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✓ 19 results · all verified · 15 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 4 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Barycentric rational and aaa approximation

1 · Prerequisites

2 · Summary

This page develops barycentric rational quotients, the Floater–Hormann family, finite-sample AAA, and exact Padé coefficient matching. The local proofs establish the denominator's nonvanishing before defining the quotient, Floater–Hormann's no-real-pole property, two supported approximation rates, and the failure of a seminorm-only rate under a uniform-mesh construction.

The AAA procedure is an exact, finite-sample algorithm with deterministic tie and singular-subspace rules, explicit failure and budget statuses, and a smallest-singular-vector least-squares guarantee on successful exact-SVD iterations. It makes no uniform off-grid or minimax claim. The Padé result gives the local Taylor order only when the full normalized moment system is solvable; it gives no pole-location or analytic-continuation guarantee. The prerequisite pages supply the exact SVD procedures, minimax context, and the real mean value theorem used in the bounded-variation result’s optional smoothness consequence. The minimax page is contextual and is not used as an item-level proof premise on this page.

3 · Logical flowchart

4 · Definitions, theorems and proofs

AlgorithmAlgorithm: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Exact Padé coefficient matching with defect statuses

Definition

Input nonnegative integers $m, n$ and coefficients $c_{0}, \dots, c_{m + n} \in C$ of a formal power series or of an analytic function at $0$ . Set $c_{k} = 0$ for negative $k$ . Seek a normalized denominator $q (z) = 1 + \sum_{j = 1}^{n} q_{j} z^{j}$ . Form the $n \times n$ moment system $\sum_{j = 1}^{n} q_{j} c_{m + ℓ - j} = - c_{m + ℓ}, 1 \leq ℓ \leq n .$ Classify it by exact Gaussian elimination on the augmented matrix. Start with pivot row $r = 1$ and scan columns $j = 1, \dots, n$ in order. In column $j$ , if all entries in rows $r$ through $n$ are zero, leave that column free and continue. Otherwise choose the first row $s \geq r$ with a nonzero entry in column $j$ , swap rows $s$ and $r$ , divide row $r$ by its column- $j$ entry, and subtract the resulting row multiple from every other row to make the other entries in column $j$ zero; then increment $r$ . After all columns, a row whose coefficient entries are all zero and whose augmented entry is nonzero makes the system inconsistent, so return status $full-order-inconsistent$ . If the system is consistent but has rank less than $n$ , return status $defective-nonunique$ together with the rank and reduced system; do not select a denominator from its affine solution set. Here rank is the number of pivot rows; consistency and rank are read from the reduced augmented matrix. If the system has rank $n$ , return status $full-order-solvable$ and the unique solution $(q_{1}, \dots, q_{n})$ and set $p_{k} = \sum_{j = 0}^{\min (k, n)} q_{j} c_{k - j}, 0 \leq k \leq m, q_{0} : = 1.$ The output is the coefficient-matching candidate $p / q$ with its full-order status. If $n = 0$ , return status $full-order-solvable$ with the empty denominator system's unique solution $q = 1$ and numerator coefficients $p_{k} = c_{k}$ for $0 \leq k \leq m$ . This procedure claims no pole-location or analytic-continuation guarantee.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Divided differences on distinct nodes

Definition

Let $q \geq 0$ , let $t_{0}, \dots, t_{q}$ be distinct real numbers, and let $f$ be real-valued at these nodes. Define the divided difference recursively by $f [t_{0}] : = f (t_{0})$ and, for $q \geq 1$ , $f [t_{0}, \dots, t_{q}] : = \frac{f [t_{1}, \dots, t_{q}] - f [t_{0}, \dots, t_{q - 1}]}{t_{q} - t_{0}} .$ All sublists in the recursion remain distinct. The denominator is nonzero because $t_{q} \neq t_{0}$ , so this recursion gives a finite real number from the node data.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

A barycentric denominator polynomial with nonzero weights is nonzero

Statement

Let $k \geq 1$ , let $z_{1}, \dots, z_{k} \in C$ be distinct, and let $w_{1}, \dots, w_{k} \in C$ be nonzero. Then the polynomial

Q (z) = \sum_{j = 1}^{k} w_{j} \prod_{ℓ \neq j} (z - z_{ℓ})

is not identically zero.

