Let's solve the problem step by step: --- ### **Step 1: Identify the Function and Interval** Given: \[ f(x) = \begin{cases} -1, & -1 \leq x < 0 \\ 1, & 0 \leq x < 1 \end{cases} \quad |x| \leq 1 \] This is an **odd periodic function** with period \(2\), defined on \([-1, 1]\). #### **Explanation:** - The function jumps from \(-1\) to \(1\) at \(x=0\), resembling a square wave. - Since \(f(-x) = -f(x)\), it's **odd**, so the Fourier series will only have **sine terms** (no cosine/aₙ terms). --- ### **Step 2: Calculate the Fourier Coefficients** Since the function is odd, the Fourier series for \(f(x)\) on \([-1, 1]\) is: \[ f(x) = \sum_{n=1}^{\infty} b_n \sin(n\pi x) \] where \[ b_n = \int_{-1}^{1} f(x) \sin(n\pi x) \, dx \] Split the integral: \[ b_n = \int_{-1}^{0} (-1)\sin(n\pi x)dx + \int_{0}^{1} (1)\sin(n\pi x)dx \] \[ = -\int_{-1}^{0} \sin(n\pi x)dx + \int_{0}^{1} \sin(n\pi x)dx \] Let’s compute each term: 1. \(\int_{-1}^{0} \sin(n\pi x)dx = \left. -\frac{1}{n\pi}\cos(n\pi x) \right|_{-1}^{0}\) 2. \(\int_{0}^{1} \sin(n\pi x)dx = \left. -\frac{1}{n\pi}\cos(n\pi x) \right|_{0}^{1}\) So, \[ b_n = -\left[-\frac{1}{n\pi}\cos(n\pi x)\right]_{-1}^{0} + \left[-\frac{1}{n\pi}\cos(n\pi x)\right]_{0}^{1} \] \[ = \frac{1}{n\pi}\left[\cos(n\pi x)\Big|_{-1}^{0} - \cos(n\pi x)\Big|_{0}^{1}\right] \] \[ = \frac{1}{n\pi}\left([\cos(0) - \cos(-n\pi)] - [\cos(n\pi) - \cos(0)]\right) \] \[ = \frac{1}{n\pi}\left([1 - \cos(-n\pi)] - [\cos(n\pi) - 1]\right) \] Since \(\cos(-n\pi) = \cos(n\pi) = (-1)^n\): \[ = \frac{1}{n\pi}\left([1 - (-1)^n] - [(-1)^n - 1]\right) = \frac{1}{n\pi}[1 - (-1)^n - (-1)^n + 1] = \frac{1}{n\pi}[2 - 2(-1)^n] = \frac{2}{n\pi}[1 - (-1)^n] \] #### **Explanation:** - For even \(n\), \(1-(-1)^n = 0\), so only **odd \(n\)** terms remain. - For odd \(n\), \(1-(-1)^n = 2\). --- ### **Final Answer: Fourier Series** \[ f(x) = \sum_{n=1,\, n \text{ odd}}^{\infty} \frac{4}{n\pi} \sin(n\pi x) \] Or, explicitly: \[ f(x) = \frac{4}{\pi} \sin(\pi x) + \frac{4}{3\pi} \sin(3\pi x) + \frac{4}{5\pi} \sin(5\pi x) + \cdots \] --- **Summary:** - **Step 1:** Identified that the function is odd and periodic on \([-1, 1]\). - **Step 2:** Calculated the sine coefficients; only odd harmonics appear. - **Final Fourier Series:** \[ f(x) = \sum_{n=1,\,n\,\text{odd}}^{\infty} \frac{4}{n\pi} \sin(n\pi x) \]