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Consider the function

\(f(x) = |x| \quad -1 \le x \le 1\)

Then the Fourier series for \(f\) is :

  • A.\(\frac{1}{2} - \frac{4}{\pi^{2}}\)\((\frac{\cos \pi x}{1^{2}} + \frac{\cos 3\pi x}{3^{2}} + \cdots)\)
  • B.\(\frac{1}{2} - \frac{4}{\pi^{2}}\)\((\frac{\sin \pi x}{1^{2}} + \frac{\sin 3\pi x}{3^{2}} + \cdots)\)
  • C.\(\frac{1}{2} - \frac{4}{\pi^{2}}\)\((\frac{\cos \pi x}{1^{2}} + \frac{\cos 3\pi x}{3^{2}} + \cdots)\) \(- \frac{4}{\pi^{2}}\)\((\frac{\sin \pi x}{1^{2}} + \frac{\sin 3\pi x}{3^{2}} + \cdots)\)
  • D.\(\frac{1}{2} + \frac{4}{\pi^{2}}\)\((\frac{\cos \pi x}{1^{2}} + \frac{\cos 3\pi x}{3^{2}} + \cdots)\) \(+ \frac{4}{\pi^{2}}\)\((\frac{\sin \pi x}{1^{2}} + \frac{\sin 3\pi x}{3^{2}} + \cdots)\)

Answer: A

The Fourier series is \(\frac{1}{2} - \frac{4}{\pi^{2}}(\cos \pi x + \frac{1}{9}\cos 3\pi x + \cdots)\), the cosine series with odd harmonics only.

\(|x|\) is even, so it has a cosine series only (which rules out the options with sine terms). On [-1, 1], with period 2:

\[ \begin{aligned} a_{0} &= \int_{-1}^{1} |x|\,dx = 1 \\ a_{n} &= 2\int_{0}^{1} x\cos n\pi x\,dx \\ &= \frac{2((-1)^{n} - 1)}{n^{2}\pi^{2}} \end{aligned} \]

So \(a_{n} = -4/(n^{2}\pi^{2})\) for odd \(n\) and 0 for even \(n\), and the constant term is \(a_{0}/2 = 1/2\). The minus sign is right: at \(x = 0\) the series must give 0, and \(1/2 - (4/\pi^{2})(\pi^{2}/8) = 0\).

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