Sine Series Eed9Df

1. **Stating the problem:** We have a piecewise function defined as: $$f(x) = \begin{cases} x, & 0 < x < 4 \\ 8 - x, & 4 < x < 8 \end{cases}$$ We want to find the sine series expansion of this function. 2. **Formula for sine series expansion:** For a function defined on $(0, L)$, the sine series expansion is given by: $$f(x) \sim \sum_{n=1}^\infty b_n \sin\left(\frac{n \pi x}{L}\right)$$ where $$b_n = \frac{2}{L} \int_0^L f(x) \sin\left(\frac{n \pi x}{L}\right) dx$$ 3. **Identify the interval and function:** Here, the function is defined on $(0,8)$, so $L=8$. 4. **Calculate the coefficients $b_n$:** We split the integral according to the piecewise definition: $$b_n = \frac{2}{8} \left( \int_0^4 x \sin\left(\frac{n \pi x}{8}\right) dx + \int_4^8 (8 - x) \sin\left(\frac{n \pi x}{8}\right) dx \right)$$ 5. **Calculate the first integral:** Let $I_1 = \int_0^4 x \sin\left(\frac{n \pi x}{8}\right) dx$. Use integration by parts: - Let $u = x$, $dv = \sin\left(\frac{n \pi x}{8}\right) dx$. - Then $du = dx$, and $$v = -\frac{8}{n \pi} \cos\left(\frac{n \pi x}{8}\right)$$ So, $$I_1 = uv \bigg|_0^4 - \int_0^4 v du = -\frac{8}{n \pi} x \cos\left(\frac{n \pi x}{8}\right) \bigg|_0^4 + \frac{8}{n \pi} \int_0^4 \cos\left(\frac{n \pi x}{8}\right) dx$$ Calculate the remaining integral: $$\int_0^4 \cos\left(\frac{n \pi x}{8}\right) dx = \frac{8}{n \pi} \sin\left(\frac{n \pi x}{8}\right) \bigg|_0^4 = \frac{8}{n \pi} \sin\left(\frac{n \pi}{2}\right)$$ Putting it all together: $$I_1 = -\frac{8}{n \pi} \cdot 4 \cos\left(\frac{n \pi}{2}\right) + \frac{8}{n \pi} \cdot \frac{8}{n \pi} \sin\left(\frac{n \pi}{2}\right) = -\frac{32}{n \pi} \cos\left(\frac{n \pi}{2}\right) + \frac{64}{n^2 \pi^2} \sin\left(\frac{n \pi}{2}\right)$$ 6. **Calculate the second integral:** Let $I_2 = \int_4^8 (8 - x) \sin\left(\frac{n \pi x}{8}\right) dx$. Rewrite as: $$I_2 = 8 \int_4^8 \sin\left(\frac{n \pi x}{8}\right) dx - \int_4^8 x \sin\left(\frac{n \pi x}{8}\right) dx$$ Calculate each separately. First, $$\int_4^8 \sin\left(\frac{n \pi x}{8}\right) dx = -\frac{8}{n \pi} \cos\left(\frac{n \pi x}{8}\right) \bigg|_4^8 = -\frac{8}{n \pi} \left[ \cos(n \pi) - \cos\left(\frac{n \pi}{2}\right) \right]$$ Second, let $J = \int_4^8 x \sin\left(\frac{n \pi x}{8}\right) dx$. Use integration by parts again: - $u = x$, $dv = \sin\left(\frac{n \pi x}{8}\right) dx$, - $du = dx$, $v = -\frac{8}{n \pi} \cos\left(\frac{n \pi x}{8}\right)$. So, $$J = uv \bigg|_4^8 - \int_4^8 v du = -\frac{8}{n \pi} x \cos\left(\frac{n \pi x}{8}\right) \bigg|_4^8 + \frac{8}{n \pi} \int_4^8 \cos\left(\frac{n \pi x}{8}\right) dx$$ Calculate the remaining integral: $$\int_4^8 \cos\left(\frac{n \pi