# Approximation using Lagrange Interpolation

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?

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.

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 $$0\leq j \leq n$$.