Facts & Assumptions

Given: A positive integer $k$ , distinct complex nodes $z_{1}, \dots, z_{k}$ , and nonzero complex weights $w_{1}, \dots, w_{k}$ .

Proof

1.1givenchoosealgebra

Fix an index $j \in {1, \dots, k}$ . At $z = z_{j}$ , each summand with index different from $j$ contains the factor $z_{j} - z_{j} = 0$ , so $Q (z_{j}) = w_{j} \prod_{ℓ \neq j} (z_{j} - z_{ℓ}) .$ This is the support-node evaluation of $Q$ .

2.1givenstep 1.1algebra∎

Every factor in the product in step 1.1 is nonzero because the nodes are distinct, and $w_{j} \neq 0$ . Thus $Q (z_{j}) \neq 0$ ; an identically zero polynomial would vanish at $z_{j}$ , so $Q$ is not identically zero.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Barycentric rational quotient with support values and reduced poles

Definition

Let $k \geq 1$ , let $z_{1}, \dots, z_{k} \in C$ be distinct support nodes, let $f_{1}, \dots, f_{k} \in C$ be data, and let each weight $w_{j} \in C$ be nonzero. Define the cleared polynomials $P (z) = \sum_{j = 1}^{k} w_{j} f_{j} \prod_{ℓ \neq j} (z - z_{ℓ}), Q (z) = \sum_{j = 1}^{k} w_{j} \prod_{ℓ \neq j} (z - z_{ℓ}) .$ The barycentric rational quotient is the rational function $r = P / Q$ . Away from the support nodes this equals $r (z) = \frac{\sum_{j = 1}^{k} w_{j} f_{j} / (z - z_{j})}{\sum_{j = 1}^{k} w_{j} / (z - z_{j})} .$ At a support node its value is given by the cleared quotient $P (z_{j}) / Q (z_{j})$ . A pole means a zero of the denominator after all common polynomial factors of numerator and denominator have been canceled.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Divided differences are symmetric leading coefficients

Statement

For distinct real nodes $t_{0}, \dots, t_{q}$ and assigned values $f (t_{j})$ , $f [t_{0}, \dots, t_{q}] = \sum_{j = 0}^{q} \frac{f (t_{j})}{\prod_{\begin{array}{c} 0 \leq k \leq q \\ k \neq j \end{array}} (t_{j} - t_{k})} .$ Consequently the divided difference is symmetric in its nodes. It is the coefficient of $t^{q}$ in the unique polynomial of degree at most $q$ that takes the assigned values at those nodes.

Facts & Assumptions

Given: $q \in N$ , distinct real nodes $t_{0}, \dots, t_{q}$ , and real data $f (t_{0}), \dots, f (t_{q})$ .

[F1]

Divided differences on distinct nodes: the order-zero value is $f [t_{0}] = f (t_{0})$ and the recursive rule is $f [t_{0}, \dots, t_{q}] = (f [t_{1}, \dots, t_{q}] - f [t_{0}, \dots, t_{q - 1}]) / (t_{q} - t_{0})$ for $q \geq 1$ .

[F2]

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

[F3]

The natural numbers $N$ (von Neumann): $0 \in N$ , so the order-zero case is included.

[F4]

The principle of mathematical induction: a property true at $0$ and preserved by successor holds for every natural number.

Proof

1.1givenF2constructalgebra

For each $j$ define $L_{j} (t) = \prod_{k \neq j} (t - t_{k}) / (t_{j} - t_{k})$ . Its denominators are nonzero, and $L_{j} (t_{k})$ equals $1$ when $k = j$ and $0$ otherwise. Sort the nodes increasingly, carrying their data along. Reindexing the finite sum and products does not change them, so [F2] applies and $P (t) = \sum_{j = 0}^{q} f (t_{j}) L_{j} (t)$ is the unique polynomial of degree at most $q$ taking the assigned values.

2.1step 1.1algebra

The coefficient of $t^{q}$ in $P$ is the displayed sum, because the leading coefficient of $L_{j}$ is $1 / \prod_{k \neq j} (t_{j} - t_{k})$ . The sum is unchanged by any permutation of the nodes, since the permutation only reorders its terms and their symmetric denominators.

