Let's solve the problem step by step: --- ### **Step 1: Write the General Fourier Series** For a \(2\pi\)-periodic function \(f(x)\), the Fourier series is: \[ f(x) = a_0 + \sum_{n=1}^{\infty} \left[ a_n \cos(nx) + b_n \sin(nx) \right] \] where: \[ a_0 = \frac{1}{2\pi} \int_0^{2\pi} f(x) dx \] \[ a_n = \frac{1}{\pi} \int_0^{2\pi} f(x) \cos(nx) dx \] \[ b_n = \frac{1}{\pi} \int_0^{2\pi} f(x) \sin(nx) dx \] Since \(f(x) = \frac{x}{2}\) for \(0 < x < 2\pi\), we can use this directly in the integrals. **Explanation:** We set up the general Fourier series and write out the formulas for the coefficients using the given \(f(x)\). --- ### **Step 2: Compute the Coefficients and Write the Series** #### **Compute \(a_0\):** \[ a_0 = \frac{1}{2\pi} \int_0^{2\pi} \frac{x}{2} dx = \frac{1}{4\pi} \int_0^{2\pi} x \, dx = \frac{1}{4\pi} \left[ \frac{x^2}{2} \right]_0^{2\pi} = \frac{1}{4\pi} \cdot \frac{(2\pi)^2}{2} = \frac{1}{4\pi} \cdot 2\pi^2 = \frac{\pi}{2} \] #### **Compute \(a_n\):** \[ a_n = \frac{1}{\pi} \int_0^{2\pi} \frac{x}{2} \cos(nx) dx \] Let’s integrate by parts: Let \(u = x\), \(dv = \cos(nx)dx\), so \(du = dx\), \(v = \frac{\sin(nx)}{n}\): \[ = \frac{1}{2\pi} \int_0^{2\pi} x \cos(nx) dx \] \[ = \frac{1}{2\pi} \left[ x \cdot \frac{\sin(nx)}{n} \Big|_0^{2\pi} - \int_0^{2\pi} \frac{\sin(nx)}{n} dx \right] \] \[ = \frac{1}{2\pi} \left[ \frac{2\pi \sin(2\pi n)}{n} - 0 - \frac{1}{n} \int_0^{2\pi} \sin(nx) dx \right] \] But \(\sin(2\pi n) = 0\), and \(\int_0^{2\pi} \sin(nx) dx = 0\), so \(a_n = 0\). #### **Compute \(b_n\):** \[ b_n = \frac{1}{\pi} \int_0^{2\pi} \frac{x}{2} \sin(nx) dx \] Again, by parts, let \(u = x, dv = \sin(nx)dx\), \(du = dx, v = -\frac{\cos(nx)}{n}\): \[ = \frac{1}{2\pi} \int_0^{2\pi} x \sin(nx) dx \] \[ = \frac{1}{2\pi} \left[ -x \frac{\cos(nx)}{n} \Big|_0^{2\pi} + \int_0^{2\pi} \frac{\cos(nx)}{n} dx \right] \] \[ = \frac{1}{2\pi} \left[ -\frac{2\pi \cos(2\pi n)}{n} + 0 + \frac{1}{n} \int_0^{2\pi} \cos(nx) dx \right] \] But \(\cos(2\pi n) = 1\), and \(\int_0^{2\pi} \cos(nx) dx = 0\) for \(n \neq 0\): \[ = \frac{1}{2\pi} \left[ -\frac{2\pi}{n} \right] = -\frac{1}{n} \] --- ### **Final Fourier Series:** \[ \boxed{ f(x) = \frac{\pi}{2} - \sum_{n=1}^{\infty} \frac{1}{n} \sin(nx) } \] --- **Summary:** - **Step 1:** Wrote the general Fourier series and set up coefficient formulas. - **Step 2:** Calculated each coefficient. \(a_0 = \frac{\pi}{2}\), all \(a_n = 0\), and \(b_n = -\frac{1}{n}\). - **Final Answer:** The Fourier series is \(f(x) = \frac{\pi}{2} - \sum_{n=1}^{\infty} \frac{1}{n} \sin(nx)\).

