Let's solve the problem step by step: ### **Step 1: Find the Fourier Series for \( f(x) \)** Given: \[ f(x) = \begin{cases} -x & \text{for } -\pi < x < 0 \\ x & \text{for } 0 < x < \pi \\ 0 & \text{for } x = 0 \end{cases} \] Notice **\( f(x) \) is an odd function** because \( f(-x) = -f(x) \), so the Fourier series will only have sine terms: \[ f(x) = \sum_{n=1}^{\infty} b_n \sin(nx) \] where \[ b_n = \frac{2}{\pi} \int_0^\pi f(x) \sin(nx) dx \] For \( 0 < x < \pi \), \( f(x) = x \): \[ b_n = \frac{2}{\pi} \int_0^\pi x \sin(nx) dx \] Use integration by parts (\( u = x, dv = \sin(nx)dx \)): \[ u = x, \quad dv = \sin(nx)dx \implies du = dx, \quad v = -\frac{1}{n}\cos(nx) \] \[ \int x \sin(nx) dx = -\frac{x}{n}\cos(nx) + \frac{1}{n}\int \cos(nx) dx = -\frac{x}{n}\cos(nx) + \frac{1}{n^2}\sin(nx) \] Evaluating from \( 0 \) to \( \pi \): \[ \int_0^\pi x \sin(nx) dx = \left[ -\frac{x}{n}\cos(nx) + \frac{1}{n^2}\sin(nx) \right]_0^\pi \] \[ = -\frac{\pi}{n}\cos(n\pi) + 0 - (0 + 0) = -\frac{\pi}{n}(-1)^n \] So, \[ b_n = \frac{2}{\pi} \times \left( -\frac{\pi}{n}(-1)^n \right) = -\frac{2}{n}(-1)^n = \frac{2}{n}(-1)^{n+1} \] **Fourier Series:** \[ f(x) = \sum_{n=1}^{\infty} \frac{2}{n}(-1)^{n+1} \sin(nx) \] --- #### **Explanation:** We calculated the Fourier sine coefficients for the given odd function and wrote its Fourier series. --- ### **Step 2: Deduce the Series \(\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \cdots = \frac{\pi^2}{8}\)** Set \( x = \frac{\pi}{2} \) in the Fourier series: \[ f\left(\frac{\pi}{2}\right) = \frac{\pi}{2} \] \[ \frac{\pi}{2} = \sum_{n=1}^{\infty} \frac{2}{n}(-1)^{n+1} \sin\left(n \frac{\pi}{2}\right) \] Now, \(\sin\left(n \frac{\pi}{2}\right)\) cycles as \(1, 0, -1, 0, 1, \ldots\) for \(n = 1, 2, 3, 4, 5, \ldots\). Only odd terms are nonzero: For \( n = 1, 3, 5, \ldots \): \[ \sin\left(n \frac{\pi}{2}\right) = (-1)^{(n-1)/2} \] So, \[ \frac{\pi}{2} = \sum_{k=0}^{\infty} \frac{2}{2k+1} (-1)^{2k+2} (-1)^k \] But \( (-1)^{2k+2} = 1 \), so: \[ \frac{\pi}{2} = 2\sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1} \] This is a known result (Leibniz series for \(\pi\)), but to relate to the sum of reciprocals of odd squares, we use Parseval's identity: Parseval's Identity: \[ \frac{1}{\pi} \int_{-\pi}^{\pi} [f(x)]^2 dx = \sum_{n=1}^{\infty} b_n^2 \] Calculate LHS: \[ f(x) = |x| \implies [f(x)]^2 = x^2 \] \[ \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 dx = \frac{2}{\pi} \int_0^{\pi} x^2 dx = \frac{2}{\pi} \left[ \frac{x^3}{3} \right]_0^\pi = \frac{2}{\pi} \cdot \frac{\pi^3}{3} = \frac{2\pi^2}{3} \] RHS: \[ b_n = \frac{2}{n}(-1)^{n+1} \] \[ b_n^2 = \frac{4}{n^2} \] So, \[ \sum_{n=1}^{\infty} b_n^2 = 4 \sum_{n=1}^{\infty} \frac{1}{n^2} \] Equate both sides: \[ \frac{2\pi^2}{3} = 4 \sum_{n=1}^{\infty} \frac{1}{n^2} \] \[ \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \] But we want only the **sum over