Multiple choice

The Fourier series expansion of a symmetric and even function, ƒ(x) where f(x) = 1 + $(2x/ \pi), - \pi$< x < 0 and = 1– $(2\pi/\pi)$0 < x < $\pi$ will be

  1. $\sum\limits_{n=1}^\infty (4 / \pi^2n^2)(1 + cos n \pi)$
  2. $\sum\limits_{n=1}^\infty (4 / \pi^2 n^2)(1 - cos n \pi)$
  3. $\sum\limits_{n=1}^\infty (4 / \pi^2 n^2)(1 - sin n \pi)$
  4. $\sum\limits_{n=1}^\infty (4 / \pi^2 n^2)(1 + sin n \pi)$
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B Correct answer
Explanation

For an even function f(x) = 1 + 2x/π for -π < x < 0 and f(x) = 1 - 2x/π for 0 < x < π, the Fourier cosine coefficients are a₀ = 2 and an = (4/π²n²)(1 - cos nπ). The term (1 - cos nπ) gives 2 for odd n and 0 for even n, resulting in only odd harmonics.

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