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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?

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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 $$ \int_{-\pi}^{\pi}|y-p|^2 = \frac{65\pi}{16\,384} $$

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    Which method did you use to integrate $\int_{-\pi}^{\pi} (sinx)^{16} dx $ ? – 2017-02-22
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    Also could you explain why $|y-p|=|y|^{2}+|p|^{2}-2yp$ why is there no absolute values in $yp$? – 2017-02-22
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    @fejz1234 Which method did **you** use to calculate $c_0$, given that it requires the evaluation of $\int_{-\pi}^{\pi}(\sin x)^{8}\,dx$ ? For $|y-p|^2$ (here $|\cdot|$ denotes the complex modulus, not the absolute value) I simply used the fact that for all $z\in\Bbb{C}$ we have $|z|^2=z\overline{z}$ where $\overline{z}$ denotes the complex conjugate of $z$. – 2017-02-22
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    To calculate $c_{0}$ I used the binomial expansion to expand $(sinx)^8$ and then by orthogonality I could find all the coefficients..But I realized now that Parsevals work for $\int_{-\pi}^{\pi} (sinx)^{16}$. – 2017-02-22
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    Okay ah its the complex modulus, but is $y\overline{p}$ the same as $p\overline{y}$ ? And shouldnt we also have the modulus of $2yp$ ? – 2017-02-22
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    @fejz1234 If you mean that you "expand" $(\sin x)^8 = \left(\frac{e^{-ix}-e^{ix}}{2i}\right)^8$ using the binomial theorem, it's OK. Then you should be able to do the same thing for $(\sin x)^{16} = \left((\sin x)^8\right)^2$. A "calculus" approach would be to reduce the powers repeatedly using the half angle formula $(\sin(x))^2 = \frac{1}{2}(1-\cos(2x))$. I don't think Parseval will help you calculating $\| y \|_2^2$ since you would have to obtain *all* Fourier coefficients of $y$ and then sum an infinite series (which is difficult). Parseval is useful for $p$ because the sum is finite. – 2017-02-22
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    @fejz1234 I wrote **in this case** in my answer because here we do have $y\overline{p}=p\overline{y}$. I chose to let you verify this in order to keep my answer succinct. First note that $\overline{p}=p$ because the Fourier coefficients $c_n$ are even, i.e. $c_n=c_{-n}$. Then note that $\overline{y}=y$ because $y$ is real-valued. Hence $y\overline{p}=p\overline{y}=yp$. The point of using the formula $|z|^2=z\overline{z}$ is to get rid of moduli which can be troublesome to manipulate. The fact that we get $|y|^2$ and $|p|^2$ back is that we have the terms $y\overline{y}$ and $p\overline{p}$. – 2017-02-22