1. **Problem statement:** Determine the parity of the piecewise function $$f(x) = \begin{cases} \frac{1}{2} & \text{if } x \in [-1,0] \\ 1 & \text{if } x \in (0,1] \end{cases}$$ 2. **Recall parity definitions:** - A function is **even** if $f(-x) = f(x)$ for all $x$ in the domain. - A function is **odd** if $f(-x) = -f(x)$ for all $x$ in the domain. - Otherwise, the function is neither even nor odd. 3. **Check parity:** - For $x \in (0,1]$, $f(x) = 1$. - Then $f(-x)$ for $x \in (0,1]$ is $f(-x) = \frac{1}{2}$ since $-x \in [-1,0]$. - Since $f(-x) \neq f(x)$ and $f(-x) \neq -f(x)$, the function is neither even nor odd. --- 4. **Fourier series expansion:** - The function is defined on $[-1,1]$ and piecewise constant. - The Fourier series on $[-1,1]$ with period 2 is: $$f(x) \sim \frac{a_0}{2} + \sum_{n=1}^\infty \left(a_n \cos(n \pi x) + b_n \sin(n \pi x)\right)$$ 5. **Calculate coefficients:** - $$a_0 = \int_{-1}^1 f(x) \, dx = \int_{-1}^0 \frac{1}{2} \, dx + \int_0^1 1 \, dx = \frac{1}{2} \times 1 + 1 \times 1 = \frac{1}{2} + 1 = \frac{3}{2}$$ - $$a_n = \int_{-1}^1 f(x) \cos(n \pi x) \, dx = \int_{-1}^0 \frac{1}{2} \cos(n \pi x) \, dx + \int_0^1 1 \cos(n \pi x) \, dx$$ - Compute each integral: - $$\int_{-1}^0 \frac{1}{2} \cos(n \pi x) \, dx = \frac{1}{2} \left[ \frac{\sin(n \pi x)}{n \pi} \right]_{-1}^0 = \frac{1}{2} \times \frac{\sin(0) - \sin(-n \pi)}{n \pi} = 0$$ - $$\int_0^1 \cos(n \pi x) \, dx = \left[ \frac{\sin(n \pi x)}{n \pi} \right]_0^1 = \frac{\sin(n \pi) - \sin(0)}{n \pi} = 0$$ - So, $a_n = 0$ for all $n$. - $$b_n = \int_{-1}^1 f(x) \sin(n \pi x) \, dx = \int_{-1}^0 \frac{1}{2} \sin(n \pi x) \, dx + \int_0^1 1 \sin(n \pi x) \, dx$$ - Compute each integral: - $$\int_{-1}^0 \frac{1}{2} \sin(n \pi x) \, dx = \frac{1}{2} \left[ -\frac{\cos(n \pi x)}{n \pi} \right]_{-1}^0 = \frac{1}{2} \times \frac{1 - \cos(-n \pi)}{n \pi} = \frac{1}{2} \times \frac{1 - \cos(n \pi)}{n \pi}$$ - $$\int_0^1 \sin(n \pi x) \, dx = \left[ -\frac{\cos(n \pi x)}{n \pi} \right]_0^1 = \frac{1 - \cos(n \pi)}{n \pi}$$ - Summing: $$b_n = \frac{1}{2} \times \frac{1 - \cos(n \pi)}{n \pi} + \frac{1 - \cos(n \pi)}{n \pi} = \frac{3}{2} \times \frac{1 - \cos(n \pi)}{n \pi}$$ - Note that $\cos(n \pi) = (-1)^n$, so: $$b_n = \frac{3}{2} \times \frac{1 - (-1)^n}{n \pi}$$ - For even $n$, $1 - (-1)^n = 0$, so $b_n = 0$. - For odd $n$, $1 - (-1)^n = 2$, so: $$b_n = \frac{3}{2} \times \frac{2}{n \pi} = \frac{3}{n \pi}$$ 6. **Final Fourier series:** $$f(x) \sim \frac{3}{4} + \sum_{\substack{n=1 \\ n \text{ odd}}}^\infty \frac{3}{n \pi} \sin(n \pi x)$$ --- **Summary:** - The function is neither even nor odd. - Its Fourier series contains only sine terms with odd indices and a constant term.