3.1F1F3F4step 2.1algebrainduction∎

We prove that the recursively defined divided difference equals this leading coefficient by induction on $q$ using [F4]. Since $0 \in N$ by [F3], for $q = 0$ both are $f (t_{0})$ . For $q \geq 1$ , substitute the induction formula for the two order- $(q - 1)$ differences in [F1] and divide by $t_{q} - t_{0}$ . For an interior index $0 < j < q$ , the resulting coefficient of $f (t_{j})$ is $\frac{1}{t_{q} - t_{0}} (\frac{1}{t_{j} - t_{q}} - \frac{1}{t_{j} - t_{0}}) \frac{1}{\prod_{\begin{array}{c} 1 \leq k \leq q - 1 \\ k \neq j \end{array}} (t_{j} - t_{k})} = \frac{1}{\prod_{k \neq j} (t_{j} - t_{k})} .$ For $j = 0$ , only the second recursive term contributes and its minus sign supplies the missing factor $t_{0} - t_{q}$ ; for $j = q$ , only the first term contributes and division supplies $t_{q} - t_{0}$ . Their coefficients are also $1 / \prod_{k \neq j} (t_{j} - t_{k})$ . This is exactly the coefficient in step 2.1.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

A solvable Padé moment system gives the stated Taylor order

Statement

Let $m, n \geq 0$ , $R > 0$ , and $f (z) = \sum_{k = 0}^{\infty} c_{k} z^{k}$ for $∣ z ∣ < R$ , with the series convergent there. Suppose $q_{0} = 1, q_{1}, \dots, q_{n}$ solve the full normalized Padé moment equations of Exact Padé coefficient matching with defect statuses. Define $p_{k} = \sum_{j = 0}^{\min (k, n)} q_{j} c_{k - j}$ for $0 \leq k \leq m$ , and $p (z) = \sum_{k = 0}^{m} p_{k} z^{k}$ , $q (z) = \sum_{j = 0}^{n} q_{j} z^{j}$ . Then

f (z) - \frac{p (z)}{q (z)} = O (z^{m + n + 1}) as z \to 0.

This is a local Taylor-matching statement only; it gives no pole-location claim and no guarantee about analytic continuation beyond the neighborhood where $f$ is given.

Here $h (z) = O (z^{M})$ as $z \to 0$ means there are constants $C, δ > 0$ such that $∣ h (z) ∣ \leq C ∣ z ∣^{M}$ whenever $0 < ∣ z ∣ < δ$ .

Facts & Assumptions

Given: Integers $m, n \geq 0$ , a radius $R > 0$ , a convergent power series $f (z) = \sum_{k = 0}^{\infty} c_{k} z^{k}$ for $∣ z ∣ < R$ , and a normalized denominator solution $q_{0} = 1, q_{1}, \dots, q_{n}$ .

[F1]

Exact Padé coefficient matching with defect statuses: the moment equations make coefficients of $q f$ vanish in degrees $m + 1, \dots, m + n$ , while $p_{k}$ equals the coefficient of $q f$ for $0 \leq k \leq m$ .

[A1]

Since $q (0) = q_{0} = 1$ , continuity gives a neighborhood of $0$ on which $∣ q (z) ∣ \geq 1 / 2$ , so $1 / q$ is bounded there.

Proof

1.1givenF1algebra

Multiplication of the Taylor series of $f$ by $q$ gives coefficient $\sum_{j = 0}^{\min (k, n)} q_{j} c_{k - j}$ at degree $k$ . By [F1], for each $0 \leq k \leq m$ this coefficient is $p_{k}$ , so the coefficients of $q f - p$ through degree $m$ vanish.

2.1givenF1step 1.1algebra

For each $m < k \leq m + n$ , write $k = m + ℓ$ with $1 \leq ℓ \leq n$ . The moment equation in [F1] says $\sum_{j = 1}^{n} q_{j} c_{k - j} = - c_{k}$ , exactly the assertion that the degree- $k$ coefficient of $q f$ is zero, because $q_{0} = 1$ . Therefore every coefficient of $q f - p$ through degree $m + n$ vanishes.

