How to find closest trigonometric polynomial?

Question

Could you give me any hints?

How to find the trigonometric polynomial of the third order closest to the function:

$f(x) \in L_{2\pi}^2, \,f(x) = \text{sgn}(x), \quad x \in (-\pi, \pi)$ in the norm $L_{2\pi}^2$ ?

Answer 1

We seek a trigonometric polynomial of the form \begin{equation} P(x)=\frac{a_0}{2} + a_1\cos(x) + a_2 \cos(2x) + a_3\cos(3x) +b_1\sin(x) +b_2\sin(2x) +b_3\sin(3x) \end{equation} such that minimizes $||f-P||_{L^2_{2\pi}}$.

Now, it's only necessary project the function $f$ in the subspace generated by the orthogonal bases $\{1/2, \cos(x), \cos(2x), \cos(3x), \sin(x), \sin(2x), \sin(3x)\}$, using the scalar product $(\cdot,\cdot)_{L^2_\pi}$. So we have that $$a_0 = \frac{(f,1/2)}{||1/2||_{L^2_{2\pi}}}$$ $$a_1 = \frac{(f,\cos(x))}{||\cos(x)||_{L^2_{2\pi}}}$$ $$b_1 = \frac{(f,\sin(x))}{||\sin(x)||_{L^2_{2\pi}}}$$ $$a_2 = \frac{(f,\cos(2x))}{||\cos(2x)||_{L^2_{2\pi}}}$$ and so on for others coefficients.