So far I've got an orthogonal base with:
$\psi_0 = 1$,
$\psi_1 = x$
$\psi_2 = x^{2}-\frac{2}{6}$
Am I supposed to calculate $p$ as: $\alpha_0\psi_0+\alpha_1\psi_1+\alpha_2\psi_2$ with:
$\alpha_i = \frac{\langle{x^3,\psi_i}\rangle}{\langle{\psi_i,\psi_i}\rangle}$
If so I got that $p = \frac{6}{10} \cdot x$.
If I am doing this completly wrong, can I have hints/a method on how I am supposed to do this?
Thanks!
Your method is correct; the norm $||x^3-p(x)||_2$ is minimized by the orthogonal projection of the function $x^3$ onto the subspace $\cal{P}_2$ of quadratic polynomials:
$$p(x)=\sum_{i=0}^2\frac{\langle{x^3,\psi_i(x)}\rangle}{\langle{\psi_i(x),\psi_i(x)}\rangle}\psi_i(x)\in\cal{P}_2,$$
where $\{\psi_0(x),\psi_1(x),\psi_2(x)\}$ is an orthogonal basis for $\cal{P}_2$.
Fortunately, the computation is not difficult here. We see that $\langle{x^3,\psi_i(x)}\rangle=0$ if $i$ is even, since the integral of an odd function over a symmetric interval is always zero. Furthermore,
$$\langle{x^3,\psi_1(x)}\rangle=\frac{2}{5}\text{ and }\langle{\psi_1(x),\psi_1(x)}\rangle=\frac{2}{3},$$
so your answer is also correct.