The function $\phi(x) = x$ on the interval $[-l,l]$ has the Fourier series $$x = \frac{2 l}{\pi}\sum_{m=1}^{\infty}\frac{(-1)^{m+1}}{m}\sin\left(\frac{m\pi x}{l} \right) = \frac{2 l}{\pi}\left(\sin\left(\frac{\pi x}{l} \right) - \frac{1}{2}\sin\left(\frac{2\pi x}{l} \right) + \frac{1}{3}\sin\left(\frac{3\pi x}{l} \right) - \ldots \right)$$ a.) Set $x = \frac{l}{2}$ to find the sum of the series $$\sum_{n=0}^{\infty}\frac{(-1)^n}{2n + 1} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \ldots$$ b.) Integrate the series term-by-term to find the Fourier series for $\frac{1}{2}x^2$, up tp a constant of integration (which is then the $\frac{1}{2}A_0$ term in the cosine series). Find the $A_0$ using the standard formula to completely determine the series. c.) Set $x = 0$ in your solution to part a.) and find the sum $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}$. d.) Integrate the series again, and find the sine series for $x^3 - l^2 x$. e.) Set $x - \frac{l}{2}$ to find the sum of the series $$\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n + 1)^3} = 1 - \frac{1}{3^3} + \frac{1}{5^3} - \frac{1}{7^3} + \ldots$$ f.) Integrate one last time to find the cosine series for $x^4 - 2l^2 x^2$. Remember, you'll need to find the constant term separately. g.) Finally, put $x = 0$ in your series in part d.) to find the sum $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^4}$
Attempted solution a.) - Letting $x = \frac{l}{2}$ then \begin{align*} \frac{l}{2} &= \frac{2l}{\pi}\sum_{m=1}^{\infty}\frac{(-1)^{m+1}}{m}\sin\left(\frac{m\pi(l/2)}{l} \right) = \frac{2l}{\pi}\left(\sin\left(\frac{\pi(l/2)}{l}\right) - \frac{1}{2}\sin\left(\frac{2\pi (l/2)}{l}\right) + \frac{1}{3}\sin\left(\frac{3\pi(l/2)}{l}\right) - \ldots \right)\\ &\Rightarrow \sum_{m=1}^{\infty}\frac{(-1)^{m+1}}{m}\sin\left(\frac{m\pi}{2}\right) = \left(\sin\left(\frac{\pi}{2}\right) - \frac{1}{2}\sin(\pi) + \frac{1}{3}\sin\left(\frac{3\pi}{2}\right) - \frac{1}{4}\sin\left(\frac{4\pi}{2}\right) + \frac{1}{5}\sin\left(\frac{5\pi}{2}\right) \right) \end{align*} Note that $$\sin\left(\frac{m\pi}{2}\right) = \begin{cases} 0, \ \ &\text{if} \ m \ \text{is even}, \ m = 2n, n=0,1,2,\ldots\\ (-1)^n \ \ &\text{if} \ m \ \text{is odd}, \ m = 2n+1, n = 0,1,2,\ldots \end{cases}$$ Then we have \begin{align*} \sum_{m=1}^{\infty}\frac{(-1)^{m+1}}{m}\sin\left(\frac{m\pi}{2}\right) = \sum_{n=0}^{\infty}\frac{(-1)^{2n+1+1\text{?}}}{2n+1}(-1)^{n} &= \sum_{n=0}^{\infty}\frac{(-1)^{3n+2}}{2n+1}\\ &=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}\\ &= 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7}\\ &= \frac{\pi}{4} \end{align*}
