Least square using orthogonal polynomial

Question

I have obtained some orthogonal polynomials, using Gram orthogonal process, and the next question says, using them (O.P.) obtain the least square approximation of second degree for $f(x)=x^{3/2}$ on $[0,1]$.

How can i approximate it when it is already of lower degree than second degree ? Like in Chebyshev case I used $$P_{n-1}=P_n-a_n\bar{T}_n$$

But how do I obtain for the above part, given that I have obtianed orthogonal polynomial as $$P_0=1 $$ $$P_1=x-\frac12$$ $$P_2=x^2-x+\frac16$$

Answer 1

First we construct a set of orthogonal polynomials on the interval $[0,1]$ obtaining with kernel $1$, the polynomials

$$ \cases{ p_0(x) = -1\\ p_1(x) = 2 \sqrt{3} x-\sqrt{3}\\ p_2(x) = -6 \sqrt{5} x^2+6 \sqrt{5} x-\sqrt{5} } $$

after that, with $f(x) = x^{\frac 32}$ we obtain

$$ c_k = \left< f, p_k\right> = \int_0^1 f(x)p_k(x)dx $$

and finally

$$ f(x)\approx \sum_{k=0}^{k=2} c_k p_k(x) $$

or

$$ f(x)\approx -\frac{2 \left(-6 \sqrt{5} x^2+6 \sqrt{5} x-\sqrt{5}\right)}{21 \sqrt{5}}+\frac{6}{35} \sqrt{3} \left(2 \sqrt{3} x-\sqrt{3}\right)+\frac{2}{5} $$

Follows a plot showing in blue $f(x)$ and in red the corresponding approximation