1. **Problem (a):** Find the Fourier series expansion of $$f(x) = \begin{cases} \pi + x, & -\pi < x < 0 \\ \pi - x, & 0 < x < \pi \end{cases}$$ 2. **Step 1: Identify the interval and period** The function is defined on $(-\pi, \pi)$, so the period is $2\pi$. 3. **Step 2: Calculate $a_0$** $$a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \, dx = \frac{1}{\pi} \left( \int_{-\pi}^0 (\pi + x) \, dx + \int_0^{\pi} (\pi - x) \, dx \right)$$ Calculate each integral: $$\int_{-\pi}^0 (\pi + x) \, dx = \left[ \pi x + \frac{x^2}{2} \right]_{-\pi}^0 = 0 - \left(-\pi^2 + \frac{\pi^2}{2}\right) = \frac{\pi^2}{2}$$ $$\int_0^{\pi} (\pi - x) \, dx = \left[ \pi x - \frac{x^2}{2} \right]_0^{\pi} = \pi^2 - \frac{\pi^2}{2} = \frac{\pi^2}{2}$$ So, $$a_0 = \frac{1}{\pi} \left( \frac{\pi^2}{2} + \frac{\pi^2}{2} \right) = \frac{1}{\pi} \pi^2 = \pi$$ 4. **Step 3: Calculate $a_n$** $$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx = \frac{1}{\pi} \left( \int_{-\pi}^0 (\pi + x) \cos(nx) \, dx + \int_0^{\pi} (\pi - x) \cos(nx) \, dx \right)$$ Calculate each integral using integration by parts: For $\int_{-\pi}^0 (\pi + x) \cos(nx) \, dx$: Let $u = \pi + x$, $dv = \cos(nx) dx$; then $du = dx$, $v = \frac{\sin(nx)}{n}$. $$\int (\pi + x) \cos(nx) dx = (\pi + x) \frac{\sin(nx)}{n} - \int \frac{\sin(nx)}{n} dx = (\pi + x) \frac{\sin(nx)}{n} + \frac{\cos(nx)}{n^2} + C$$ Evaluate from $-\pi$ to $0$: $$= \left[(\pi + 0) \frac{\sin(0)}{n} + \frac{\cos(0)}{n^2}\right] - \left[(\pi - \pi) \frac{\sin(-n\pi)}{n} + \frac{\cos(-n\pi)}{n^2}\right] = \frac{1}{n^2} - \frac{\cos(n\pi)}{n^2} = \frac{1 - (-1)^n}{n^2}$$ Similarly for $\int_0^{\pi} (\pi - x) \cos(nx) \, dx$: Let $u = \pi - x$, $dv = \cos(nx) dx$; then $du = -dx$, $v = \frac{\sin(nx)}{n}$. $$\int (\pi - x) \cos(nx) dx = (\pi - x) \frac{\sin(nx)}{n} + \int \frac{\sin(nx)}{n} dx = (\pi - x) \frac{\sin(nx)}{n} - \frac{\cos(nx)}{n^2} + C$$ Evaluate from $0$ to $\pi$: $$= \left[(\pi - \pi) \frac{\sin(n\pi)}{n} - \frac{\cos(n\pi)}{n^2}\right] - \left[(\pi - 0) \frac{\sin(0)}{n} - \frac{\cos(0)}{n^2}\right] = - \frac{(-1)^n}{n^2} + \frac{1}{n^2} = \frac{1 - (-1)^n}{n^2}$$ So, $$a_n = \frac{1}{\pi} \left( \frac{1 - (-1)^n}{n^2} + \frac{1 - (-1)^n}{n^2} \right) = \frac{2}{\pi} \frac{1 - (-1)^n}{n^2}$$ Since $1 - (-1)^n = 0$ for even $n$ and $2$ for odd $n$, only odd $n$ terms survive: $$a_n = \begin{cases} \frac{4}{\pi n^2}, & n \text{ odd} \\ 0, & n \text{ even} \end{cases}$$ 5. **Step 4: Calculate $b_n$** $$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx = \frac{1}{\pi} \left( \int_{-\pi}^0 (\pi + x) \sin(nx) \, dx + \int_0^{\pi} (\pi - x) \sin(nx) \, dx \right)$$ Calculate each integral: For $\int_{-\pi}^0 (\pi + x) \sin(nx) dx$: Use integration by parts: Let $u = \pi + x$, $dv = \sin(nx) dx$; then $du = dx$, $v = -\frac{\cos(nx)}{n}$. $$\int (\pi + x) \sin(nx) dx = -(\pi + x) \frac{\cos(nx)}{n} + \int \frac{\cos(nx)}{n} dx = -(\pi + x) \frac{\cos(nx)}{n} + \frac{\sin(nx)}{n^2} + C$$ Evaluate from $-\pi$ to $0$: $$= -\pi \frac{\cos(0)}{n} + \frac{\sin(0)}{n^2} - \left[-\pi \frac{\cos(-n\pi)}{n} + \frac{\sin(-n\pi)}{n^2}\right] = -\frac{\pi}{n} + \frac{\pi (-1)^n}{n} = \frac{\pi}{n}((-1)^n - 1)$$ For $\int_0^{\pi} (\pi - x) \sin(nx) dx$: Let $u = \pi - x$, $dv = \sin(nx) dx$; then $du = -dx$, $v = -\frac{\cos(nx)}{n}$. $$\int (\pi - x) \sin(nx) dx = -(\pi - x) \frac{\cos(nx)}{n} - \int -\frac{\cos(nx)}{n} dx = -(\pi - x) \frac{\cos(nx)}{n} - \frac{\sin(nx)}{n^2} + C$$ Evaluate from $0$ to $\pi$: $$= -0 + 0 - \left[-\pi \frac{\cos(0)}{n} - 0\right] = \frac{\pi}{n}$$ So, $$b_n = \frac{1}{\pi} \left( \frac{\pi}{n}((-1)^n - 1) + \frac{\pi}{n} \right) = \frac{1}{n} \left( (-1)^n - 1 + 1 \right) = \frac{(-1)^n}{n}$$ 6. **Step 5: Write the Fourier series** $$f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty \left( a_n \cos(nx) + b_n \sin(nx) \right) = \frac{\pi}{2} + \sum_{n=1}^\infty \left( a_n \cos(nx) + \frac{(-1)^n}{n} \sin(nx) \right)$$ where $$a_n = \begin{cases} \frac{4}{\pi n^2}, & n \text{ odd} \\ 0, & n \text{ even} \end{cases}$$ --- 1. **Problem (b):** $$f(x) = \begin{cases} -\pi - x, & -\pi < x < 0 \\ \pi - x, & 0 < x < \pi \end{cases}$$ 2. **Step 1: Calculate $a_0$** $$a_0 = \frac{1}{\pi} \left( \int_{-\pi}^0 (-\pi - x) dx + \int_0^{\pi} (\pi - x) dx \right)$$ Calculate each integral: $$\int_{-\pi}^0 (-\pi - x) dx = \left[-\pi x - \frac{x^2}{2}\right]_{-\pi}^0 = 0 - \left(\pi^2 - \frac{\pi^2}{2}\right) = -\frac{\pi^2}{2}$$ $$\int_0^{\pi} (\pi - x) dx = \frac{\pi^2}{2}$$ Sum: $$a_0 = \frac{1}{\pi} \left(-\frac{\pi^2}{2} + \frac{\pi^2}{2}\right) = 0$$ 3. **Step 2: Calculate $a_n$** $$a_n = \frac{1}{\pi} \left( \int_{-\pi}^0 (-\pi - x) \cos(nx) dx + \int_0^{\pi} (\pi - x) \cos(nx) dx \right)$$ Using integration by parts similarly as in (a), the results are: $$\int_{-\pi}^0 (-\pi - x) \cos(nx) dx = -\frac{1 - (-1)^n}{n^2}$$ $$\int_0^{\pi} (\pi - x) \cos(nx) dx = \frac{1 - (-1)^n}{n^2}$$ Sum: $$a_n = \frac{1}{\pi} \left(-\frac{1 - (-1)^n}{n^2} + \frac{1 - (-1)^n}{n^2}\right) = 0$$ 4. **Step 3: Calculate $b_n$** $$b_n = \frac{1}{\pi} \left( \int_{-\pi}^0 (-\pi - x) \sin(nx) dx + \int_0^{\pi} (\pi - x) \sin(nx) dx \right)$$ Calculate each integral: $$\int_{-\pi}^0 (-\pi - x) \sin(nx) dx = -\frac{\pi}{n}((-1)^n + 1)$$ $$\int_0^{\pi} (\pi - x) \sin(nx) dx = \frac{\pi}{n}$$ Sum: $$b_n = \frac{1}{\pi} \left(-\frac{\pi}{n}((-1)^n + 1) + \frac{\pi}{n}\right) = \frac{1}{n} \left(1 - ((-1)^n + 1)\right) = -\frac{(-1)^n}{n}$$ 5. **Step 4: Write the Fourier series** $$f(x) = \sum_{n=1}^\infty b_n \sin(nx) = -\sum_{n=1}^\infty \frac{(-1)^n}{n} \sin(nx)$$ --- 1. **Problem (c):** $$f(x) = \begin{cases} x, & -\pi < x < 0 \\ \pi - x, & 0 < x < \pi \end{cases}$$ 2. **Step 1: Calculate $a_0$** $$a_0 = \frac{1}{\pi} \left( \int_{-\pi}^0 x dx + \int_0^{\pi} (\pi - x) dx \right) = \frac{1}{\pi} \left( \frac{x^2}{2} \Big|_{-\pi}^0 + \left[ \pi x - \frac{x^2}{2} \right]_0^{\pi} \right)$$ Calculate: $$\int_{-\pi}^0 x dx = 0 - \frac{(-\pi)^2}{2} = -\frac{\pi^2}{2}$$ $$\int_0^{\pi} (\pi - x) dx = \pi^2 - \frac{\pi^2}{2} = \frac{\pi^2}{2}$$ Sum: $$a_0 = \frac{1}{\pi} \left(-\frac{\pi^2}{2} + \frac{\pi^2}{2}\right) = 0$$ 3. **Step 2: Calculate $a_n$** $$a_n = \frac{1}{\pi} \left( \int_{-\pi}^0 x \cos(nx) dx + \int_0^{\pi} (\pi - x) \cos(nx) dx \right)$$ Calculate each integral by parts: $$\int_{-\pi}^0 x \cos(nx) dx = \frac{\pi}{n} (-1)^n$$ $$\int_0^{\pi} (\pi - x) \cos(nx) dx = \frac{1 - (-1)^n}{n^2}$$ Sum: $$a_n = \frac{1}{\pi} \left( \frac{\pi}{n} (-1)^n + \frac{1 - (-1)^n}{n^2} \right) = \frac{(-1)^n}{n} + \frac{1 - (-1)^n}{\pi n^2}$$ 4. **Step 3: Calculate $b_n$** $$b_n = \frac{1}{\pi} \left( \int_{-\pi}^0 x \sin(nx) dx + \int_0^{\pi} (\pi - x) \sin(nx) dx \right)$$ Calculate each integral: $$\int_{-\pi}^0 x \sin(nx) dx = -\frac{\pi}{n}$$ $$\int_0^{\pi} (\pi - x) \sin(nx) dx = \frac{\pi}{n}$$ Sum: $$b_n = \frac{1}{\pi} ( -\frac{\pi}{n} + \frac{\pi}{n} ) = 0$$ 5. **Step 4: Write the Fourier series** $$f(x) = \sum_{n=1}^\infty a_n \cos(nx) = \sum_{n=1}^\infty \left( \frac{(-1)^n}{n} + \frac{1 - (-1)^n}{\pi n^2} \right) \cos(nx)$$ --- 1. **Problem (d):** $$f(x) = x^2, \quad 0 < x < 2\pi$$ 2. **Step 1: Period and interval** Assuming period $2\pi$, redefine interval to $(-\pi, \pi)$ by shifting or consider Fourier series on $(0, 2\pi)$ with period $2\pi$. 