Error of lagrange interpolation

Question

If the original function I want to approximate using Lagrange interpolation is a polynomial the error function $(x-x_{0})...(x-x_{n})\frac{f^{(n+1)}(\xi)}{(n+1)!}$ is not working because the $n+1$ derivative of $f$ is zero. Do any one know another error formula for Lagrange??

Answer 1

Let $(x_0;y_0)\dots. (x_n;y_n)$ be a set of $n+1$ data points one wants to interpolate. Lagrange interpolation gives the way to build the only one polynomial $L(x)$ of degree $d\leq n$ with : $$L(x_i) = y_i.$$ By using Lagrange interpolation one has : $$L(x) = \sum_{j = 0}^n y_j l_j(x),\quad l_j(x)=\prod_{i = 0, i\neq j}^n \frac{x-x_i}{x_j-x_i}.$$ $l_j(x)$ is a polynomial of degree $n$ with $l_j(x_i)=\delta_{ij}$. It is easy to prove the uniqueness of $L(x)$ with linear algebra.

Let $Q(x)$ be another polynomial of degree $d'\leq n$ verifying : $$Q(x_i) = y_i,$$ then $L-Q$ is a polynomial of degree $d'' \leq n$ vanishing in $n+1$ points. This is possible iff : $$L = Q.$$ So if the function you want to interpolate is a polynomial of degree $\leq n$ then $L(x)$ is exactly this function and the error generated by the interpolation method vanishes and the formula you give for the interpolation error is still correct. In this case, the only error term you may take into account is the numerical error generated by your computer. This is why this method is widely used to interpolate functions behaving like polynomials or more precisely like $exp(x)$.