I want to determine $p(x)=\sum_{|n|\le4}a_{n}e^{inx}$ so that i can minimize $\int_{-\pi}^{\pi} |y(x)-p(x)|^{2} dx$.
$y(x)=(sinx)^{8}$ and im also asked to state the min.value.
I have a theorem in my book that says that it will attain its min. when $a_{n}$ are equal to the fourier coefficients $c_{n}$ of y(x). So by computing $c_{n}$ for $|n|\le4$ I get that: $p(x)=\frac{1}{128}(-28e^{2ix}+14e^{4ix}+35+14e^{-4ix}-28e^{-2ix})$
But how do I now calculate the min. value of the integral, when I plug in my p(x)? I will have the absolute value of some expressions that are not friendly squared. Any help?
Note that in this case \begin{align} |y-p|^2&=(y-p)(\overline{y}-\overline{p})\\ &=|y|^2+|p|^2-2yp \end{align} You can calculate, just like you did for $c_0$, $$ \int_{-\pi}^{\pi}|y|^2 = \int_{-\pi}^{\pi}(\sin x)^{16} \,dx = \frac{6435\pi}{16\,384} $$ By Parseval's formula, we have $$ \int_{-\pi}^{\pi}|p|^2 = 2\pi\sum_{|n|\leq4}|c_n|^2 = \frac{6370\pi}{16\,384} $$ Finally, observe that \begin{align} -2\int_{-\pi}^{\pi}yp &= -4\pi \sum_{|n|\leq4} c_n\frac{1}{2\pi}\int_{-\pi}^{\pi}y(x)e^{inx}\,dx\\ &=-4\pi\sum_{|n|\leq4}c_n^2\\ &=-\frac{12\,740\pi}{16\,384} \end{align} Adding everything, we obtain $$ \int_{-\pi}^{\pi}|y-p|^2 = \frac{65\pi}{16\,384} $$