Given $n + 1$ samples of a $n+1$ times continuously differentiable function $f \in C^{k + 1}$:
\begin{equation}
(x_0, f(x_0)), (x_1, f(x_1)), \dots, (x_n, f(x_n))
\end{equation}
Lagrange polynomial is a unique polynomial $p(x)$ interpolating these points:
\begin{equation}
p(x_i) = f(x_i)
\end{equation}
with the interpolation error $x \in [x_0, x_n]$:
\begin{equation}
f(x_i) – p(x_i) = \frac{f^{(n + 1)}(\zeta)}{(n+1)!}\pi_{n+1}(x)
\end{equation}
where $\pi(x) = (x – x_0)\dots(x – x_n)$ and $\zeta \in (a, b)$.
The following implies there exists a constant $C$ for which the interpolation error $x \in [x_0, x_n]$:
\begin{equation}
|p(x_i) – f(x_i)| \leqslant C|x_n – x_0|
\end{equation}
My question is whether the same can be said for the extrapolation error $x \in [x_0, x_n + h]$? I usually find the theorems regarding the interpolation error and few statements that extrapolation is really error prone, but without any lower or upper bounds.
Can you, please, point me to a reference dealing with the error of polynomial extrapolation?
Best Answer
After reviewing the expression for error of the Lagrange interpolation polynomial in:
Suli, Mayers - An Introduction to Numerical Analysis (2003)
I realized I was for some reason to restrictive in the interval. The statement of the theorem says that support abscissas are in the closed interval $x_i \in [a, b]$, the polynomial interpolation error for $x \in [a, b]$ is equal to: \begin{equation} f(x_i) - p(x_i) = \frac{f^{(n + 1)}(\zeta)}{(n+1)!}\pi_{n+1}(x) \end{equation} The proof is also available at proofwiki. I think before, I missed the fact that the function $g$ (defined in the previous link) has $n + 2$ roots where $n + 1$ roots are support abscissas $x_0, \dots, x_n$, and the last one is an arbitrary one $x \in [a, b]$.