for what values of $a,b$, $\int_{-1}^{1}((x^{2}+3 x+1)-(a x+b))^{2} \sqrt{1-x^{2}} d x$ is minimal?

Question

Let $V=\mathbb{R}_{\leq 3}[X]$ I need to find $a,b \in \mathbb{R}$ such that the below expression is minimal.

$$\int_{-1}^{1}\left(\left(x^{2}+3 x+1\right)-(a x+b)\right)^{2} \sqrt{1-x^{2}} d x$$

I got a hint to show that $$\langle f(x), g(x)\rangle=\int_{-1}^{1} f(x) g(x) \sqrt{\left(1-x^{2}\right)} d x$$ is an inner product space and so I did but I am not sure how to continue from here

Answer 1

Idea:

You can work in $V' = \Bbb R_{\le 2}[x]$ for the purpose of this question. In the following, we fix the inner product as given in the hint.

1. Take the (ordered) basis $B = (1, x, x^2)$ of $V'$.
2. Using Gram Schmidt (GS), obtain an orthonormal (ordered) basis $B' = (p_0(x), p_1(x), p_2(x))$ of $V'$.
3. Since GS preserves the span of the first $k$ vectors, we see that $\{p_0(x), p_1(x)\}$ is a basis of $\Bbb R_{\le 1}[x]$.
4. Express $x^2 + 3x + 1$ as $$\alpha p_0(x) + \beta p_1(x) + \gamma p_2(x)$$ for appropriate choice of $\alpha, \beta, \gamma \in \Bbb R$.
   (You can find these by taking the inner product of $x^2 + 3x +1$ with the appropriate basis vectors. You don't even need to find $\gamma$.)
5. The desired polynomial $ax + b$ will be given as $$\alpha p_0(x) + \beta p_1(x).$$

Answer 2

Using calculus, the problem does not seem too difficult since $$\left((x^{2}+3 x+1)-(a x+b)\right)^{2}=$$ $$ \left(b-1\right)^2+ 2( a b- a-3 b+3)x+ \left(a^2-6 a-2 b+11\right)x^2+2(3- a)x^3+x^4$$ Now $$I_n=\int_{-1}^1 x^n \sqrt{1-x^2}\,dx$$ $$I_{2n+1}=0 \qquad \text{and} \qquad I_{2n}=\frac{\sqrt{\pi }\, \Gamma \left(n+\frac{1}{2} \right)}{2\, \Gamma (n+2)}$$ making $$I_0=\frac \pi 2\qquad I_2=\frac \pi 8 \qquad I_4=\frac \pi {16}$$

Compute the whole stuff and take derivatives with respect to $a$ and $b$; set them equal to $0$. The solution is simple.