Let's break down the solution step by step: --- ### **Step 1: Sketch the even periodic extension of \( f(x) \) on \([-\pi, \pi]\)** Given: \[ f(x) = \begin{cases} x, & 0 < x < \frac{\pi}{2} \\ \frac{\pi}{2}, & \frac{\pi}{2} < x < \pi \end{cases} \] For the **even extension**, reflect \( f(x) \) about the y-axis: \[ f(-x) = f(x) \] So, for \(-\pi < x < 0\): - For \( -\frac{\pi}{2} < x < 0 \): \( f(x) = -x \) - For \( -\pi < x < -\frac{\pi}{2} \): \( f(x) = \frac{\pi}{2} \) **Explanation:** We create the even extension by reflecting the function about the y-axis. For negative \( x \), use \( f(-x) \). --- ### **Step 2: Find the Fourier Cosine Coefficients** For the **Fourier Cosine Series** on \([0, \pi]\): \[ f(x) = a_0 + \sum_{n=1}^{\infty} a_n \cos(nx) \] where \[ a_0 = \frac{1}{\pi} \int_0^{\pi} f(x) dx \] \[ a_n = \frac{2}{\pi} \int_0^{\pi} f(x) \cos(nx) dx \] **Calculate \( a_0 \):** \[ a_0 = \frac{1}{\pi} \left( \int_0^{\frac{\pi}{2}} x dx + \int_{\frac{\pi}{2}}^{\pi} \frac{\pi}{2} dx \right) \] \[ = \frac{1}{\pi} \left( \left[ \frac{x^2}{2} \right]_0^{\frac{\pi}{2}} + \frac{\pi}{2} \left[ x \right]_{\frac{\pi}{2}}^{\pi} \right) \] \[ = \frac{1}{\pi} \left( \frac{(\pi/2)^2}{2} + \frac{\pi}{2} (\pi - \frac{\pi}{2}) \right) = \frac{1}{\pi} \left( \frac{\pi^2}{8} + \frac{\pi}{2} \cdot \frac{\pi}{2} \right) = \frac{1}{\pi} \left( \frac{\pi^2}{8} + \frac{\pi^2}{4} \right) = \frac{1}{\pi} \left( \frac{3\pi^2}{8} \right) = \frac{3\pi}{8} \] **Calculate \( a_n \):** \[ a_n = \frac{2}{\pi} \left( \int_0^{\frac{\pi}{2}} x \cos(nx) dx + \int_{\frac{\pi}{2}}^{\pi} \frac{\pi}{2} \cos(nx) dx \right) \] For the first term, use integration by parts: Let \( u = x, dv = \cos(nx) dx \Rightarrow du = dx, v = \frac{\sin(nx)}{n} \). \[ \int x \cos(nx) dx = x \frac{\sin(nx)}{n} - \int \frac{\sin(nx)}{n} dx = x \frac{\sin(nx)}{n} + \frac{\cos(nx)}{n^2} \] Evaluate from \( 0 \) to \( \frac{\pi}{2} \): \[ \int_0^{\frac{\pi}{2}} x \cos(nx) dx = \left[ x \frac{\sin(nx)}{n} + \frac{\cos(nx)}{n^2} \right]_0^{\frac{\pi}{2}} \] \[ = \frac{\pi}{2} \cdot \frac{\sin(n\pi/2)}{n} + \frac{\cos(n\pi/2) - 1}{n^2} \] Second term: \[ \int_{\frac{\pi}{2}}^{\pi} \frac{\pi}{2} \cos(nx) dx = \frac{\pi}{2} \left[ \frac{\sin(nx)}{n} \right]_{\frac{\pi}{2}}^{\pi} = \frac{\pi}{2n} (\sin(n\pi) - \sin(n\pi/2)) = -\frac{\pi}{2n} \sin(n\pi/2) \quad (\text{since } \sin(n\pi) = 0) \] Combine both: \[ a_n = \frac{2}{\pi} \left( \frac{\pi}{2} \cdot \frac{\sin(n\pi/2)}{n} + \frac{\cos(n\pi/2) - 1}{n^2} - \frac{\pi}{2n} \sin(n\pi/2) \right) \] \[ = \frac{2}{\pi} \left( 0 + \frac{\cos(n\pi/2) - 1}{n^2} \right) = \frac{2}{\pi} \cdot \frac{\cos(n\pi/2) - 1}{n^2} \] **Explanation:** We compute the Fourier cosine coefficients by integrating over the intervals, using integration by parts and properties of sine and cosine at integer multiples of \( \pi/2 \). --- ### **Step 3: Write the Corresponding Fourier Cosine Series** \[ f(x) = \frac{3\pi}{8} + \sum_{n=1}^{\infty} \left[ \frac{2}{\pi} \cdot \frac{\cos(n\pi/2) - 1}{n^2} \right] \cos(nx) \] **Explanation:** The series is written using the calculated coefficients, representing the even periodic extension of the original function. --- ## **Final Answer** \[ \boxed{ f(x) = \frac{3\pi}{8} + \sum_{n=1}^{\infty} \left[ \frac{2}{\pi} \cdot \frac{\cos(n\pi/2) - 1}{n^2} \right] \cos(nx) } \] This is the Fourier Cosine Series for the even periodic extension of the given function.

Let's approach the problem step-by-step: --- ### **a) Sketch the even periodic extension of \( f(x) \) on \([-\pi, \pi]\)** **Given function:** \[ f(x) = \begin{cases} x, & 0 < x < \frac{\pi}{2} \\ \frac{\pi}{2}, & \frac{\pi}{2} < x < \pi \end{cases} \] **Step 1: Understand the original function** - For \(x\) in \((0, \pi/2)\), \(f(x) = x\). - For \(x\) in \((\pi/2, \pi)\), \(f(x) = \pi/2\). **Step 2: Construct the even extension** - The even extension is achieved by reflecting \(f(x)\) about the y-axis: \[ f(-x) = f(x) \] - For \(-\pi < x < 0\): - When \(-\pi/2 < x < 0\), \(f(x) = -x\). - When \(-\pi < x < -\pi/2\), \(f(x) = \pi/2\). **Step 3: Sketch** - On \([0, \pi]\): - Linearly increasing from 0 to \(\pi/2\) on \((0, \pi/2)\). - Constant at \(\pi/2\) on \((\pi/2, \pi)\). - On \([- \pi, 0]\): - Symmetric to the above: - Decreasing from 0 at \(x=0\) to \(-\pi/2\) at \(x=-\pi/2\) (since \(f(-x)=f(x)\)). - Constant at \(\pi/2\) for \(-\pi < x < -\pi/2\). **Summary:** - The even periodic extension combines these parts, producing a symmetric, piecewise function with the described behavior over \([- \pi, \pi]\). --- ### **b) Find the Fourier coefficients corresponding to the Fourier Cosine Series** **Step 1: Recall the series form** Since the extension is even, the Fourier series reduces to a cosine series: \[ f(x) = a_0 + \sum_{n=1}^{\infty} a_n \cos(nx) \] where: \[ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \, dx \] \[ a_n = \frac{2}{\pi} \int_{0}^{\pi} f(x) \cos(nx) \, dx \] **Step 2: Calculate \(a_0\)** \[ a_0 = \frac{1}{\pi} \left[ \int_{-\pi}^0 f(x) dx + \int_0^{\pi} f(x) dx \right] \] Using symmetry: \[ a_0 = \frac{2}{\pi} \int_0^{\pi} f(x) dx \] Split the integral according to the piecewise definition: \[ a_0 = \frac{2}{\pi} \left( \int_0^{\pi/2} x \, dx + \int_{\pi/2}^{\pi} \frac{\pi}{2} \, dx \right) \] Compute each: - \(\int_0^{\pi/2} x \, dx = \left[ \frac{x^2}{2} \right]_0^{\pi/2} = \frac{(\pi/2)^2}{2} = \frac{\pi^2}{8}\) - \(\int_{\pi/2}^{\pi} \frac{\pi}{2} \, dx = \frac{\pi}{2} (\pi - \pi/2) = \frac{\pi}{2} \times \frac{\pi}{2} = \frac{\pi^2}{4}\) Sum: \[ a_0 = \frac{2}{\pi} \left( \frac{\pi^2}{8} + \frac{\pi^2}{4} \right) = \frac{2}{\pi} \times \left( \frac{\pi^2}{8} + \frac{2 \pi^2}{8} \right) = \frac{2}{\pi} \times \frac{3 \pi^2}{8} = \frac{2}{\pi} \times \frac{3 \pi^2}{8} \] Simplify: \[ a_0 = \frac{2 \times 3 \pi^2}{8 \pi} = \frac{6 \pi^2}{8 \pi} = \frac{3 \pi}{4} \] **Answer:** \[ \boxed{a_0 = \frac{3 \pi}{4}} \] --- ### **c) Write the Fourier Cosine Series** **Step 1: Compute \(a_n\) for \(n \ge 1\)** \[ a_n = \frac{2}{\pi} \left( \int_0^{\pi/2} x \cos(nx) dx + \int_{\pi/2}^{\pi} \frac{\pi}{2} \cos(nx) dx \right) \] **First integral:** \[ I_1 = \int_0^{\pi/2} x \cos(nx) dx \] Use integration by parts: - Let \( u = x \Rightarrow du = dx \) - \( dv = \cos(nx) dx \Rightarrow v = \frac{\sin(nx)}{n} \) \[ I_1 = x \frac{\sin(nx)}{n} \bigg|_0^{\pi/2} - \int_0^{\pi/2} \frac{\sin(nx)}{n} dx \] Calculate: \[ I_1 = \frac{\pi/2 \times \sin(n \pi/2)}{n} - \frac{1}{n} \left[ - \frac{\cos(nx)}{n} \right]_0^{\pi/2} = \frac{\pi/2 \times \sin(n \pi/2)}{n} + \frac{1}{n^2} \left( \cos(n \pi/2) - 1 \right) \] **Second integral:** \[ I_2 = \int_{\pi/2}^{\pi} \frac{\pi}{2} \cos(nx) dx = \frac{\pi}{2} \left[ \frac{\sin(nx)}{n} \right]_{\pi/2}^{\pi} = \frac{\pi}{2n} \left( \sin(n \pi) - \sin(n \pi/2) \right) \] Since \(\sin(n \pi) = 0\): \[ I_2 = - \frac{\pi}{2n} \sin(n \pi/2) \] **Combine:** \[ a_n = \frac{2}{\pi} \left( I_1 + I_2 \right) = \frac{2}{\pi} \left( \frac{\pi/2 \times \sin(n \pi/2)}{n} + \frac{\cos(n \pi/2) - 1}{n^2} - \frac{\pi}{2n} \sin(n \pi/2) \right) \] Note that: \[ \frac{\pi/2 \times \sin(n \pi/2)}{n} - \frac{\pi/2 n} \sin(n \pi/2) = 0 \] They cancel out because they are equal in magnitude and opposite in sign. Thus: \[ a_n = \frac{2}{\pi} \times \frac{\cos(n \pi/2) - 1}{n^2} \] --- ### **Final Fourier Cosine Series:** \[ \boxed{ f(x) = \frac{3 \pi}{4} + \sum_{n=1}^\infty \left[ \frac{2}{\pi} \times \frac{\cos(n \pi/2) - 1}{n^2} \right] \cos(nx) } \] --- ## **Summary:** - **a)** The even periodic extension of \(f(x)\) on \([- \pi, \pi]\) is symmetric, with linear and constant parts as described. - **b)** The zeroth Fourier cosine coefficient is \(a_0 = \frac{3 \pi}{4}\). - **c)** The Fourier Cosine Series is: \[ \boxed{ f(x) = \frac{3 \pi}{4} + \sum_{n=1}^\infty \left[ \frac{2}{\pi} \times \frac{\cos(n \pi/2) - 1}{n^2} \right] \cos(nx) } \]