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Floater–Hormann interpolants have no real poles

Statement

For any $n \geq 1$ , any distinct ordered real nodes $x_{0} < \dots < x_{n}$ , any real data $f_{0}, \dots, f_{n}$ , and any integer $0 \leq d < n$ , the exact Floater–Hormann rational function in Floater–Hormann barycentric interpolant has no real poles.

Facts & Assumptions

Given: Ordered real nodes $x_{0} < \dots < x_{n}$ , real data, and $0 \leq d < n$ .

[F1]

Floater–Hormann weights equal the local-polynomial blend weights: clearing the node factors in the barycentric denominator gives $(- 1)^{n - d} s (x)$ , where $s (x) = \sum_{i = 0}^{n - d} μ_{i} (x)$ and $μ_{i} (x) = \prod_{j < i} (x - x_{j}) \prod_{k > i + d} (x_{k} - x)$ .

[F2]

Floater–Hormann blend partial sums are positive in each node gap: in every open node gap the nonempty partial sums from the three index blocks $I_{1}, I_{2}, I_{3}$ are positive.

Proof

1.1givenF1F2construct

At a node $x = x_{α}$ , a term $μ_{i} (x_{α})$ vanishes if $i > α$ or $i + d < α$ . Otherwise every factor in $μ_{i} (x_{α})$ is strictly positive. The index set $\max (0, α - d) \leq i \leq \min (α, n - d)$ is nonempty, so $s (x_{α}) > 0$ .

1.2givenF1F2algebra

Suppose $x_{α} < x < x_{α + 1}$ . By [F2], every nonempty block sum over $I_{1}, I_{2}, I_{3}$ is positive. Their index sets partition ${0, \dots, n - d}$ , so $s (x) > 0$ throughout each interior node gap.

1.3givenF1algebra

If $x < x_{0}$ , then $μ_{0} > 0$ and the terms in increasing index order alternate in sign with decreasing magnitudes, since $μ_{i + 1} / μ_{i} = (x - x_{i}) / (x_{i + d + 1} - x) < 0$ and $∣ μ_{i + 1} / μ_{i} ∣ = (x_{i} - x) / (x_{i + d + 1} - x) < 1$ . Pairing from the first term gives $s (x) > 0$ . If $x > x_{n}$ , then $μ_{n - d} > 0$ and reading backward gives alternating decreasing magnitudes: $μ_{i - 1} / μ_{i} = (x_{i + d} - x) / (x - x_{i - 1}) < 0$ and $∣ μ_{i - 1} / μ_{i} ∣ = (x - x_{i + d}) / (x - x_{i - 1}) < 1$ . Pairing from the last term again gives $s (x) > 0$ .

2.1F1step 1.1step 1.2step 1.3algebra∎

The cleared polynomial denominator of the barycentric quotient is a nonzero constant multiple of $s$ by [F1]. Steps 1.1–1.3 show that this polynomial has no real zero. Cancelling common polynomial factors cannot create a real denominator zero, so the reduced rational function has no real poles.

Depends on

Used by

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Sources

Michael S. Floater and Kai Hormann, Barycentric rational interpolation with no poles and high rates of approximation, Theorem 1 (standard reference, not scraped)