Let's approach this problem in two steps: --- ### **Step 1: Understand the problem and set up the Fourier series** Since \(f(x)\) is periodic with period \(\frac{\pi}{2}\), the Fourier series for \(f(x)\) over one period \([0, \pi/2]\) is given by: \[ f(x) = a_0 + \sum_{n=1}^{\infty} \left[ a_n \cos \left( \frac{2\pi n}{T} x \right) + b_n \sin \left( \frac{2\pi n}{T} x \right) \right] \] where \(T = \frac{\pi}{2}\). Therefore, \[ f(x) = a_0 + \sum_{n=1}^\infty \left[ a_n \cos (4 n x) + b_n \sin (4 n x) \right] \] The coefficients are: \[ a_0 = \frac{2}{T} \int_0^{T} f(x) dx \] \[ a_n = \frac{2}{T} \int_0^{T} f(x) \cos (4 n x) dx \] \[ b_n = \frac{2}{T} \int_0^{T} f(x) \sin (4 n x) dx \] **Explanation:** We set up the Fourier series for a function with period \(T = \pi/2\). The fundamental frequency becomes \(2\pi/T = 4\), leading to \(\cos(4nx)\) and \(\sin(4nx)\) terms. --- ### **Step 2: Compute the Fourier coefficients for \(f(x) = \frac{x}{2}\) over \([0, \pi/2]\)** Given \(f(x) = \frac{x}{2}\), compute the coefficients: **a. Compute \(a_0\):** \[ a_0 = \frac{2}{\pi/2} \int_0^{\pi/2} \frac{x}{2} dx = \frac{4}{\pi} \times \frac{1}{2} \int_0^{\pi/2} x dx = \frac{2}{\pi} \left[ \frac{x^2}{2} \right]_0^{\pi/2} \] \[ = \frac{2}{\pi} \times \frac{1}{2} \left( \frac{\pi^2}{4} \right) = \frac{2}{\pi} \times \frac{\pi^2}{8} = \frac{\pi}{4} \] **b. Compute \(a_n\):** \[ a_n = \frac{2}{\pi/2} \int_0^{\pi/2} \frac{x}{2} \cos(4 n x) dx = \frac{4}{\pi} \times \frac{1}{2} \int_0^{\pi/2} x \cos(4 n x) dx = \frac{2}{\pi} \int_0^{\pi/2} x \cos(4 n x) dx \] Use integration by parts: Let \(u = x\), \(dv = \cos(4 n x) dx\), then \(du = dx\), \(v = \frac{\sin(4 n x)}{4 n}\): \[ a_n = \frac{2}{\pi} \left[ x \frac{\sin(4 n x)}{4 n} \Big|_0^{\pi/2} - \int_0^{\pi/2} \frac{\sin(4 n x)}{4 n} dx \right] \] Calculate boundary term: \[ x \frac{\sin(4 n x)}{4 n} \Big|_0^{\pi/2} = \frac{\pi/2 \times \sin(2 n \pi)}{4 n} = \frac{\pi/2 \times 0}{4 n} = 0 \] Now, integral: \[ \int_0^{\pi/2} \sin(4 n x) dx = - \frac{\cos(4 n x)}{4 n} \Big|_0^{\pi/2} = - \frac{\cos(2 n \pi) - \cos(0)}{4 n} = - \frac{1 - 1}{4 n} = 0 \] So, \[ a_n = 0 \] **c. Compute \(b_n\):** \[ b_n = \frac{2}{\pi} \int_0^{\pi/2} \frac{x}{2} \sin(4 n x) dx = \frac{1}{\pi} \int_0^{\pi/2} x \sin(4 n x) dx \] Again, integrate by parts: Let \(u = x\), \(dv = \sin(4 n x) dx\), then \(du = dx\), \(v = - \frac{\cos(4 n x)}{4 n}\): \[ b_n = \frac{1}{\pi} \left[ - x \frac{\cos(4 n x)}{4 n} \Big|_0^{\pi/2} + \int_0^{\pi/2} \frac{\cos(4 n x)}{4 n} dx \right] \] Evaluate boundary term: \[ - \frac{\pi/2 \times \cos(2 n \pi)}{4 n} + 0 = - \frac{\pi/2 \times 1}{4 n} = - \frac{\pi}{8 n} \] Integral: \[ \int_0^{\pi/2} \cos(4 n x) dx = \frac{\sin(4 n x)}{4 n} \Big|_0^{\pi/2} = \frac{\sin(2 n \pi)}{4 n} = 0 \] Thus, \[ b_n = - \frac{\pi}{8 n} \] --- ### **Final Fourier series:** Putting it all together: \[ f(x) = a_0 + \sum_{n=1}^\infty \left[ a_n \cos(4 n x) + b_n \sin(4 n x) \right] \] \[ \boxed{ f(x) = \frac{\pi}{4} - \sum_{n=1}^\infty \frac{\pi}{8 n} \sin(4 n x) } \] --- **Summary:** - The Fourier series for \(f(x) = \frac{x}{2}\) with period \(\pi/2\) is: \[ \boxed{ f(x) = \frac{\pi}{4} - \sum_{n=1}^\infty \frac{\pi}{8 n} \sin(4 n x) } \] which is as simplified as possible. --- **Note:** The series contains only sine terms (since \(a_n = 0\) for all \(n\)) and a constant term.

