Numerical Methods Qualification Exam Problems and Solutions (University of Maryland)/August 2002

All nodes lies in (a,b)
Let $$ \{x_i\}_{i=1}^l \!\, $$ be the nodes that lie in the interval $$(a,b) \!\, $$.

Let $$q_l(x)=\prod_{i=1}^l(x-x_i) \!\, $$ which is a polynomial of degree $$ l\!\, $$.

Let $$p_n(x)=\prod_{i=1}^n(x-x_i)=q_l(x)\prod_{i=1}^{n-l}(x-x_i) \!\, $$ which is a polynomial of degree $$n >l \!\, $$.

Then

$$ \langle p_n, q_l \rangle = \int_a^b q_l^2(x) \underbrace{\prod_{i=1}^{n-l}(x-x_i)}_{r(x)} \neq 0 \!\, $$

since $$r(x) \!\, $$ is of one sign in the interval $$(a,b) \!\, $$ since for $$i=1,2,\ldots n -l \!\, $$, $$x_i \not \in (a,b).\!\, $$

This implies $$q_l \!\, $$ is of degree $$n \!\, $$ since otherwise

$$\langle p_n, q_l \rangle = 0\!\, $$

from the orthogonality of $$p_n \!\, $$.