1. **Problem:** Find the Fourier series of the function $$f(t) = \begin{cases} 1, & -\pi < t < 0 \\ 0, & 0 < t < \pi \end{cases}$$ 2. **Fourier Series Formula:** For a function with period $2\pi$, the Fourier series is $$f(t) \sim \frac{a_0}{2} + \sum_{n=1}^\infty \left(a_n \cos(nt) + b_n \sin(nt)\right)$$ where $$a_0 = \frac{1}{\pi} \int_{-\pi}^\pi f(t) dt$$ $$a_n = \frac{1}{\pi} \int_{-\pi}^\pi f(t) \cos(nt) dt$$ $$b_n = \frac{1}{\pi} \int_{-\pi}^\pi f(t) \sin(nt) dt$$ 3. **Calculate $a_0$:** $$a_0 = \frac{1}{\pi} \left( \int_{-\pi}^0 1 dt + \int_0^\pi 0 dt \right) = \frac{1}{\pi} (0 - (-\pi)) = 1$$ 4. **Calculate $a_n$:** $$a_n = \frac{1}{\pi} \left( \int_{-\pi}^0 1 \cdot \cos(nt) dt + \int_0^\pi 0 \cdot \cos(nt) dt \right) = \frac{1}{\pi} \int_{-\pi}^0 \cos(nt) dt$$ Evaluate the integral: $$\int_{-\pi}^0 \cos(nt) dt = \left[ \frac{\sin(nt)}{n} \right]_{-\pi}^0 = \frac{\sin(0) - \sin(-n\pi)}{n} = \frac{0 - 0}{n} = 0$$ So, $$a_n = 0$$ 5. **Calculate $b_n$:** $$b_n = \frac{1}{\pi} \left( \int_{-\pi}^0 1 \cdot \sin(nt) dt + \int_0^\pi 0 \cdot \sin(nt) dt \right) = \frac{1}{\pi} \int_{-\pi}^0 \sin(nt) dt$$ Evaluate the integral: $$\int_{-\pi}^0 \sin(nt) dt = \left[ -\frac{\cos(nt)}{n} \right]_{-\pi}^0 = -\frac{\cos(0) - \cos(-n\pi)}{n} = -\frac{1 - \cos(n\pi)}{n}$$ Recall $\cos(n\pi) = (-1)^n$, so $$b_n = \frac{1}{\pi} \cdot \left(-\frac{1 - (-1)^n}{n}\right) = -\frac{1 - (-1)^n}{n\pi}$$ 6. **Simplify $b_n$:** - For even $n$, $(-1)^n = 1$, so $b_n = 0$ - For odd $n$, $(-1)^n = -1$, so $$b_n = -\frac{1 - (-1)}{n\pi} = -\frac{2}{n\pi}$$ 7. **Final Fourier series:** Only odd $n$ terms contribute: $$f(t) \sim \frac{1}{2} + \sum_{\substack{n=1 \\ n \text{ odd}}}^\infty -\frac{2}{n\pi} \sin(nt) = \frac{1}{2} - \frac{2}{\pi} \sum_{k=0}^\infty \frac{\sin((2k+1)t)}{2k+1}$$ **Explanation:** - The function is a step function equal to 1 on $(-\pi,0)$ and 0 on $(0,\pi)$. - The Fourier series captures this discontinuity using sine terms only (odd function behavior). - The constant term $a_0/2 = 1/2$ represents the average value over one period. This completes the solution for problem 1.