1. **Problem Statement:** Find the Fourier series of the piecewise function $f(x)$ defined as $f(x) = -1$ for $-2 < x < 0$ and $f(x) = 1$ for $0 < x < 2$, with period $p = 4$. 2. **Fourier Series Formula:** For a function with period $p$, the Fourier series is given by: $$ f(x) = a_0 + \sum_{n=1}^\infty \left(a_n \cos \frac{2\pi n x}{p} + b_n \sin \frac{2\pi n x}{p}\right) $$ where $$ a_0 = \frac{1}{p} \int_{-p/2}^{p/2} f(x) \, dx, \quad a_n = \frac{2}{p} \int_{-p/2}^{p/2} f(x) \cos \frac{2\pi n x}{p} \, dx, \quad b_n = \frac{2}{p} \int_{-p/2}^{p/2} f(x) \sin \frac{2\pi n x}{p} \, dx. $$ 3. **Calculate $a_0$:** $$ a_0 = \frac{1}{4} \left( \int_{-2}^0 (-1) \, dx + \int_0^2 1 \, dx \right) = \frac{1}{4} \left( -2 + 2 \right) = 0. $$ 4. **Calculate $a_n$:** $$ a_n = \frac{2}{4} \left( \int_{-2}^0 (-1) \cos \frac{\pi n x}{2} \, dx + \int_0^2 1 \cos \frac{\pi n x}{2} \, dx \right) = \frac{1}{2} \left( - \int_{-2}^0 \cos \frac{\pi n x}{2} \, dx + \int_0^2 \cos \frac{\pi n x}{2} \, dx \right). $$ Since cosine is even, $\cos(\theta)$ is symmetric, so these integrals cancel out: $$ a_n = 0. $$ 5. **Calculate $b_n$:** $$ b_n = \frac{1}{2} \left( - \int_{-2}^0 \sin \frac{\pi n x}{2} \, dx + \int_0^2 \sin \frac{\pi n x}{2} \, dx \right). $$ Since sine is odd, $\sin(-x) = -\sin x$, the first integral becomes: $$ \int_{-2}^0 \sin \frac{\pi n x}{2} \, dx = - \int_0^2 \sin \frac{\pi n x}{2} \, dx. $$ Therefore, $$ b_n = \frac{1}{2} \left( - (- \int_0^2 \sin \frac{\pi n x}{2} \, dx) + \int_0^2 \sin \frac{\pi n x}{2} \, dx \right) = \int_0^2 \sin \frac{\pi n x}{2} \, dx. $$ 6. **Evaluate the integral:** $$ \int_0^2 \sin \frac{\pi n x}{2} \, dx = \left[ - \frac{2}{\pi n} \cos \frac{\pi n x}{2} \right]_0^2 = - \frac{2}{\pi n} \left( \cos \pi n - 1 \right). $$ Since $\cos \pi n = (-1)^n$, $$ b_n = - \frac{2}{\pi n} \left( (-1)^n - 1 \right). $$ 7. **Simplify $b_n$:** - For even $n$, $(-1)^n = 1$, so $b_n = 0$. - For odd $n$, $(-1)^n = -1$, so $$ b_n = - \frac{2}{\pi n} (-1 - 1) = \frac{4}{\pi n}. $$ 8. **Final Fourier series:** Only odd terms contribute: $$ f(x) = \sum_{n=1,3,5,\ldots} \frac{4}{\pi n} \sin \frac{\pi n x}{2}. $$ **Answer:** $$ f(x) = \frac{4}{\pi} \left( \sin \frac{\pi x}{2} + \frac{1}{3} \sin \frac{3 \pi x}{2} + \frac{1}{5} \sin \frac{5 \pi x}{2} + \cdots \right). $$