Why the zeros of the orthogonal polynomials are symmetric about the origin if the weight function is even?
t's killing me and every book i've seen is left as an exercise. I'm studying Gauss Legendre.
And also i would like to know why then in the quadrature(the sum of Aj*f(xj), xj the zeros of Pn) the Aj will be the same.
Let $\bigl(p_n(x)\bigr)$ be a sequence of orthogonal polynomials on $[-a,a]$ wrt. an even weight function $w(x)$. Try to prove that $\bigl(p_n(-x)\bigr)$ is also such sequence. This is an immediate consequence of integration by substitution ($t=-x$). Hint: prove that $$\int_{-a}^a p_n(-x)p_m(-x)w(x)\,\text{d}x=0,$$ whenever $n\ne m.$
This concludes the desired assertion. Namely, if $x$ is a zero of $p_n(x)$, then $-x$ is a zero of $p_n(-x)$, but $p_n(-x)=\alpha p_n(x)$ for some $\alpha\ne 0$, since the orthogonal polynomials are unique up to such constant.
Thinking about coefficients of the interpolatory quadrature rule with given nodes (i.e. zeros of orthogonal polynomials, which means that you consider the Gauss quadrature), they are given by some integral formulae, so they are the same for all functions.
