Let's solve **Problem 9:** \( f(x) = x, \quad -\pi < x < \pi \) We are to **find the Fourier series** for \( f(x) \) on \([-\pi, \pi]\). ## **Step 1: Recognize the Fourier Series Form** For a function \( f(x) \) defined on \( [-L, L] \), the Fourier Series is: \[ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left[ a_n \cos \frac{n \pi x}{L} + b_n \sin \frac{n \pi x}{L} \right] \] Here \( L = \pi \), so it becomes: \[ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left[ a_n \cos(nx) + b_n \sin(nx) \right] \] ## **Step 2: Compute the Coefficients** ### **(a) Compute \( a_0 \):** \[ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} x \, dx \] Since \( x \) is an odd function over symmetric interval: \[ a_0 = 0 \] --- ### **(b) Compute \( a_n \):** \[ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x \cos(nx) \, dx \] Here, \( x \cos(nx) \) is an **odd function** (since \( x \) is odd, \( \cos(nx) \) is even, so the product is odd), so: \[ a_n = 0 \] --- ### **(c) Compute \( b_n \):** \[ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x \sin(nx) \, dx \] Here, \( x \sin(nx) \) is **even** (since both \( x \) and \( \sin(nx) \) are odd, product is even), so this integral is generally nonzero. Compute: \[ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x \sin(nx) \, dx \] Let’s compute the integral: Let’s use **integration by parts**: Let \( u = x \), \( dv = \sin(nx) dx \). Then \( du = dx \), \( v = -\frac{1}{n} \cos(nx) \). So, \[ \int x\sin(nx)dx = -\frac{x}{n}\cos(nx) + \int\frac{1}{n}\cos(nx)dx = -\frac{x}{n}\cos(nx) + \frac{1}{n^2}\sin(nx) \] So, \[ \int_{-\pi}^{\pi} x\sin(nx)dx = \left[ -\frac{x}{n}\cos(nx) + \frac{1}{n^2}\sin(nx) \right]_{-\pi}^{\pi} \] Now compute at the limits: - \(\sin(n\pi) = 0\), \(\sin(-n\pi) = 0\) - \(\cos(n\pi) = (-1)^n\), \(\cos(-n\pi) = (-1)^n\) So, At \( x = \pi \): \[ -\frac{\pi}{n}\cos(n\pi) + 0 = -\frac{\pi}{n}(-1)^n \] At \( x = -\pi \): \[ +\frac{\pi}{n}\cos(-n\pi) + 0 = +\frac{\pi}{n}(-1)^n \] So, subtract: \[ \left[ -\frac{x}{n}\cos(nx) \right]_{-\pi}^{\pi} = -\frac{\pi}{n}(-1)^n - \left( +\frac{\pi}{n}(-1)^n \right) = -\frac{2\pi}{n}(-1)^n \] Therefore, \[ b_n = \frac{1}{\pi} \left( -\frac{2\pi}{n}(-1)^n \right) = -\frac{2}{n}(-1)^n \] --- ## **Step 3: Write the Fourier Series** So, \[ f(x) = x = \sum_{n=1}^{\infty} b_n \sin(nx) \] with \[ b_n = -\frac{2}{n}(-1)^n \] So, \[ \boxed{ x = 2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \sin(nx) } \] **Final Answer:** \[ \boxed{ x = 2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \sin(nx) } \] on \( -\pi < x < \pi \). --- Let me know if you want the solution for another problem!

Certainly! Let's carefully solve each problem from 9 to 16, providing step-by-step solutions and explanations. --- ### **Problem 9: \(f(x) = x\), \(-\pi < x < \pi\)** **Objective:** Find the Fourier series expansion. --- **Step 1:** Recognize the symmetry of \(f(x)\): - \(f(x) = x\) is an **odd** function. - Its Fourier series on \([- \pi, \pi]\) will contain only sine terms (since cosine terms correspond to even functions). --- **Step 2:** Fourier series form: \[ f(x) = \sum_{n=1}^\infty b_n \sin(n x) \] with \[ b_n = \frac{1}{\pi} \int_{-\pi}^\pi x \sin(n x) dx \] --- **Step 3:** Compute \(b_n\): Integral: \[ b_n = \frac{1}{\pi} \int_{-\pi}^\pi x \sin(n x) dx \] Since \(x \sin(n x)\) is an **even** function (\(x\) odd, \(\sin(n x)\) odd, product even), we can write: \[ b_n = \frac{2}{\pi} \int_0^{\pi} x \sin(n x) dx \] --- **Step 4:** Compute the integral: \[ I = \int_0^{\pi} x \sin(n x) dx \] Use **integration by parts**: - \(u = x \Rightarrow du = dx\) - \(dv = \sin(n x) dx \Rightarrow v = -\frac{1}{n} \cos(n x)\) Then: \[ I = uv - \int v du = -\frac{x}{n} \cos(n x) + \frac{1}{n} \int \cos(n x) dx \] \[ I = -\frac{x}{n} \cos(n x) + \frac{1}{n^2} \sin(n x) \] Evaluate from 0 to \(\pi\): At \(x = \pi\): \[ -\frac{\pi}{n} \cos(n \pi) + \frac{1}{n^2} \sin(n \pi) \] \(\sin(n \pi) = 0\), \(\cos(n \pi) = (-1)^n\), so: \[ -\frac{\pi}{n} (-1)^n \] At \(x=0\): \[ 0 + 0 = 0 \] Thus, \[ I = -\frac{\pi}{n} (-1)^n \] **Therefore:** \[ b_n = \frac{2}{\pi} \times \left( -\frac{\pi}{n} (-1)^n \right) = -\frac{2}{n} (-1)^n \] --- **Final Fourier series:** \[ f(x) = \sum_{n=1}^\infty b_n \sin(n x) = \sum_{n=1}^\infty \left( -\frac{2}{n} (-1)^n \right) \sin(n x) \] Alternatively, \[ f(x) = 2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \sin(n