1. **Problem Statement:** Find the Fourier series of the function $f(x) = x$ defined on the interval $-\pi < x \leq \pi$. Then use the series to show that $$1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots = \frac{\pi}{4}.$$ 2. **Fourier Series Formula:** For a function $f(x)$ defined on $(-\pi, \pi]$, the Fourier series is given by: $$f(x) \sim \frac{a_0}{2} + \sum_{n=1}^\infty \left(a_n \cos(nx) + b_n \sin(nx)\right)$$ where $$a_0 = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \, dx, \quad a_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \cos(nx) \, dx, \quad b_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \sin(nx) \, dx.$$ 3. **Calculate coefficients:** Since $f(x) = x$ is an odd function, $a_0 = 0$ and $a_n = 0$ because cosine is even and the product of odd and even is odd, integral over symmetric interval is zero. 4. Calculate $b_n$: $$b_n = \frac{1}{\pi} \int_{-\pi}^\pi x \sin(nx) \, dx.$$ Since $x \sin(nx)$ is even (odd times odd), we can write: $$b_n = \frac{2}{\pi} \int_0^\pi x \sin(nx) \, dx.$$ 5. Use integration by parts: Let $u = x$, $dv = \sin(nx) dx$, then $du = dx$, $v = -\frac{\cos(nx)}{n}$. $$\int x \sin(nx) dx = -\frac{x \cos(nx)}{n} + \frac{1}{n} \int \cos(nx) dx = -\frac{x \cos(nx)}{n} + \frac{\sin(nx)}{n^2} + C.$$ 6. Evaluate definite integral: $$\int_0^\pi x \sin(nx) dx = \left[-\frac{x \cos(nx)}{n} + \frac{\sin(nx)}{n^2}\right]_0^\pi = -\frac{\pi \cos(n\pi)}{n} + \frac{\sin(n\pi)}{n^2} - 0.$$ Since $\sin(n\pi) = 0$, and $\cos(n\pi) = (-1)^n$, we get: $$\int_0^\pi x \sin(nx) dx = -\frac{\pi (-1)^n}{n}.$$ 7. Substitute back: $$b_n = \frac{2}{\pi} \times \left(-\frac{\pi (-1)^n}{n}\right) = \frac{2 (-1)^{n+1}}{n}.$$ 8. **Fourier series of $f(x) = x$ is:** $$x \sim \sum_{n=1}^\infty \frac{2 (-1)^{n+1}}{n} \sin(nx).$$ 9. **Show the series for $\frac{\pi}{4}$:** Set $x = \frac{\pi}{2}$ in the Fourier series: $$\frac{\pi}{2} = \sum_{n=1}^\infty \frac{2 (-1)^{n+1}}{n} \sin\left(n \frac{\pi}{2}\right).$$ 10. Note that $\sin\left(n \frac{\pi}{2}\right)$ equals: - $1$ for $n=1,5,9,\ldots$ (i.e., $n=4k+1$) - $-1$ for $n=3,7,11,\ldots$ (i.e., $n=4k+3$) - $0$ for even $n$ 11. So only odd $n$ contribute, and the series reduces to: $$\frac{\pi}{2} = 2 \sum_{k=0}^\infty \frac{(-1)^{2k+1}}{2k+1} \sin\left((2k+1) \frac{\pi}{2}\right) = 2 \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} \times (-1)^k = 2 \sum_{k=0}^\infty \frac{1}{2k+1} (-1)^k.$$ 12. Divide both sides by 2: $$\frac{\pi}{4} = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots.$$ **Final answer:** $$1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots = \frac{\pi}{4}.$$