Let’s solve each part step by step: --- ### **(a) Determine the Fourier coefficients \( \hat{f}_n \).** Given: \[ f(x) = \begin{cases} 1, & 0 < x < \pi \\ -1, & -\pi < x < 0 \end{cases} \] with period \( 2\pi \). The complex Fourier coefficients are: \[ \hat{f}_n = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) e^{-inx} dx \] Split the integral: \[ \hat{f}_n = \frac{1}{2\pi} \left( \int_{-\pi}^0 (-1) e^{-inx} dx + \int_0^\pi (1) e^{-inx} dx \right) \] Calculate each part: - \( \int_{-\pi}^0 (-1) e^{-inx} dx = -\int_{-\pi}^0 e^{-inx} dx \) - \( \int_0^\pi (1) e^{-inx} dx = \int_0^\pi e^{-inx} dx \) Evaluate: \[ \int e^{-inx} dx = \frac{e^{-inx}}{-in} \] So, \[ \int_{-\pi}^0 e^{-inx} dx = \left[ \frac{e^{-inx}}{-in} \right]_{-\pi}^0 = \frac{1 - e^{in\pi}}{in} \] \[ \int_0^{\pi} e^{-inx} dx = \left[ \frac{e^{-inx}}{-in} \right]_0^{\pi} = \frac{e^{-in\pi} - 1}{-in} \] So, \[ \hat{f}_n = \frac{1}{2\pi} \left( -\frac{1 - e^{in\pi}}{in} + \frac{e^{-in\pi} - 1}{-in} \right) \] \[ = \frac{1}{2\pi} \cdot \frac{-(1 - e^{in\pi}) - (e^{-in\pi} - 1)}{in} \] \[ = \frac{1}{2\pi} \cdot \frac{-(1 - e^{in\pi}) - e^{-in\pi} + 1}{in} \] \[ = \frac{1}{2\pi} \cdot \frac{-1 + e^{in\pi} - e^{-in\pi} + 1}{in} \] \[ = \frac{1}{2\pi} \cdot \frac{e^{in\pi} - e^{-in\pi}}{in} \] \[ = \frac{1}{2\pi} \cdot \frac{2i \sin(n\pi)}{in} \] \[ = \frac{1}{\pi} \cdot \frac{\sin(n\pi)}{n} \] But \( \sin(n\pi) = 0 \) for all integers except for odd \( n \), for which it is 0, but let's check the sine form. However, the real Fourier series (sine-cosine form) for an odd function over \( (-\pi, \pi) \) is: \[ f(x) = \sum_{n=1}^{\infty} b_n \sin(nx) \] where \[ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) dx \] But \( f(x) \) is odd, so the cosine coefficients vanish. Calculate \( b_n \): \[ b_n = \frac{1}{\pi} \left( \int_{-\pi}^0 (-1) \sin(nx) dx + \int_0^\pi (1) \sin(nx) dx \right) \] Evaluate: \[ \int \sin(nx) dx = -\frac{1}{n} \cos(nx) \] So, \[ \int_{-\pi}^0 \sin(nx) dx = \left[ -\frac{1}{n} \cos(nx) \right]_{-\pi}^0 = -\frac{1}{n} [\cos(0) - \cos(-n\pi)] = -\frac{1}{n} [1 - \cos(n\pi)] \] Similarly, \[ \int_0^\pi \sin(nx) dx = \left[ -\frac{1}{n} \cos(nx) \right]_0^\pi = -\frac{1}{n} [\cos(n\pi) - \cos(0)] = -\frac{1}{n} [\cos(n\pi) - 1] \] Putting together: \[ b_n = \frac{1}{\pi} \left( -\int_{-\pi}^0 \sin(nx) dx + \int_0^\pi \sin(nx) dx \right) \] \[ = \frac{1}{\pi} \left( -\left(-\frac{1}{n} [1 - \cos(n\pi)] \right) -\frac{1}{n} [\cos(n\pi) - 1] \right) \] \[ = \frac{1}{\pi} \left( \frac{1}{n} [1 - \cos(n\pi)] -\frac{1}{n} [\cos(n\pi) - 1] \right) \] \[ = \frac{1}{\pi} \cdot \frac{1}{n} (1 - \cos(n\pi) - \cos(n\pi) + 1) \] \[ = \frac{1}{\pi} \cdot \frac{2}{n} (1 - \cos(n\pi)) \] But \( \cos(n\pi) = (-1)^n \), so: \[ b_n = \frac{2}{\pi n} (1 - (-1)^n) \] Thus, - For even \( n \): \( (-1)^n = 1 \implies b_n = 0 \) - For odd \( n \): \( (-1)^n = -1 \implies b_n = \frac{4}{\pi n} \) **Final answer for (a):** \[ \boxed{ b_n = \begin{cases} \frac{4}{\pi n}, & \text{if } n \text{ is odd} \\ 0, & \text{if } n \text{ is even} \end{cases} } \] --- Would you like to continue step-by-step with the next parts?