x) \] --- ### **Answer for Problem 9:** \[ \boxed{ x = 2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \sin(n x), \quad -\pi < x < \pi } \] --- ### **Problem 10: \(f(x) = \left\{\begin{cases} 1, & 0 < x < 2 \\ 0, & \text{otherwise} \end{cases}\)\)** **Interval:** \( [0, 2] \). Assuming the function is extended periodically outside this interval with period \(T\). **Objective:** Find the Fourier series. --- **Step 1:** Clarify the period: - Since the interval is from 0 to 2, the period is likely \(2\). - We'll assume the function is defined on \([0,2]\) and extended periodically. --- **Step 2:** Fourier series over \([0, 2]\): \[ f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty \left[ a_n \cos\left( \frac{n \pi x}{L} \right) + b_n \sin \left( \frac{n \pi x}{L} \right) \right] \] with \(L = 1\) (since \(T=2\), \(L=1\)): \[ f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty \left[ a_n \cos(n \pi x) + b_n \sin(n \pi x) \right] \] --- **Step 3:** Compute coefficients: - \(a_0 = \frac{1}{L} \int_0^{2} f(x) dx = \int_0^{2} 1 dx = 2\) - \(a_n = \frac{1}{L} \int_0^{2} f(x) \cos(n \pi x) dx = \int_0^{2} \cos(n \pi x) dx\) - \(b_n = \frac{1}{L} \int_0^{2} f(x) \sin(n \pi x) dx = \int_0^{2} \sin(n \pi x) dx\) --- **Step 4:** Calculate: \[ a_0 = 2 \] \[ a_n = \int_0^{2} \cos(n \pi x) dx = \left[ \frac{\sin(n \pi x)}{n \pi} \right]_0^2 = \frac{\sin(2 n \pi)}{n \pi} - 0 = 0 \] (since \(\sin(2 n \pi) = 0\)). \[ b_n = \int_0^{2} \sin(n \pi x) dx = \left[ - \frac{\cos(n \pi x)}{n \pi} \right]_0^2 = - \frac{\cos(2 n \pi)}{n \pi} + \frac{\cos 0}{n \pi} = - \frac{1}{n \pi} + \frac{1}{n \pi} = 0 \] since \(\cos(2 n \pi) = 1\). --- **Conclusion:** All Fourier coefficients except \(a_0\) are zero. **Final Fourier series:** \[ f(x) = \frac{a_0}{2} = 1 \] which is just a constant function. But since the function is 1 on \((0, 2)\), and zero elsewhere, the Fourier series approximates the periodic extension of this step function. --- ### **Problem 11: \(f(x) = \left\{\begin{cases} 25, & |x| < a \\ 0, & \text{otherwise} \end{cases}\)** Assuming the period \(T\), and that \(f(x)\) is periodic. Suppose period \(T\), and function is symmetric. --- **Step 1:** The problem is similar to a rectangular pulse. --- **Step 2:** For simplicity, assume \(f(x)\) is even (since it's symmetric in \(|x|<a\)), and period \(T\). --- **Step 3:** Fourier coefficients: - \(a_0 = \frac{2}{T} \int_0^{a} 25 dx = \frac{2}{T} \times 25 a = \frac{50 a}{T}\) - \(a_n = \frac{2}{T} \int_0^{a} 25 \cos\left( \frac{2 \pi n x}{T} \right) dx\) - \(b_n = 0\) (due to even symmetry) --- **Step 4:** Calculate \(a_n\): \[ a_n = \frac{2}{T} \times 25 \int_0^{a} \cos \left( \frac{2 \pi n x}{T} \right) dx \] \[ = \frac{50}{T} \left[ \frac{\sin \left( \frac{2 \pi n x}{T} \right)}{\frac{2 \pi n}{T}} \right]_0^{a} = \frac{50}{T} \times \frac{T}{2 \pi n} \left[ \sin \left( \frac{2 \pi n a}{T} \right) - 0 \right] \] \[ = \frac{50}{2 \pi n} \sin \left( \frac{2 \pi n a}{T} \right) = \frac{25}{\pi n} \sin \left( \frac{2 \pi n a}{T} \right) \] --- **Final Fourier series:** \[ f(x) \approx \frac{50 a}{T} + \sum_{n=1}^\infty \frac{25}{\pi n} \sin \left( \frac{2 \pi n a}{T} \right) \cos \left( \frac{2 \pi n x}{T} \right) \] --- ### **Problems 12 to 16:** Due to the partial data and formatting issues in the image, precise interpretation is difficult. However, based on typical Fourier series problems, I will outline general approaches: --- ### **Problem 12:** \(f(x) = \text{(some function)}\) - Identify symmetry (even/odd). - Compute Fourier coefficients similarly. - Use integrals over the period or specified interval. - Write the sum. --- ### **Problem 13-16:** - Follow the same steps: identify the function, symmetry, period. - Compute the Fourier coefficients using integrals. - Write the Fourier series. - Use computational tools to plot partial sums. --- ### **Summary:** - For each function, identify the symmetry to simplify calculations. - Compute coefficients via integrals. - Write the Fourier series. - Use graphing tools to visualize partial sums. --- **If you provide clearer descriptions of Problems 12-16 or specific functions, I can give detailed solutions for each.** --- **Final note:** For plotting partial sums, use graphing calculators or software like Desmos, GeoGebra, or WolframAlpha. --- **Let me know if you'd like detailed solutions for specific problems or functions!**