x}{8}\right) dx = \frac{8}{n \pi} \sin\left(\frac{n \pi x}{8}\right) \bigg|_4^8 = \frac{8}{n \pi} \left[ \sin(n \pi) - \sin\left(\frac{n \pi}{2}\right) \right] = -\frac{8}{n \pi} \sin\left(\frac{n \pi}{2}\right)$$ Evaluate the boundary term: $$-\frac{8}{n \pi} x \cos\left(\frac{n \pi x}{8}\right) \bigg|_4^8 = -\frac{8}{n \pi} \left(8 \cos(n \pi) - 4 \cos\left(\frac{n \pi}{2}\right) \right) = -\frac{64}{n \pi} \cos(n \pi) + \frac{32}{n \pi} \cos\left(\frac{n \pi}{2}\right)$$ Putting it all together: $$J = -\frac{64}{n \pi} \cos(n \pi) + \frac{32}{n \pi} \cos\left(\frac{n \pi}{2}\right) - \frac{64}{n^2 \pi^2} \sin\left(\frac{n \pi}{2}\right)$$ 7. **Combine to find $I_2$:** $$I_2 = 8 \left(-\frac{8}{n \pi} \left[ \cos(n \pi) - \cos\left(\frac{n \pi}{2}\right) \right]\right) - J$$ $$= -\frac{64}{n \pi} \cos(n \pi) + \frac{64}{n \pi} \cos\left(\frac{n \pi}{2}\right) - \left(-\frac{64}{n \pi} \cos(n \pi) + \frac{32}{n \pi} \cos\left(\frac{n \pi}{2}\right) - \frac{64}{n^2 \pi^2} \sin\left(\frac{n \pi}{2}\right)\right)$$ Simplify: $$I_2 = -\frac{64}{n \pi} \cos(n \pi) + \frac{64}{n \pi} \cos\left(\frac{n \pi}{2}\right) + \frac{64}{n \pi} \cos(n \pi) - \frac{32}{n \pi} \cos\left(\frac{n \pi}{2}\right) + \frac{64}{n^2 \pi^2} \sin\left(\frac{n \pi}{2}\right)$$ $$I_2 = \left(-\frac{64}{n \pi} \cos(n \pi) + \frac{64}{n \pi} \cos(n \pi)\right) + \left(\frac{64}{n \pi} - \frac{32}{n \pi}\right) \cos\left(\frac{n \pi}{2}\right) + \frac{64}{n^2 \pi^2} \sin\left(\frac{n \pi}{2}\right)$$ $$I_2 = 0 + \frac{32}{n \pi} \cos\left(\frac{n \pi}{2}\right) + \frac{64}{n^2 \pi^2} \sin\left(\frac{n \pi}{2}\right)$$ 8. **Sum $I_1$ and $I_2$ to get $b_n$ numerator:** $$I_1 + I_2 = \left(-\frac{32}{n \pi} \cos\left(\frac{n \pi}{2}\right) + \frac{64}{n^2 \pi^2} \sin\left(\frac{n \pi}{2}\right)\right) + \left(\frac{32}{n \pi} \cos\left(\frac{n \pi}{2}\right) + \frac{64}{n^2 \pi^2} \sin\left(\frac{n \pi}{2}\right)\right)$$ Simplify: $$I_1 + I_2 = 0 + \frac{128}{n^2 \pi^2} \sin\left(\frac{n \pi}{2}\right) = \frac{128}{n^2 \pi^2} \sin\left(\frac{n \pi}{2}\right)$$ 9. **Calculate $b_n$:** $$b_n = \frac{2}{8} \times \frac{128}{n^2 \pi^2} \sin\left(\frac{n \pi}{2}\right) = \frac{32}{n^2 \pi^2} \sin\left(\frac{n \pi}{2}\right)$$ 10. **Final sine series expansion:** $$f(x) \sim \sum_{n=1}^\infty \frac{32}{n^2 \pi^2} \sin\left(\frac{n \pi}{2}\right) \sin\left(\frac{n \pi x}{8}\right)$$ Note: Since $\sin\left(\frac{n \pi}{2}\right)$ is zero for even $n$, only odd terms contribute. **Answer:** $$f(x) = \sum_{n=1,3,5,\dots} \frac{32}{n^2 \pi^2} (-1)^{\frac{n-1}{2}} \sin\left(\frac{n \pi x}{8}\right)$$ This is the sine series expansion of the given piecewise function on $(0,8)$.