1. **Find the Fourier series for** $f(x) = x$, $-\pi < x < \pi$. The function $f(x) = x$ is odd, so its Fourier series contains only sine terms. The general Fourier series for a function with period $2\pi$ is: $$f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty \left(a_n \cos(nx) + b_n \sin(nx)\right)$$ Since $f(x)$ is odd, $a_0 = 0$ and $a_n = 0$ for all $n$. The sine coefficients are: $$b_n = \frac{1}{\pi} \int_{-\pi}^\pi x \sin(nx) \, dx$$ Because $x \sin(nx)$ is even (odd * odd = even), we can write: $$b_n = \frac{2}{\pi} \int_0^\pi x \sin(nx) \, dx$$ Use integration by parts: Let $u = x$, $dv = \sin(nx) dx$. Then $du = dx$, $v = -\frac{\cos(nx)}{n}$. So, $$\int_0^\pi x \sin(nx) dx = \left. -\frac{x \cos(nx)}{n} \right|_0^\pi + \frac{1}{n} \int_0^\pi \cos(nx) dx$$ Evaluate the boundary term: $$-\frac{\pi \cos(n\pi)}{n} + 0 = -\frac{\pi (-1)^n}{n}$$ Evaluate the integral: $$\int_0^\pi \cos(nx) dx = \left. \frac{\sin(nx)}{n} \right|_0^\pi = 0$$ So, $$\int_0^\pi x \sin(nx) dx = -\frac{\pi (-1)^n}{n}$$ Therefore, $$b_n = \frac{2}{\pi} \left(-\frac{\pi (-1)^n}{n}\right) = -\frac{2 (-1)^n}{n}$$ The Fourier series is: $$f(x) = \sum_{n=1}^\infty -\frac{2 (-1)^n}{n} \sin(nx) = 2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \sin(nx)$$ --- 2. **Find the Fourier sine series for the square 2\pi-periodic wave defined by:** $$f(x) = \begin{cases} 0, & -\pi \leq x \leq 0 \\ 1, & 0 < x \leq \pi \end{cases}$$ Since $f(x)$ is defined on $[-\pi, \pi]$ and is zero on the negative half, the sine series is: $$b_n = \frac{2}{\pi} \int_0^\pi 1 \cdot \sin(nx) dx = \frac{2}{\pi} \left[-\frac{\cos(nx)}{n}\right]_0^\pi = \frac{2}{\pi} \frac{1 - (-1)^n}{n}$$ Note that $1 - (-1)^n$ is zero for even $n$ and 2 for odd $n$. So, $$b_n = \begin{cases} \frac{4}{\pi n}, & n \text{ odd} \\ 0, & n \text{ even} \end{cases}$$ The Fourier sine series is: $$f(x) = \sum_{\substack{n=1 \\ n \text{ odd}}}^\infty \frac{4}{\pi n} \sin(nx)$$ --- 3. **Determine the Fourier cosine series for the function:** $$f(x) = \begin{cases} -1, & -\pi < x < -\frac{\pi}{2} \\ 1, & -\frac{\pi}{2} < x < \frac{\pi}{2} \\ -1, & \frac{\pi}{2} < x < \pi \end{cases}$$ This function is even, so the Fourier cosine series is: $$a_0 = \frac{1}{\pi} \int_{-\pi}^\pi f(x) dx = \frac{2}{\pi} \int_0^\pi f(x) dx$$ Calculate $a_0$: $$a_0 = \frac{2}{\pi} \left( \int_0^{\frac{\pi}{2}} 1 \, dx + \int_{\frac{\pi}{2}}^\pi (-1) \, dx \right) = \frac{2}{\pi} \left( \frac{\pi}{2} - \left( \pi - \frac{\pi}{2} \right) \right) = \frac{2}{\pi} (\frac{\pi}{2} - \frac{\pi}{2}) = 0$$ Calculate $a_n$ for $n \geq 1$: $$a_n = \frac{2}{\pi} \int_0^\pi f(x) \cos(nx) dx = \frac{2}{\pi} \left( \int_0^{\frac{\pi}{2}} 1 \cdot \cos(nx) dx + \int_{\frac{\pi}{2}}^\pi (-1) \cdot \cos(nx) dx \right)$$ Evaluate each integral: $$\int_0^{\frac{\pi}{2}} \cos(nx) dx = \left. \frac{\sin(nx)}{n} \right|_0^{\frac{\pi}{2}} = \frac{\sin(\frac{n\pi}{2})}{n}$$ $$\int_{\frac{\pi}{2}}^\pi \cos(nx) dx = \left. \frac{\sin(nx)}{n} \right|_{\frac{\pi}{2}}^\pi = \frac{\sin(n\pi) - \sin(\frac{n\pi}{2})}{n} = -\frac{\sin(\frac{n\pi}{2})}{n}$$ So, $$a_n = \frac{2}{\pi} \left( \frac{\sin(\frac{n\pi}{2})}{n} - \frac{\sin(\frac{n\pi}{2})}{n} \right) = \frac{2}{\pi} \cdot \frac{2 \sin(\frac{n\pi}{2})}{n} = \frac{4}{\pi n} \sin\left(\frac{n\pi}{2}\right)$$ Note that $\sin(\frac{n\pi}{2})$ is: - 1 for $n=1,5,9,...$ (i.e., $n \equiv 1 \pmod{4}$) - -1 for $n=3,7,11,...$ (i.e., $n \equiv 3 \pmod{4}$) - 0 for even $n$ Therefore, $$a_n = \begin{cases} \frac{4}{\pi n} (-1)^{\frac{n-1}{2}}, & n \text{ odd} \\ 0, & n \text{ even} \end{cases}$$ The Fourier cosine series is: $$f(x) = \sum_{\substack{n=1 \\ n \text{ odd}}}^\infty \frac{4}{\pi n} (-1)^{\frac{n-1}{2}} \cos(nx)$$