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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?

Slaven Glumac
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  • Taylor polynomials are the limit of Lagrange polynomials as you move your points closer and closer together, and Taylor's theorem gives an explicit bound on the extrapolation error in that case. I would guess that something similar can be said for your problem. – Arthur Jun 14 '19 at 13:50
  • @Arthur Thank you. If you expand the comment with a reference or two, and copy it as an answer, I will accept it. Otherwise, I will type it in on my own. – Slaven Glumac Jun 14 '19 at 14:00
  • I honestly don't think it answers your question, though. It was just an idea thrown out there to see if it would stick. Taylor's theorem doesn't apply directly to Lagrange polynomials in any way, they are just related concepts. And I don't actually know the proof of Taylors theorem, and especially, I don't know it well enough to try to adapt it to Lagrange polynomials myself. – Arthur Jun 14 '19 at 14:03
  • Judging by this question https://math.stackexchange.com/questions/753580/interpolation-and-taylors-theorem you gave a good hint. I just need to do a lot more reading to make sure I completely understand everything. – Slaven Glumac Jun 14 '19 at 14:36

2 Answers2

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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 $x \in [a, b]$.

Slaven Glumac
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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 $x \notin I_s$.

Laurent90
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