Orthogonal polynomials equations

Question

Let ${Q_n(X)}_{n \in N_0}$ sequence of orthogonal polynomials with respect to the weight function $p(x)$ on interval $(a,b)$ . Let $x_i$, $1 \leq i \leq n$ be zeros of $Q_n$. How to prove that $$ \sum_{i=0}^{n} \frac {Q_i(x_k)}{\|Q_i\|^{2}} \int_{a}^bp(x)Q_i(x)dx=1 $$ for fixed $k$, $1 \leq k\leq n$. My proffesor used for Cristoffel-Gaus numerical integration, but I can't figure out why it is true.

Answer 1

If $\{Q_k\}_{0\leq k\leq n}$ is an orthogonal basis of the space of polynomial of degree equal or less than $n$, then $\left\{\frac{Q_k}{\|Q_k\|}\right\}_{0\leq k\leq n}$ is an orthonormal basis of the same space.

Thus any polynomial $P$ of degree $\leq n$, can be decomposed in that basis: $$P(u)=\sum_{i=0}^n \frac{Q_i(u)}{\|Q_i\|}\int_a^bP(x)\frac{Q_i(x)}{\|Q_i\|}p(x)dx$$ Applying this for $P=1$, and plugging in $u=x_k$ yields the result.