# Bounds of Error in Higher Order Lagrange Interpolating Polynomials

Recall from the Higher Order Lagrange Interpolating Polynomials page that if $(x_0, y_0)$, $(x_1, y_1)$, …, $(x_n, y_n)$ are $n + 1$ points where $x_0$, $x_1$, …, $x_n$ are distinct numbers, then the $n^{\mathrm{th}}$ order Lagrange interpolating polynomial is given by the following formula:

(1)The functions $L_0$, $L_1$, …, $L_n$ are given by the formula $L_k(x) = \frac{(x - x_0)(x - x_1)…(x - x_{k-1})(x - x_{k+1})…(x - x_n)}{(x_k - x_0)(x_k - x_1)…(x_k - x_{k-1})(x_k - x_{k+1})…(x_k - x_n)}$ for each $k = 0, 1, …, n$. Of course, if the $n + 1$ points above lie on the graph of $y = f(x)$ for some function $f$, then $P_n(x)$ approximates $f(x)$. Of course, $P_n(x)$ does not equal $f(x)$ and so a remainder/error term exists. The following theorem will tell us what that remainder is.

Theorem 1: Let $f$ be a function that has $n + 1$ continuous derivatives on the interval $[a, b]$, and let $(x_0, y_0)$, $(x_1, y_1)$, …, $(x_n, y_n)$ be $n + 1$ points where $x_0$, $x_1$, …, $x_n$ are distinct numbers. Let $P_n$ be the $n^{\mathrm{th}}$ order Lagrange interpolating polynomial for these points. Then for each $x \in [a, b]$ there exists a $\xi_n$ between $m = \min \{ x_0, x_1, …, x_n, x\}$ and $M = \mathrm{max} \{ x_0, x_1, …, x_n, x\}$ such that $f(x) - P_n(x) = \frac{f^{(n+1)}(\xi_n)}{(n + 1)!} (x - x_0)(x - x_1)…(x - x_n)$. |

Note that we can bound this error formula above by maximizing the error on the interval $[a, b]$, that is find $x \in [a, b]$ such that:

(2)