3.1givenA1step 2.1algebra∎

Analyticity implies $q f - p$ has a Taylor expansion whose first possibly nonzero term has degree at least $m + n + 1$ , so $q f - p = O (z^{m + n + 1})$ . By [A1], divide by the nonvanishing $q$ near zero to obtain the asserted local order for $f - p / q$ . This proof uses only the solvable coefficient equations; it makes no claim about zeros of $q$ away from $0$ .

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Floater–Hormann barycentric interpolant

Definition

Let $n \geq 1$ , let $x_{0} < \dots < x_{n}$ be distinct real nodes, let $f_{0}, \dots, f_{n} \in R$ be data, and let $d$ be an integer with $0 \leq d < n$ . For $j = 0, \dots, n$ , define $w_{j} = (- 1)^{j - d} \sum_{i = \max (0, j - d)}^{\min (j, n - d)} \prod_{\begin{array}{c} i \leq k \leq i + d \\ k \neq j \end{array}} ∣ x_{j} - x_{k} ∣^{- 1} .$ The index range is nonempty and every product is positive, so every $w_{j}$ is nonzero. The Floater–Hormann interpolant of degree parameter $d$ is the barycentric rational quotient of these nodes, data, and weights from Barycentric rational quotient with support values and reduced poles. Away from the nodes, exact evaluation uses the quotient with this denominator; a zero of the reduced denominator is reported as pole status. The no-real-pole theorem below proves that this status never occurs for admissible exact real inputs.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Adjacent-window divided-difference recurrence

Statement

Let $n \in N$ , $n \geq 1$ , and $d \in N$ with $d < n$ ; let $x_{0} < \dots < x_{n}$ be real nodes, let $x \in R$ be distinct from these nodes, and let $f : {x_{0}, \dots, x_{n}, x} \to R$ . For every $0 \leq i \leq n - d$ , put $G_{i} : = f [x_{i}, \dots, x_{i + d}, x] .$ Then, for every $0 \leq i < n - d$ , $G_{i} - G_{i + 1} = - (x_{i + d + 1} - x_{i}) f [x_{i}, \dots, x_{i + d + 1}, x] .$

Facts & Assumptions

Given: $n \in N$ , $n \geq 1$ , $d \in N$ with $d < n$ , ordered distinct real nodes $x_{0}, \dots, x_{n}$ , an additional distinct real point $x$ , and real-valued data $f$ on these points.

[F1]

Divided differences on distinct nodes: the divided difference satisfies the recursive endpoint rule.

[F2]

Divided differences are symmetric leading coefficients: divided differences are symmetric in all listed nodes.

[F3]

The natural numbers $N$ (von Neumann): $N$ contains $0$ and is closed under successor, so $d + 2 \in N$ .

Proof

1.1F1F2F3algebra

Apply the recursive endpoint rule [F1] to the order- $(d + 2)$ divided difference on the ordered endpoints $x_{i}, x_{i + d + 1}$ and the intermediate list $x_{i + 1}, \dots, x_{i + d}, x$ . Symmetry [F2] identifies its two numerator terms as $G_{i + 1}$ and $G_{i}$ , respectively, so $f [x_{i}, \dots, x_{i + d + 1}, x] = \frac{G_{i + 1} - G_{i}}{x_{i + d + 1} - x_{i}} .$

2.1givenstep 1.1algebra∎

The denominator in step 1.1 is positive because the nodes are strictly ordered. Rearranging that identity gives the stated identity.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Rolle bound for a divided difference

Statement

Let $a \leq b$ , $q \in N$ , let $f \in C^{q} ([a, b])$ , and let $t_{0}, \dots, t_{q}$ be distinct points of $[a, b]$ . Then

∣ f [t_{0}, \dots, t_{q}] ∣ \leq \frac{∥ f^{(q)} ∥_{\infty, [a, b]}}{q!},

where $0! : = 1$ , $f^{(0)} : = f$ , and $∥ g ∥_{\infty, [a, b]} : = \sup_{x \in [a, b]} ∣ g (x) ∣$ .

Facts & Assumptions

Given: $a \leq b$ , $q \in N$ , $f \in C^{q} ([a, b])$ , and distinct nodes $t_{0}, \dots, t_{q} \in [a, b]$ .

[F1]

Divided differences on distinct nodes and Divided differences are symmetric leading coefficients: the divided difference is the leading coefficient of the degree-at-most- $q$ Lagrange interpolant through the node data.