Attempted solution b.) - We want to integrate $$x = \frac{2l}{\pi}\sum_{m=1}^{\infty}\frac{(-1)^{m+1}}{m}\sin\left(\frac{m\pi x}{l}\right)$$ to find the Fourier series for $\frac{1}{2}x^2$. Thus, \begin{align*} \frac{1}{2}x^2 + \tau &= \frac{2l}{\pi}\int \sum_{m=1}^{\infty}\frac{(-1)^{m+1}}{m}\sin\left(\frac{m\pi x}{l}\right)dx\\ &= \frac{2l}{\pi}\sum_{m=1}^{\infty}\left(\frac{(-1)^{m+1}}{m}\frac{-1l}{m\pi}\cos\left(\frac{m\pi x}{l}\right) + \tau_m \right)\\ &= \frac{2l^2}{\pi^2}\sum_{m=1}^{\infty}\left(\frac{(-1)^m}{m^2}\cos\left(\frac{m\pi x}{l} \right) + \tau_m \right)\\ \Rightarrow \frac{1}{2}x^2 &= \frac{2l^2}{\pi^2}\sum_{m=1}^{\infty}\frac{(-1)^m}{m^2}\cos\left(\frac{m\pi x}{l}\right) + \tilde{\tau} \end{align*} where $\tilde{\tau} = \frac{2l^2}{\pi^2}\sum_{m=1}^{\infty}\tau_m - \tau$. Since this is a Fourier cosine series for $\frac{1}{2}x^2$ we get $\tilde{\tau}$ by using the standard formula to completely determine the series. Thus $$\tilde{\tau} = \frac{1}{2}A_0 = \frac{1}{2}\frac{1}{l}\int_{-l}^{l}\frac{1}{2}x^2 dx = \frac{l^2}{6}$$ Therefore, $$\frac{1}{2}x^2 = \frac{2l^2}{\pi^2}\sum_{m=1}^{\infty}\frac{(-1)^m}{m^2}\cos\left(\frac{m\pi x}{l}\right) + \frac{l^2}{6}$$
Attempted solution c.) - Let $x = 0$ for our solution to part a.). Then we have \begin{align*} 0 &= \frac{2l^2}{\pi^2}\sum_{m=1}^{\infty}\frac{(-1)^m}{m^2} + \frac{l^2}{6}\\ &\Rightarrow \frac{2l^2}{\pi^2}\sum_{m=1}^{\infty}\frac{(-1)^m}{m^2} = -\frac{l^2}{6}\\ &\Rightarrow \sum_{m=1}^{\infty}\frac{(-1)^{m+1}}{m^2} = \frac{\pi^2}{12} \end{align*}
Attempted solution d.) - We know that $$\frac{1}{2}x^2 = \frac{2l^2}{\pi^2}\sum_{m=1}^{\infty}\frac{(-1)^m}{m^2}\cos\left(\frac{m\pi x}{l}\right) + \frac{l^2}{6}$$ or that $$\frac{1}{2}x^2 - \frac{l^2}{6} = \frac{4l^2}{\pi^2}\sum_{m=1}^{\infty}\frac{(-1)^m}{m^2}\cos\left(\frac{m\pi x}{l}\right)$$ Thus integrating the series we have \begin{align*} \frac{1}{6}x^3 - \frac{l^2}{6}x &= \frac{2l^2}{\pi^2}\sum_{m=1}^{\infty}\frac{(-1)^{m}}{m^2}\frac{l}{m\pi}\sin\left(\frac{m\pi x}{l}\right) + \xi\\ &= \frac{2l^3}{\pi^3}\sum_{m=1}^{\infty}\frac{(-1)^{m}}{m^3}\sin\left(\frac{m\pi x}{l}\right) + \xi \end{align*} Then $$\xi = \frac{1}{2l}\int_{-l}^{l}\frac{1}{6}x^3 - \frac{l^2}{6}x dx 0$$ Therefore $$x^3 - l^2x = \frac{12l^3}{\pi^3}\sum_{m=1}^{\infty}\frac{(-1)^{m}}{m^3}\sin\left(\frac{m\pi x}{l}\right)$$
