We are tasked with finding the Fourier series expansion for the following periodic function with a period of \( \pi \): \[ f(x) = \begin{cases} 1, & 0 \leq x < \frac{\pi}{2} \\ -1, & \frac{\pi}{2} \leq x < \pi \end{cases} \] ### Step 1: Determine the Fourier Series Formula The Fourier series expansion of a periodic function \( f(x) \) with period \( T = \pi \) is given by: \[ f(x) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos\left(\frac{2n\pi x}{T}\right) + b_n \sin\left(\frac{2n\pi x}{T}\right) \right) \] Since \( T = \pi \), the formula simplifies to: \[ f(x) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos(2nx) + b_n \sin(2nx) \right) \] ### Step 2: Compute \( a_0 \) The constant term \( a_0 \) is the average value of \( f(x) \) over one period: \[ a_0 = \frac{1}{\pi} \int_{0}^{\pi} f(x) \, dx \] Split the integral over the two intervals: \[ a_0 = \frac{1}{\pi} \left( \int_{0}^{\frac{\pi}{2}} 1 \, dx + \int_{\frac{\pi}{2}}^{\pi} (-1) \, dx \right) \] Compute the integrals: \[ a_0 = \frac{1}{\pi} \left( \left[ x \right]_0^{\frac{\pi}{2}} - \left[ x \right]_{\frac{\pi}{2}}^{\pi} \right) \] \[ a_0 = \frac{1}{\pi} \left( \frac{\pi}{2} - \left( \pi - \frac{\pi}{2} \right) \right) \] \[ a_0 = \frac{1}{\pi} \left( \frac{\pi}{2} - \frac{\pi}{2} \right) = 0 \] ### Step 3: Compute \( a_n \) The coefficients \( a_n \) are