odd n**: \[ \sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} \] It's known that: \[ \sum_{n=1}^{\infty} \frac{1}{n^2} = \sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} + \sum_{k=1}^{\infty} \frac{1}{(2k)^2} \] \[ = \sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} + \frac{1}{4}\sum_{k=1}^{\infty} \frac{1}{k^2} \] \[ = S + \frac{1}{4} \cdot \frac{\pi^2}{6} \] So, \[ \frac{\pi^2}{6} = S + \frac{\pi^2}{24} \implies S = \frac{\pi^2}{6} - \frac{\pi^2}{24} = \frac{\pi^2}{8} \] --- #### **Explanation:** By using Parseval's identity and separating the odd and even terms, we deduced the sum of reciprocals of odd squares. --- ## **Final Answer:** \[ \boxed{ \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \cdots = \frac{\pi^2}{8} } \]

Let's analyze the problem step by step: ### **Step 1: Determine the Fourier Series for \(f(x)\)** Given: \[ f(x) = \begin{cases} -\pi & \text{for } -\pi < x < 0 \\ x & \text{for } 0 < x < \pi \\ -\pi/2 & \text{for } x=0 \end{cases} \] **Step 1 Explanation:** First, observe the nature of \(f(x)\): - For \(-\pi < x < 0\), \(f(x) = -\pi\) - For \(0 < x < \pi\), \(f(x) = x\) - At \(x=0\), \(f(0) = -\pi/2\) Check whether \(f(x)\) is even, odd, or neither: - For \(x \neq 0\), compare \(f(-x)\): \[ f(-x) = \begin{cases} -\pi & \text{if } x > 0 \Rightarrow -x < 0 \Rightarrow f(-x) = -\pi \\ (-x) & \text{if } -x > 0 \Rightarrow x < 0 \Rightarrow f(-x) = -x \end{cases} \] - For \(x > 0\): \[ f(-x) = -\pi \] \[ f(x) = x \] - For \(x < 0\): \[ f(-x) = -x \] \[ f(x) = -\pi \] Since \(f(-x) \neq -f(x)\) in general, \(f(x)\) is neither purely even nor odd. But the function has a jump at zero, so it's a piecewise function. --- **Fourier Series Expression:** Given the general Fourier series on \([- \pi, \pi]\): \[ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left[ a_n \cos(nx) + b_n \sin(nx) \right] \] - Since \(f(x)\) is neither symmetric nor antisymmetric, we need to compute both \(a_n\) and \(b_n\). --- ### **Step 2: Calculate Fourier coefficients \(a_n\) and \(b_n\)** --- **Calculate \(a_0\):** \[ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) dx \] Break into parts: \[ a_0 = \frac{1}{\pi} \left[ \int_{-\pi}^0 (-\pi) dx + \int_0^{\pi} x dx \right] \] Compute: \[ \int_{-\pi}^0 (-\pi) dx = -\pi \times (0 - (-\pi)) = -\pi \times \pi = -\pi^2 \] \[ \int_0^{\pi} x dx = \left[ \frac{x^2}{2} \right]_0^{\pi} = \frac{\pi^2}{2} \] So, \[ a_0 = \frac{1}{\pi} (-\pi^2 + \frac{\pi^2}{2}) = \frac{1}{\pi} \left( -\frac{\pi^2}{2} \right) = - \frac{\pi}{2} \] --- **Calculate \(a_n\):** \[ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) dx \] Break into two parts: \[ a_n = \frac{1}{\pi} \left[ \int_{-\pi}^0 (-\pi) \cos(nx) dx + \int_0^{\pi} x \cos(nx) dx \right] \] Calculate each: 1. \(\int_{-\pi}^0 -\pi \cos(nx) dx\): \[ -\pi \int_{-\pi}^0 \cos(nx) dx = -\pi \left[ \frac{\sin(nx)}{n} \right]_{x=-\pi}^{0} = -\pi \left( \frac{\sin(0)}{n} - \frac{\sin(-n\pi)}{n} \right) \] Note that: \[ \sin(0) = 0, \quad \sin(-n\pi) = - \sin(n\pi) = 0 \] Thus, \[ \int_{-\pi}^0 -\pi \cos(nx) dx = -\pi \times (0 - 0) = 0 \] 2. \(\int_0^{\pi} x \cos(nx) dx\): Use integration by parts: Let \(u = x \Rightarrow du = dx\) \(dv = \cos(nx) dx \Rightarrow v = \frac{\sin(nx)}{n}\) \[ \int x \cos(nx) dx = x \frac{\sin(nx)}{n} - \int \frac{\sin(nx)}{n} dx \] Calculate the second integral: \[ \int \frac{\sin(nx)}{n} dx = - \frac{\cos(nx)}{n^2} \] Putting it all together: \[ \int_0^{\pi} x \cos(nx) dx = \left[ x \frac{\sin(nx)}{n} \right]_0^{\pi} + \frac{\cos(nx)}{n^2} \Big|_0^{\pi} \] Evaluate: \[ \left[ x \frac{\sin(nx)}{n} \right]_0^{\pi} = \frac{\pi \sin(n\pi)}{n} - 0 = 0 \] (since \(\sin(n\pi) = 0\)) And \[ \frac{\cos(n\pi) - \cos(0)}{n^2} = \frac{\cos(n\pi) - 1}{n^2} \] Recall: \[ \cos(n\pi) = (-1)^n \] Therefore, \[ a_n = \frac{1}{\pi} \times \left( 0 + \frac{(-1)^n - 1}{n^2} \right) = \frac{(-1)^n - 1}{\pi n^2} \] --- **Calculate \(b_n\):** \[ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) dx \] Break into two parts: \[ b_n = \frac{1}{\pi} \left[ \int_{-\pi}^0 (-\pi) \sin(nx) dx + \int_0^{\pi} x \sin(nx) dx \right] \] 1. \(\int_{-\pi}^0 -\pi \sin(nx) dx\): \[ -\pi \left[ - \frac{\cos(nx)}{n} \right]_{x=-\pi}^{0} = -\pi \left( - \frac{\cos(0)}{n} + \frac{\cos(-n\pi)}{n} \right) = -\pi \left( - \frac{1}{n} + \frac{\cos(n\pi)}{n} \right) \] \[ = -\pi \left( - \frac{1}{n} + \frac{(-1)^n}{n} \right) = -\pi \left( \frac{-1 + (-1)^n}{n} \right) = \frac{\pi}{n} \left( 1 - (-1)^n \right) \] 2. \(\int_0^{\pi} x \sin(nx) dx\): Again, integrate by parts: \(u = x \Rightarrow du = dx\) \(dv = \sin(nx) dx \Rightarrow v = - \frac{\cos(nx)}{n}\) \[ \int x \sin(nx) dx = -x \frac{\cos(nx)}{n} + \int \frac{\cos(nx)}{n} dx \] \[ = -x \frac{\cos(nx)}{n} + \frac{\sin(nx)}{n^2} \] Evaluate from \(0\) to \(\pi\): \[ \left[ -x \frac{\cos(nx)}{n} + \frac{\sin(nx)}{n^2} \right]_0^{\pi} = \left( -\pi \frac{\cos(n\pi)}{n} + \frac{\sin(n\pi)}{n^2} \right) - \left( 0 + 0 \right) \] \[ = -\frac{\pi}{n} (-1)^n + 0 = - \frac{\pi}{n} (-1)^n \] Thus, \[ b_n = \frac{1}{\pi} \left[ \frac{\pi}{n} (1 - (-1)^n) - \frac{\pi}{n} (-1)^n \right] \] Simplify: \[ b_n = \frac{1}{\pi} \left( \frac{\pi}{n} (1 - (-1)^n) - \frac{\pi}{n} (-1)^n \right) = \frac{1}{n} (1 - (-1)^n) - \frac{1}{n} (-1)^n \] \[ b_n = \frac{1 - (-1)^n - (-1)^n}{n} = \frac{1 - 2(-1)^n}{n} \] --- ### **Final Fourier Series Expression:** \[ f(x) = - \frac{\pi}{2} + \sum_{n=1}^{\infty} \left[ \frac{(-1)^n - 1}{\pi n^2} \cos(nx) + \frac{1 - 2(-1)^n}{n} \sin(nx) \right] \] --- **Summary:** - The constant term: \[ a_0/2 = -\frac{\pi}{4} \] - Cosine coefficients: \[ a_n = \frac{(-1)^n - 1}{\pi n^2} \] - Sine coefficients: \[ b_n = \frac{1 - 2(-1)^n}{n} \] --- **This completes the Fourier series expansion for the given function \(f(x)\).**