Approximation using Lagrange Interpolation

Question

I am aware of the formula of this method. However, is it true that the method produces more accurate polynomial when the $x$ points are closer to each other? if so or not, why?

Moreover, If I am given a set of $n$ points i.e. $n_1,n_2,n_3,\cdots, n_n$ and I want to approximate the value of the function of a point laying between $n_1,n_2$ using second order Lagrange polynomial. Does the best interpolating polynomial is the one which brackets this point? i.e the one which takes as input points $n_1,n_2$? if so or not, Why?

Thanks in advance.

Answer 1

After a little research, I have figured out the following:

The error bound for Lagrange polynomial $l(x)$ interpolating the true function is:

$$f(x) - l(x) = \prod_{j=0}^n (x-x_j) \frac{f^{n+1}(\zeta (x))}{(n+1)!}$$

• Now, if points are closer, the term $(x-x_j)$ get smaller leading to a tighter error bound. This answers the first part.

• For the second part, If the polynomial is bracketing the point, then $(x-x_j)$ gets also smaller when substituting $x$ by the point in demand.

I hope my thoughts are correct.

Answer 2

I suppose that the answer depends on what do mean with 'the "best" interpolating polynomial'. If you want to minimize the supremum norm of the polynomial $\prod_{j=0}^n(x-x_j)$ the answer is that you have to take the zeros of the Chebyschev's polynomial of degree $n$ as the interpolation points $x_j$ with $0\leq j \leq n$.

See: https://en.wikipedia.org/wiki/Chebyshev_nodes