Let's solve the problem step by step: --- ### **Step 1: Identify the Type of Function and Fourier Series Expansion** Given: \[ f(x) = \begin{cases} -1 & \text{for } -\pi < x < 0 \\ 1 & \text{for } 0 < x < \pi \end{cases} \] with period \(2\pi\): \(f(x+2\pi) = f(x)\). This is an **odd function** (since \(f(-x) = -f(x)\)), so its Fourier series will only have sine terms: \[ f(x) = \sum_{n=1}^{\infty} b_n \sin(nx) \] where \[ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx \] **Explanation:** Since the function is odd, all cosine (even) terms and the constant term \(a_0\) will be zero. --- ### **Step 2: Calculate the Fourier Coefficients \(b_n\)** \[ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx \] Split the integral: \[ b_n = \frac{1}{\pi} \left( \int_{-\pi}^{0} (-1)\sin(nx) dx + \int_{0}^{\pi} (1)\sin(nx) dx \right) \] \[ = \frac{1}{\pi} \left( -\int_{-\pi}^{0} \sin(nx) dx + \int_{0}^{\pi} \sin(nx) dx \right) \] Calculate each part: \[ \int \sin(nx) dx = -\frac{1}{n}\cos(nx) \] So: \[ \int_{-\pi}^{0} \sin(nx) dx = [-\frac{1}{n} \cos(nx)]_{-\pi}^{0} = -\frac{1}{n}(\cos(0) - \cos(-n\pi)) = -\frac{1}{n}(1 - (-1)^n) \] \[ \int_{0}^{\pi} \sin(nx) dx = [-\frac{1}{n} \cos(nx)]_{0}^{\pi} = -\frac{1}{n}(\cos(n\pi) - \cos(0)) = -\frac{1}{n}((-1)^n - 1) \] Plug into \(b_n\): \[ b_n = \frac{1}{\pi} \left( -\left(-\frac{1}{n}(1 - (-1)^n)\right) + \left(-\frac{1}{n}((-1)^n - 1)\right) \right) \] \[ = \frac{1}{\pi} \left( \frac{1}{n}(1 - (-1)^n) - \frac{1}{n}((-1)^n - 1) \right) \] \[ = \frac{1}{\pi} \left( \frac{1}{n}(1 - (-1)^n + 1 - (-1)^n) \right) \] \[ = \frac{1}{\pi} \left( \frac{2}{n}(1 - (-1)^n) \right) \] Now, for **even \(n\), \(1 - (-1)^n = 0\)**. For **odd \(n\), \(1 - (-1)^n = 2\)**. So, for **odd \(n\)**: \[ b_n = \frac{4}{n\pi} \] for **even \(n\)**: \[ b_n = 0 \] **Explanation:** The only nonzero Fourier coefficients are for the sine terms with odd \(n\). --- ## **Final Answer: Fourier Series** \[ \boxed{ f(x) = \frac{4}{\pi} \left( \sin x + \frac{1}{3} \sin 3x + \frac{1}{5} \sin 5x + \cdots \right) } \] where the sum runs over odd values of \(n\).