I'm not sure how to type math symbols on here so I'll try to be as clear as possible. My homework problem wants me to show that the Fourier cosine series for $\sin\left(x\right)$ on $\left(0,\pi\right)$ is $$ \frac{2}{\pi} - \frac{4}{\pi}\sum_{n = 1}^{\infty} \frac{\cos\left(2nx\right)}{4n^{2} - 1} $$ I spent an hour on this problem the other day and filled up an entire board at my school and I got close to the answer, but I was still off by quite a bit. I think my problem is that I don't know how to find my $a_{0}$ term. My book is horrible and doesn't explain anything. I'm using the Springer Applied PDE book by J. David Logan if that is any indication to you, lol. I know I have to do several integration by parts as well. If someone can just start me on the right track I can roll with it. Thanks.
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I just add some $\LaTeX$. You can check them. There is a nice tutorial overhere http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Felix Marin May 21 '16 at 21:43
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The Fourier Cosine series for $\sin(x)$ on $[0,\pi]$ is given by
$$\sin(x)=\frac{a_0}{2}+\sum_{n=1}^\infty a_n \cos(nx)$$
where $a_n=\frac{2}{\pi}\int_0^\pi \sin(x)\cos(nx)\,dx$.
Note that $a_0=\frac{4}{\pi}$ and $a_1=0$. For $n\ge 2$, we have
$$\begin{align} a_n&=\frac{2}{\pi}\int_0^\pi \sin(x)\cos(nx)\,dx\\\\ &=\frac{1}{\pi}\int_0^\pi (\sin((n+1)x)-\sin((n-1)x))\,dx\\\\ &=-\frac{1}{\pi}\left. \left(\frac{\cos((n+1)x)}{n+1}-\frac{\cos((n-1)x)}{n-1}\right)\right|_{0}^{\pi}\\\\ &=-\frac{1}{\pi}\left(\frac{(-1)^{n+1}-1}{n+1}-\frac{(-1)^{n-1}-1}{n-1}\right)\\\\ &=\frac{1}{\pi}\left((-1)^{n+1}-1\right)\frac{2}{n^2-1}\\\\ \end{align}$$
For $n$ odd, $a_n=0$. For $n=2m$ we have
$$a_n=-\frac{4}{\pi}\frac{1}{4m^2-1}$$
Mark Viola
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