Fourier Constant 41E5Bd

1. **Problem Statement:** Find the Fourier series of the function \(f(x) = -1\) defined on the interval \(-2 < x < 2\) with period \(p = 4\). 2. **Fourier Series Formula:** For a function with period \(p\), the Fourier series is given by: $$ a_0 = \frac{2}{p} \int_{-p/2}^{p/2} f(x) \, dx $$ $$ a_n = \frac{2}{p} \int_{-p/2}^{p/2} f(x) \cos\left(\frac{2\pi n x}{p}\right) \, dx $$ $$ b_n = \frac{2}{p} \int_{-p/2}^{p/2} f(x) \sin\left(\frac{2\pi n x}{p}\right) \, dx $$ 3. **Given:** \(f(x) = -1\) for \(-2 < x < 2\), period \(p=4\). 4. **Calculate \(a_0\):** $$ a_0 = \frac{2}{4} \int_{-2}^{2} (-1) \, dx = \frac{1}{2} \times (-1) \times (2 - (-2)) = \frac{1}{2} \times (-1) \times 4 = -2 $$ 5. **Calculate \(a_n\):** $$ a_n = \frac{2}{4} \int_{-2}^{2} (-1) \cos\left(\frac{2\pi n x}{4}\right) \, dx = \frac{1}{2} \int_{-2}^{2} -\cos\left(\frac{\pi n x}{2}\right) \, dx = -\frac{1}{2} \int_{-2}^{2} \cos\left(\frac{\pi n x}{2}\right) \, dx $$ 6. **Evaluate the integral:** $$ \int_{-2}^{2} \cos\left(\frac{\pi n x}{2}\right) \, dx = \left[ \frac{2}{\pi n} \sin\left(\frac{\pi n x}{2}\right) \right]_{-2}^{2} = \frac{2}{\pi n} \left( \sin(\pi n) - \sin(-\pi n) \right) $$ 7. Since \(\sin(\pi n) = 0\) for all integers \(n\), and \(\sin(-\pi n) = -\sin(\pi n) = 0\), the integral is zero. 8. Therefore, \(a_n = -\frac{1}{2} \times 0 = 0\) for all \(n \geq 1\). 9. **Calculate \(b_n\):** $$ b_n = \frac{2}{4} \int_{-2}^{2} (-1) \sin\left(\frac{2\pi n x}{4}\right) \, dx = \frac{1}{2} \int_{-2}^{2} -\sin\left(\frac{\pi n x}{2}\right) \, dx = -\frac{1}{2} \int_{-2}^{2} \sin\left(\frac{\pi n x}{2}\right) \, dx $$ 10. **Evaluate the integral:** $$ \int_{-2}^{2} \sin\left(\frac{\pi n x}{2}\right) \, dx = \left[ -\frac{2}{\pi n} \cos\left(\frac{\pi n x}{2}\right) \right]_{-2}^{2} = -\frac{2}{\pi n} \left( \cos(\pi n) - \cos(-\pi n) \right) $$ 11. Since \(\cos(\pi n) = \cos(-\pi n)\), the difference is zero, so the integral is zero. 12. Therefore, \(b_n = -\frac{1}{2} \times 0 = 0\) for all \(n \geq 1\). 13. **Final Fourier series:** $$ f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty \left( a_n \cos\left(\frac{2\pi n x}{p}\right) + b_n \sin\left(\frac{2\pi n x}{p}\right) \right) = \frac{-2}{2} + 0 = -1 $$ 14. **Interpretation:** The Fourier series of a constant function \(-1\) is just the constant itself, with no sine or cosine terms. **Answer:** \(f(x) = -1\) for all \(x\).