Let's solve this problem step by step: --- ### **Step 1: Set up the Fourier series for a \(2\pi\)-periodic function** Given: - \(f(x) = \frac{x}{2}\) - Period: \(2\pi\) - Interval: \([0, 2\pi]\) The Fourier series expansion for a function with period \(2\pi\) is: \[ f(x) = a_0 + \sum_{n=1}^{\infty} \left[ a_n \cos(nx) + b_n \sin(nx) \right] \] where the coefficients are: \[ a_0 = \frac{1}{2\pi} \int_0^{2\pi} f(x) dx \] \[ a_n = \frac{1}{\pi} \int_0^{2\pi} f(x) \cos(nx) dx \] \[ b_n = \frac{1}{\pi} \int_0^{2\pi} f(x) \sin(nx) dx \] **Explanation:** We start with the general Fourier series formula for period \(2\pi\) and plan to compute the coefficients using the given \(f(x) = x/2\). --- ### **Step 2: Calculate the coefficients** --- **a. Compute \(a_0\):** \[ a_0 = \frac{1}{2\pi} \int_0^{2\pi} \frac{x}{2} dx = \frac{1}{4\pi} \int_0^{2\pi} x dx \] \[ = \frac{1}{4\pi} \left[ \frac{x^2}{2} \right]_0^{2\pi} = \frac{1}{4\pi} \times \frac{(2\pi)^2}{2} = \frac{1}{4\pi} \times \frac{4\pi^2}{2} = \frac{1}{4\pi} \times 2\pi^2 = \frac{\pi}{2} \] **Answer for \(a_0\):** \[ \boxed{ a_0 = \frac{\pi}{2} } \] --- **b. Compute \(a_n\):** \[ a_n = \frac{1}{\pi} \int_0^{2\pi} \frac{x}{2} \cos(nx) dx \] \[ = \frac{1}{2\pi} \int_0^{2\pi} x \cos(nx) dx \] Use integration by parts: - \(u = x \Rightarrow du = dx\) - \(dv = \cos(nx) dx \Rightarrow v = \frac{\sin(nx)}{n}\) \[ a_n = \frac{1}{2\pi} \left[ x \frac{\sin(nx)}{n} \Big|_0^{2\pi} - \int_0^{2\pi} \frac{\sin(nx)}{n} dx \right] \] Evaluate boundary term: \(\sin(n \times 2\pi) = 0\), so: \[ x \frac{\sin(nx)}{n} \Big|_0^{2\pi} = 0 \] And the integral: \[ \int_0^{2\pi} \sin(nx) dx = 0 \] Thus: \[ a_n = 0 \] **Answer for \(a_n\):** \[ \boxed{ a_n = 0 } \] --- **c. Compute \(b_n\):** \[ b_n = \frac{1}{\pi} \int_0^{2\pi} \frac{x}{2} \sin(nx) dx = \frac{1}{2\pi} \int_0^{2\pi} x \sin(nx) dx \] Use integration by parts: - \(u = x \Rightarrow du = dx\) - \(dv = \sin(nx) dx \Rightarrow v = -\frac{\cos(nx)}{n}\) \[ b_n = \frac{1}{2\pi} \left[ - x \frac{\cos(nx)}{n} \Big|_0^{2\pi} + \int_0^{2\pi} \frac{\cos(nx)}{n} dx \right] \] Evaluate boundary term: \[ - \frac{2\pi \cos(n \times 2\pi)}{n} + 0 = - \frac{2\pi \times 1}{n} = - \frac{2\pi}{n} \] Integral: \[ \int_0^{2\pi} \cos(nx) dx = \frac{\sin(nx)}{n} \Big|_0^{2\pi} = 0 \] Therefore: \[ b_n = \frac{1}{2\pi} \left( - \frac{2\pi}{n} \right) = - \frac{1}{n} \] **Answer for \(b_n\):** \[ \boxed{ b_n = - \frac{1}{n} } \] --- ### **Final Fourier Series:** Putting all together: \[ f(x) = \frac{\pi}{2} + \sum_{n=1}^\infty \left[ 0 \times \cos(nx) - \frac{1}{n} \sin(nx) \right] = \frac{\pi}{2} - \sum_{n=1}^\infty \frac{1}{n} \sin(nx) \] **Thus, the simplified Fourier series is:** \[ \boxed{ f(x) = \frac{\pi}{2} - \sum_{n=1}^\infty \frac{1}{n} \sin(nx) } \] --- **Summary:** - The Fourier series for \(f(x) = x/2\), period \(2\pi\), over \([0, 2\pi]\), is: \[ \boxed{ f(x) = \frac{\pi}{2} - \sum_{n=1}^\infty \frac{1}{n} \sin(nx) } \] which is as simplified as possible.