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Newton remainder for a Floater–Hormann window polynomial

Statement

Let $x \notin {x_{i}, \dots, x_{i + d}}$ . For any real data at these nodes and at $x$ , the local polynomial from Local polynomial in a Floater–Hormann window satisfies $f (x) - p_{i} (x) = (\prod_{k = i}^{i + d} (x - x_{k})) f [x_{i}, \dots, x_{i + d}, x] .$

Facts & Assumptions

Given: Distinct real nodes $x_{i}, \dots, x_{i + d}, x$ , real data at them, and the local window polynomial $p_{i}$ .

[F1]

Local polynomial in a Floater–Hormann window and Floater–Hormann window polynomials interpolate their local data: $p_{i}$ has degree at most $d$ and agrees with the data on the window nodes.

[F2]

Divided differences on distinct nodes and Divided differences are symmetric leading coefficients: the divided difference on $d + 2$ nodes is the leading coefficient of their Lagrange interpolant.

[F3]

Cardinal-polynomial interpolation and Existence and uniqueness of the cardinal-polynomial interpolant: for $r \in N$ , after sorting $r + 1$ distinct real nodes and carrying their data along, the cardinal Lagrange sum is the unique polynomial of degree at most $r$ taking the assigned values.

[F4]

The natural numbers $N$ (von Neumann): $N$ contains $0$ and is closed under successor, so $d + 1 \in N$ .

Proof

1.1givenF1F2F3F4construct

By [F4], $d + 1 \in N$ , so sort the combined node/data list increasingly and apply [F3], which supplies the degree-at-most- $(d + 1)$ Lagrange interpolant $q_{i}$ through the $d + 2$ nodes $x_{i}, \dots, x_{i + d}, x$ . By [F2] its leading coefficient is $c = f [x_{i}, \dots, x_{i + d}, x]$ . By [F1], $q_{i} - p_{i}$ has degree at most $d + 1$ , has leading coefficient $c$ , and vanishes at each of the $d + 1$ window nodes.

2.1givenF3F4step 1.1algebra∎

Let $R (t) = q_{i} (t) - p_{i} (t) - c \prod_{k = i}^{i + d} (t - x_{k})$ . The degree- $(d + 1)$ leading terms cancel by step 1.1, so $R$ has degree at most $d$ ; it vanishes at all $d + 1$ window nodes. By [F4], $d \in N$ , and [F3] says the only degree-at-most- $d$ polynomial with these zero values is $0$ . Thus $R = 0$ . Evaluating $q_{i} - p_{i} = c \prod_{k = i}^{i + d} (t - x_{k})$ at $t = x$ , where $q_{i} (x) = f (x)$ , proves the stated remainder.

Depends on

Used by

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Michael S. Floater and Kai Hormann, Barycentric rational interpolation with no poles and high rates of approximation, proof of Theorem 2 (standard reference, not scraped)