3. **Step 2: Calculate $a_0$** $$a_0 = \frac{1}{\pi} \int_0^{2\pi} x^2 dx = \frac{1}{\pi} \left[ \frac{x^3}{3} \right]_0^{2\pi} = \frac{(2\pi)^3}{3\pi} = \frac{8\pi^2}{3}$$ 4. **Step 3: Calculate $a_n$** $$a_n = \frac{1}{\pi} \int_0^{2\pi} x^2 \cos(nx) dx$$ Use integration by parts twice: First integration by parts: Let $u = x^2$, $dv = \cos(nx) dx$; then $du = 2x dx$, $v = \frac{\sin(nx)}{n}$. $$\int x^2 \cos(nx) dx = x^2 \frac{\sin(nx)}{n} - \int 2x \frac{\sin(nx)}{n} dx = \frac{x^2 \sin(nx)}{n} - \frac{2}{n} \int x \sin(nx) dx$$ Second integration by parts for $\int x \sin(nx) dx$: Let $u = x$, $dv = \sin(nx) dx$; then $du = dx$, $v = -\frac{\cos(nx)}{n}$. $$\int x \sin(nx) dx = -x \frac{\cos(nx)}{n} + \int \frac{\cos(nx)}{n} dx = -\frac{x \cos(nx)}{n} + \frac{\sin(nx)}{n^2} + C$$ So, $$\int x^2 \cos(nx) dx = \frac{x^2 \sin(nx)}{n} - \frac{2}{n} \left(-\frac{x \cos(nx)}{n} + \frac{\sin(nx)}{n^2} \right) + C = \frac{x^2 \sin(nx)}{n} + \frac{2x \cos(nx)}{n^2} - \frac{2 \sin(nx)}{n^3} + C$$ Evaluate from $0$ to $2\pi$: Since $\sin(n \cdot 0) = 0$, $\sin(2 n \pi) = 0$, $\cos(2 n \pi) = 1$, $\cos(0) = 1$: $$a_n = \frac{1}{\pi} \left[ \frac{2\pi^2 \sin(2 n \pi)}{n} + \frac{2 \cdot 2\pi \cos(2 n \pi)}{n^2} - \frac{2 \sin(2 n \pi)}{n^3} - \left(0 + 0 - 0\right) \right] = \frac{1}{\pi} \cdot \frac{4\pi}{n^2} = \frac{4\pi}{n^2}$$ 5. **Step 4: Calculate $b_n$** $$b_n = \frac{1}{\pi} \int_0^{2\pi} x^2 \sin(nx) dx$$ Since $x^2$ is even and $\sin(nx)$ is odd about $\pi$, the integral over $0$ to $2\pi$ is zero: $$b_n = 0$$ 6. **Step 5: Write the Fourier series** $$f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos(nx) = \frac{4\pi^2}{3} + \sum_{n=1}^\infty \frac{4\pi}{n^2} \cos(nx)$$ --- **Summary:** - (a) $f(x) = \frac{\pi}{2} + \sum_{n=1}^\infty \left( a_n \cos(nx) + \frac{(-1)^n}{n} \sin(nx) \right)$ with $a_n = \frac{4}{\pi n^2}$ for odd $n$, 0 otherwise. - (b) $f(x) = -\sum_{n=1}^\infty \frac{(-1)^n}{n} \sin(nx)$. - (c) $f(x) = \sum_{n=1}^\infty \left( \frac{(-1)^n}{n} + \frac{1 - (-1)^n}{\pi n^2} \right) \cos(nx)$. - (d) $f(x) = \frac{4\pi^2}{3} + \sum_{n=1}^\infty \frac{4\pi}{n^2} \cos(nx)$.