Question

Consider the \( 2\pi \)-periodic function defined by \( f(x) = \begin{cases} -1, & \text{if } -\pi < x \leq 0, \\ 1, & \text{if } 0 < x \leq \pi. \end{cases} \) 
Which of the following is/are correct about its Fourier series expansion, \( \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos nx + b_n \sin nx \)?

• A. \( a_n = \frac{1}{n} \, \forall n = 1, 2, \dots \)
• B. \( a_0 = 0 \)
• C. \( b_n = \frac{4}{n\pi} \) if \( n \) is odd
• D. \( b_n = -\frac{4}{n\pi} \) if \( n \) is even

Hint:

For Fourier series, use the symmetry properties of the function (even or odd) to simplify calculations. Odd functions contribute only to \( b_n \), while even functions contribute only to \( a_n \).

Answer 1

The Correct Option is B

Step 1: Fourier coefficients. The Fourier series coefficients for a periodic function \( f(x) \) with period \( 2\pi \) are given by: \[ a_0 = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \, dx, \quad a_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \cos(nx) \, dx, \quad b_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \sin(nx) \, dx. \] Step 2: Calculation of \( a_0 \). \[ a_0 = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \, dx = \frac{1}{\pi} \left[ \int_{-\pi}^0 (-1) \, dx + \int_0^\pi 1 \, dx \right] = \frac{1}{\pi} \left[ -\pi + \pi \right] = 0. \] 
Step 3: Calculation of \( a_n \). \[ a_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \cos(nx) \, dx = \frac{1}{\pi} \left[ \int_{-\pi}^0 (-1) \cos(nx) \, dx + \int_0^\pi 1 \cos(nx) \, dx \right]. \] Using the properties of cosine (even function), the integrals cancel out for all \( n \). Hence, \( a_n = 0 \) for all \( n \). 
Step 4: Calculation of \( b_n \). \[ b_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \sin(nx) \, dx = \frac{1}{\pi} \left[ \int_{-\pi}^0 (-1) \sin(nx) \, dx + \int_0^\pi 1 \sin(nx) \, dx \right]. \] Simplifying: \[ b_n = \frac{1}{\pi} \left[ \int_{-\pi}^0 -\sin(nx) \, dx + \int_0^\pi \sin(nx) \, dx \right] = \frac{1}{\pi} \left[ \frac{2}{n} \cos(nx) \Big|_{0}^\pi \right]. \] Evaluating for \( n \), we find: - \( b_n = \frac{4}{n\pi} \) for \( n \) odd. - \( b_n = 0 \) for \( n \) even. 
Conclusion: - \( a_0 = 0 \), confirming \( \mathbf{(B)} \). - \( b_n = \frac{4}{n\pi} \) for \( n \) odd, confirming \( \mathbf{(C)} \). - \( b_n = 0 \) for \( n \) even, so \( \mathbf{(D)} \) is incorrect. - \( a_n = 0 \) for all \( n \), so \( \mathbf{(A)} \) is incorrect.