So I have to find the following $\min_{a,b,c\in\mathbb{R}}\int_{-1}^{1} |x^3-a-bx-cx^2|^2dx$
I have a hint at a solution which says to consider $X=\{\mbox{polynomials of degree} \leq 2\}$.
So then we have $\min_{a,b,c\in\mathbb{R}}\int_{-1}^{1} |x^3-a-bx-cx^2|^2dx=\inf_{g\in X} ||x^3-g||$ for some $g\in X$
Where the norm $||.||$ is defined using the inner product $
So then I think I'm supposed to use an orthogonal projection somehow I think (and maybe find some orthonormal basis?) but I'm a bit lost as to how to do any of this.
Thanks for any help.
For completeness I am putting my solution (from the answers below).
The problem reduces to finding a basis for $X$ and then orthonarmalizing it. We then use the fact that the orthohgonal projection onto the space $X$ will give the minimal distance to it so we need to calculate: $||x^3-P(x^3)||$ where $P(x^3)=\sum_{i=1}^{3}
Noting that the set $\{1,x,x^2\}$ spans the space $X$ we then apply the gram-schmidt process to this set to give a set of orthogonal vectors: $\{1,x,(x^2-\frac{1}{3})\}$.
This basis now needs to be normalized but if we notice that in :
$\sum_{i=1}^{3}
The first and third terms will cancel as the integrand will be odd, so we only need to normailze the middle vector which gives $\{\sqrt{\frac{3}{2}\}}$.
And so $P(x^3)=
So we now have $||x^3-\frac{3x}{5}||=\sqrt{\int_{-1}^{1} (x^3-\frac{3x}{5})^2dx}=\sqrt{\frac{8}{175}}$
Which is the desired distance