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The function \(f(x) = \begin{cases} x, & -\pi \lt x \lt 0 \\ -x, & 0 \lt x \lt \pi \end{cases}\) is expanded as a Fourier series of the form \(a_{0} + \sum_{n=1}^{\infty} a_{n}\cos(nx) + \sum_{n=1}^{\infty} b_{n}\sin(nx)\). Which of the following is true?

  • A.\(a_{0} \neq 0, b_{n} = 0\)
  • B.\(a_{0} \neq 0, b_{n} \neq 0\)
  • C.\(a_{0} = 0, b_{n} = 0\)
  • D.\(a_{0} = 0, b_{n} \neq 0\)

Answer: A

The answer is \(a_{0} \neq 0, b_{n} = 0\).

The function equals \(-|x|\): it is even, since \(f(-x) = f(x)\). An even function has no sine terms, so every \(b_{n} = 0\).

The constant term is the mean value over the period:

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

Treating the function as odd (because each piece is linear in x) would wrongly give \(a_{0} = 0\) and nonzero \(b_{n}\).

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