n − 1 ) 2 1 = 8 π 2 .
Turn this into a strength. Explore AI-powered practice and doubt support with Success Tracker. Review answer and solution without JavaScript Interactive answer checking needs JavaScript. The published solution is available below.
Correct answer (C) Σₙ₌₁^(∞) (1)/((2n-1)²) = (π²)/(8)
Solution 1. According to Dirichlet's theorem, at a point of discontinuity x = x 0 x = x_0 x = x 0 f ( x 0 − ) + f ( x 0 + ) 2 \frac{f(x_0^-) + f(x_0^+)}{2} 2 f ( x 0 − ) + f ( x 0 + ) 2. At x = 0 x = 0 x = 0 f ( 0 − ) = 0 f(0^-) = 0 f ( 0 − ) = 0 f ( 0 + ) = π − 0 = π f(0^+) = \pi - 0 = \pi f ( 0 + ) = π − 0 = π 3. The series converges to 0 + π 2 = π 2 \frac{0 + \pi}{2} = \frac{\pi}{2} 2 0 + π = 2 π 4. Substituting x = 0 x = 0 x = 0 f ( 0 ) = π 4 + 2 π [ cos 0 1 2 + cos 0 3 2 + … ] + [ sin 0 1 + sin 0 2 + … ] f(0) = \frac{\pi}{4} + \frac{2}{\pi} \left[ \frac{\cos 0}{1^2} + \frac{\cos 0}{3^2} + \dots \right] + \left[ \frac{\sin 0}{1} + \frac{\sin 0}{2} + \dots \right] f ( 0 ) = 4 π + π 2 [ 1 2 c o s 0 + 3 2 c o s 0 + … ] + [ 1 s i n 0 + 2 s i n 0 + … ] π 2 = π 4 + 2 π ∑ n = 1 ∞ 1 ( 2 n − 1 ) 2 + 0 \frac{\pi}{2} = \frac{\pi}{4} + \frac{2}{\pi} \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} + 0 2 π = 4 π + π 2 ∑ n = 1 ∞ ( 2 n − 1 ) 2 1 + 0 5. Rearranging the terms:
π 2 − π 4 = 2 π ∑ n = 1 ∞ 1 ( 2 n − 1 ) 2 \frac{\pi}{2} - \frac{\pi}{4} = \frac{2}{\pi} \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} 2 π − 4 π = π 2 ∑ n = 1 ∞ ( 2 n − 1 ) 2 1 π 4 = 2 π ∑ n = 1 ∞ 1 ( 2 n − 1 ) 2 \frac{\pi}{4} = \frac{2}{\pi} \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} 4 π = π 2 ∑ n = 1 ∞ ( 2 n − 1 ) 2 1 ∑ n = 1 ∞ 1 ( 2 n − 1 ) 2 = π 2 8 \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \frac{\pi^2}{8} ∑ n = 1 ∞ ( 2 n − 1 ) 2 1 = 8 π 2 .
Understand the concept, then try another question Revisit Engineering Mathematics with concept notes, common mistakes and an original worked example before your next attempt.
More questions on Partial Differential Equations ← Q12 Full paper Q14 → 