Attempted solution e.) - Setting $x = \frac{l}{2}$ we get \begin{align*} &\left(\frac{l}{2} \right)^3 - l^2\left(\frac{l}{2}\right) = \frac{12l^2}{\pi^3}\sum_{m=1}^{\infty}\frac{(-1)^{m}}{m^3}\sin\left(\frac{m\pi (l/2)}{l}\right)\\ &= \frac{12l^2}{\pi^3}\sum_{m=1}^{\infty}\frac{(-1)^{m}}{m^3}\sin\left(\frac{m\pi (l/2)}{l}\right) = \frac{12l^2}{\pi^3}\left(-\sin\left(\frac{\pi(l/2)}{l}\right) + \frac{1}{8}\sin\left(\frac{2\pi (l/2)}{l}\right) - \frac{1}{27}\sin\left(\frac{3\pi(l/2)}{l}\right) + \ldots \right)\\ &\Rightarrow \sum_{m=1}^{\infty}\frac{(-1)^m}{m^3}\sin\left(\frac{m\pi}{2}\right) = \left(-\sin\left(\frac{\pi}{2}\right) + \frac{1}{8}\sin\left(\frac{2\pi}{2}\right) - \frac{1}{27}\sin\left(\frac{3\pi}{2}\right) + \ldots \right) \end{align*} Note that $$\sin\left(\frac{m\pi}{2}\right) = \begin{cases} 0, \ \ &\text{if} \ m \ \text{is even}, \ m = 2n, n=0,1,2,\ldots\\ (-1)^n \ \ &\text{if} \ m \ \text{is odd}, \ m = 2n+1, n = 0,1,2,\ldots \end{cases}$$ Then we have \begin{align*} \sum_{m=1}^{\infty}\frac{(-1)^m}{m^3}\sin\left(\frac{m\pi}{2}\right) &= \sum_{n=0}^{\infty}\frac{(-1)^{2n+2}}{(2n+1)^3}(-1)^n\\ &= \sum_{n=0}^{\infty}\frac{(-1)^{3n+2}}{(2n+1)^3}\\ &= \sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^3}\\ &= 1 - \frac{1}{3^3} + \frac{1}{5^3} - \frac{1}{7^3} + \ldots\\ &= \frac{\pi^3}{32} \end{align*}
Attempted solution f.) - We know that $$x^3 - l^2x = \frac{12l^3}{\pi^3}\sum_{m=1}^{\infty}\frac{(-1)^{m}}{m^3}\sin\left(\frac{m\pi x}{l}\right)$$ Integrating this series we get \begin{align*} \frac{x^4}{4} - \frac{l^2 x^2}{2} &= \frac{12l^3}{\pi^3}\sum_{m=1}^{\infty}\frac{(-1)^{m}}{m^3}\left(-\frac{l}{m\pi} \right)\cos\left(\frac{m\pi x}{l}\right) + \eta\\ &= \frac{12l^4}{\pi^4}\sum_{m=1}^{\infty}\frac{(-1)^{m+1}}{m^4}\cos\left(\frac{m\pi x}{l}\right) + \eta \end{align*} Then, $$\eta = \frac{1}{2l}\int_{-l}^{l}\frac{x^4}{4} - \frac{l^2 x^2}{2}dx = -\frac{7l^4}{60}$$ Therefore, $$x^4 - 2l^2 x^2 = \frac{48l^4}{\pi^3}\sum_{m=1}^{\infty}\frac{(-1)^{m+1}}{m^4}\cos\left(\frac{m\pi x}{l}\right) - \frac{7l^4}{15}$$
Attempted solution g.) - Letting $x = 0$ we have \begin{align*} \frac{48l^4}{\pi^3}\sum_{m=1}^{\infty}\frac{(-1)^{m+1}}{m^4} - \frac{7l^4}{15} &= 0\\ \frac{48l^4}{\pi^3}\sum_{m=1}^{\infty}\frac{(-1)^{m+1}}{m^4} &= \frac{7l^4}{15}\\ \Rightarrow \sum_{m=1}^{\infty}\frac{(-1)^{m+1}}{m^4} &= \frac{7\pi^4}{720} \end{align*}
These are all my attempted solutions. Please let me know if there is anything wrong.