Let $V=\mathbb{R}_7[x]$. Then define the inner product$$\langle f,g\rangle =\int \limits _0^1f(x)g(x)\,dx.$$A) Apply the Gram-Schmidt procedure to $\{1,x,x^2\}$
B) For $a,b,c\in \mathbb{R}$, minimize the integral:$$\int \limits _0^1\left (x^3-ax^2-bx-c\right )^2\,dx.$$My attempts
A) Easy, found an orthogonal base, is: $\left \{1,x-\dfrac{1}{2},x^2-x+\dfrac{1}{6}\right \}$.
B) If I am doing the formula for it, I will get:$$\left \langle x^3-ax^2-bx-c,1 \right \rangle$$and like that for $x-\dfrac{1}{2}$ and for $x^2-x+\dfrac{1}{6}$, which seems like a lot of algebra.
Is there any other way to do it?
Its not realistic to do it this way... It will take me over 3 hours just to calculate it all...
Let $t:=t(x):=ax^2+bx+c$, then we want to minimize $\int_0^1(x^3-t(x))^2\,dx$.
Observe that, by definition of the given inner product, it equals to $\langle x^3-t,\, x^3-t\rangle\ =\ \|x^3-t\|^2$ with the induced norm.
So we want to minimize the distance of $x^3$ and $t$ where $t$ runs through all (at most) quadratic polynomials $t\in\Bbb R_2[x]$.
For classical inner product, geometrically it would mean to take the orthogonal projection of the vector $x^3$ to the subspace $\Bbb R_2[x]$.
Since you already have an orthogonal basis of $\Bbb R_2[x]$, you basically just need to take the 3 inner products of $x^3$ with the basis elements and do some algebra to identify $a,b,c$.