Why Newton Coates is exact for polynomial of degree at most $n$

Question

Let us suppose $x_1,...,x_n$ be $n$ nodes and we interpolate the functions $f$ with lagrange polynomial

Then my book says $$\int_a^b f(x) dx=\sum A_i f(x_i), A_i=\int_a^b l_i(x)dx $$where $ l_i(x_j) = \delta_{ij}$

for all polynomial for degree at most n. Certainly this is true for polynomial of degree $n$, because there is unique polynomial of degree n passing through these points, but why it has to be true for polynomial of degree less than $n$?

Answer 1

If $f(x)$ is a polynomial with degree $\leq n$, it coincides with its interpolating polynomial over $n+1$ distinct nodes. So, is this case,

$$ \int_a^b f(x) dx = \int_a^b \mathop{\Large \Sigma}_{i=0}^n f(x_i)l_i(x) dx = \mathop{\Large \Sigma}_{i=0}^n f(x_i) \underbrace{\int_a^bl_i(x) dx}_{A_i} = \mathop{\Large \Sigma}_{i=0}^n A_i f(x_i). $$

If the degree is less than $n$, nothing changes.

Note: for some reason the \sum commando was not working... hence the strange sum symbols.

Answer 2

Note that we can write a (technically not a polynomial) expression for any lower degree polynomial that appears to be of degree $n$. For instance, $$ 0 x^2 + 5x - 3 $$ is a degree one polynomial written as a degree two expression. Then the manipulations you would use to show that Newton-Coates is exact when the leading coefficient is not zero work to show exactness when the leading coefficient is zero. Of course, this can be generalized to more than one leading zero coefficient.

The most extreme version of this is the zero polynomial. Do you believe Newton-Coates is exact for the polynomial $p(x) = 0$? (You can prove this with much less effort than for general polynomials.)