Polynomial Extrapolation Error Resource

Question

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?

Answer 1

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 $x \in [a, b]$.

Answer 2

The proof remains valid in the case of an extrapolation, i.e. when $x$ is outside of the interval $I_s$ delimited by the sampling points: $I_s = [\min(x_0, \ldots, x_n) ; \max(x_0, \ldots, x_n)]$. For the case of extrapolation, the interval $I=[\min(x_0, \ldots, x_n, x) ; \max(x_0, \ldots, x_n,x )]$, which is used in the proof, is such that $I_s \subset I$. Therefore, it is possible that $\xi$ is outside of $I_s$.

The approximation error can be bounded: $$|f(x) - p_n(x)| = \frac{\max\limits_{\xi \in I(x)} |f^{(n+1)}(\xi)|}{(n+1)!}\prod_{i=0}^n |x-x_i|$$ where I have stressed the dependency of $I$ on $x$. For the interpolation case, $\max\limits_{\xi \in I(x)} |f^{(n+1)}(\xi)| = \max\limits_{\xi \in I_d} |f^{(n+1)}(\xi)|$. In the extrapolation case, the error will grow fast due to the term $\prod_{i=0}^n |x-x_i|$, which monotonously increases when $x \notin I_s$.