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GATE 2026 CE (CE2) – Question 13

Engineering Mathematics · PDE: Fourier series, separation of variables, diffusion, wave and Laplace equations · 1 mark · Multiple choice

Periodic function $f(x)$ is given below.

$$f(x) = \begin{cases} -1, & -\pi < x < 0 \\ 1, & 0 < x < \pi \end{cases}; \qquad f(x + 2\pi) = f(x)$$

The CORRECT option representing the Fourier series expansion of $f(x)$ is:

  1. $f(x) = \frac{4}{\pi}\left[\sin x + \frac{\sin 3x}{3} + \frac{\sin 5x}{5} + \cdots\right]$
  2. $f(x) = -\frac{4}{\pi}\left[\sin x + \frac{\sin 3x}{3} + \frac{\sin 5x}{5} + \cdots\right]$
  3. $f(x) = 1 + \frac{4}{\pi}\left[\sin x + \frac{\sin 3x}{3} + \frac{\sin 5x}{5} + \cdots\right]$
  4. $f(x) = 1 - \frac{4}{\pi}\left[\sin x + \frac{\sin 3x}{3} + \frac{\sin 5x}{5} + \cdots\right]$

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Correct answer: (A) $f(x) = \frac{4}{\pi}\left[\sin x + \frac{\sin 3x}{3} + \frac{\sin 5x}{5} + \cdots\right]$

Explanation

The function is odd, so only sine terms appear and the mean is zero. Then $b_n = \frac{2}{\pi}\int_0^{\pi} \sin nx\,dx = \frac{2}{n\pi}(1 - \cos n\pi)$, which is $\frac{4}{n\pi}$ for odd $n$ and 0 for even $n$. So $f(x) = \frac{4}{\pi}\left[\sin x + \frac{\sin 3x}{3} + \frac{\sin 5x}{5} + \cdots\right]$.