1. **Problem Statement:** Apply Parseval's Identity to the function $f(x) = |x|$ defined on the interval $-\pi < x < \pi$, and deduce the value of the series $1 + \frac{1}{3^4} + \frac{1}{5^4} + \cdots$. 2. **Recall Parseval's Identity:** For a function with Fourier series coefficients $a_0$, $a_n$, and $b_n$, Parseval's Identity states: $$\frac{1}{\pi} \int_{-\pi}^{\pi} |f(x)|^2 dx = \frac{a_0^2}{2} + \sum_{n=1}^\infty (a_n^2 + b_n^2)$$ 3. **Fourier Series of $f(x) = |x|$:** Since $|x|$ is even, all $b_n = 0$. The Fourier series is: $$f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos(nx)$$ Where: $$a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} |x| dx = \frac{2}{\pi} \int_0^{\pi} x dx = \frac{2}{\pi} \cdot \frac{\pi^2}{2} = \pi$$ $$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} |x| \cos(nx) dx = \frac{2}{\pi} \int_0^{\pi} x \cos(nx) dx$$ 4. **Calculate $a_n$:** Using integration by parts: Let $u = x$, $dv = \cos(nx) dx$, then $du = dx$, $v = \frac{\sin(nx)}{n}$. $$a_n = \frac{2}{\pi} \left[ x \frac{\sin(nx)}{n} \Big|_0^{\pi} - \int_0^{\pi} \frac{\sin(nx)}{n} dx \right] = \frac{2}{\pi} \left[ \frac{\pi \sin(n\pi)}{n} - \frac{1}{n} \int_0^{\pi} \sin(nx) dx \right]$$ Since $\sin(n\pi) = 0$, $$a_n = - \frac{2}{\pi n} \int_0^{\pi} \sin(nx) dx = - \frac{2}{\pi n} \left[ -\frac{\cos(nx)}{n} \Big|_0^{\pi} \right] = \frac{2}{\pi n^2} (\cos(0) - \cos(n\pi))$$ Since $\cos(n\pi) = (-1)^n$ and $\cos(0) = 1$, $$a_n = \frac{2}{\pi n^2} (1 - (-1)^n)$$ 5. **Simplify $a_n$:** For even $n$, $1 - 1 = 0$, so $a_n = 0$. For odd $n$, $1 - (-1) = 2$, so $$a_n = \frac{4}{\pi n^2}$$ 6. **Apply Parseval's Identity:** Calculate the left side: $$\frac{1}{\pi} \int_{-\pi}^{\pi} |x|^2 dx = \frac{2}{\pi} \int_0^{\pi} x^2 dx = \frac{2}{\pi} \cdot \frac{\pi^3}{3} = \frac{2 \pi^2}{3}$$ Calculate the right side: $$\frac{a_0^2}{2} + \sum_{n=1}^\infty a_n^2 = \frac{\pi^2}{2} + \sum_{\substack{n=1 \\ n \text{ odd}}}^\infty \left( \frac{4}{\pi n^2} \right)^2 = \frac{\pi^2}{2} + \frac{16}{\pi^2} \sum_{\substack{n=1 \\ n \text{ odd}}}^\infty \frac{1}{n^4}$$ 7. **Equate both sides:** $$\frac{2 \pi^2}{3} = \frac{\pi^2}{2} + \frac{16}{\pi^2} \sum_{\substack{n=1 \\ n \text{ odd}}}^\infty \frac{1}{n^4}$$ Rearranged: $$\sum_{\substack{n=1 \\ n \text{ odd}}}^\infty \frac{1}{n^4} = \frac{\pi^4}{16} \left( \frac{2}{3} - \frac{1}{2} \right) = \frac{\pi^4}{16} \cdot \frac{1}{6} = \frac{\pi^4}{96}$$ 8. **Final result:** $$1 + \frac{1}{3^4} + \frac{1}{5^4} + \cdots = \frac{\pi^4}{96}$$