[F2]

Higher derivatives and the classes $C^{k}$ and $C^{\infty}$ : $f \in C^{q} ([a, b])$ means $f^{(j)}$ exists and is continuous on $[a, b]$ for every $0 \leq j \leq q$ , with one-sided endpoint derivatives.

[F3]

Rolle's theorem: if $a < b$ , $f$ is continuous on $[a, b]$ , differentiable at every point of $(a, b)$ , and $f (a) = f (b)$ , then $f^{'} (c) = 0$ for some $c \in (a, b)$ : a function continuous on a closed interval, differentiable on its interior, and equal at the endpoints has an interior point where its derivative is zero.

[F4]

The natural numbers $N$ (von Neumann): $0 \in N$ , so the order-zero case is part of the stated range.

[F5]

Extreme value theorem: a continuous real function on a nonempty compact subset of $R$ attains a greatest and a least value: a continuous real function on a nonempty compact subset of $R$ has a finite attained maximum and minimum.

Proof

1.1givenF1F4F5

If $q = 0$ , which is allowed by [F4], then $f [t_{0}] = f (t_{0})$ , so the claim is $∣ f (t_{0}) ∣ \leq ∥ f ∥_{\infty, [a, b]}$ . The norm is finite by [F5].

2.1givenF1F2F3F5algebra∎

Suppose $q \geq 1$ , and let $P$ be the Lagrange interpolant in [F1], with leading coefficient $c = f [t_{0}, \dots, t_{q}]$ . The function $g = f - P$ has a zero at each of the $q + 1$ distinct ordered nodes. Order those nodes increasingly and apply [F3] on each consecutive interval to obtain $q$ distinct zeros of $g^{'}$ . Applying [F3] between consecutive zeros of $g^{'}$ gives $q - 1$ distinct zeros of $g^{''}$ ; continuing, after $r$ rounds $g^{(r)}$ has $q + 1 - r$ distinct zeros. Because $f \in C^{q}$ by [F2] and $P$ is a polynomial, each derivative through order $q$ is continuous and its predecessor is differentiable, so every application satisfies [F3]. After $q$ rounds one zero $ξ$ of $g^{(q)}$ remains. Since $P$ has degree at most $q$ and leading coefficient $c$ , its $q$ th derivative is the constant $q! c$ . Hence $f^{(q)} (ξ) = q! c$ , and the claimed bound follows by taking absolute values. The derivative norm is finite by [F5].

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Hermite–Genocchi simplex formula for divided differences

Statement

Let $q \in N$ , let $f \in C^{q} ([a, b])$ , and let $t_{0}, \dots, t_{q} \in [a, b]$ be distinct real nodes. For $q \geq 1$ , define $Δ_{q} = {(u_{1}, \dots, u_{q}) \in [0, \infty)^{q} : u_{1} + \dots + u_{q} \leq 1}, u_{0} = 1 - \sum_{j = 1}^{q} u_{j} .$ Then $f [t_{0}, \dots, t_{q}] = \int_{Δ_{q}} f^{(q)} ⁣ (\sum_{j = 0}^{q} u_{j} t_{j}) d u_{1} \dots d u_{q}, vol (Δ_{q}) = \frac{1}{q!} .$ For $q = 0$ , the integral is by convention evaluation at $t_{0}$ , and $vol (Δ_{0}) : = 1$ . For $q \geq 1$ , the integral over $Δ_{q}$ is the iterated Riemann integral with $0 \leq u_{1} \leq 1$ and, successively, $0 \leq u_{j} \leq 1 - \sum_{k = 1}^{j - 1} u_{k}$ for $2 \leq j \leq q$ .

Facts & Assumptions

Given: $q \in N$ , $f \in C^{q} ([a, b])$ , and distinct nodes $t_{0}, \dots, t_{q} \in [a, b]$ .

[F1]

Divided differences on distinct nodes: divided differences satisfy the recursive endpoint rule.

[F2]

Divided differences are symmetric leading coefficients: divided differences are symmetric in their nodes.

[F3]

Higher derivatives and the classes $C^{k}$ and $C^{\infty}$ : $f \in C^{q}$ means its derivatives through order $q$ exist and are continuous, including their one-sided endpoint extensions.

[F4]

Newton–Leibniz needs only continuity on $[a, b]$ , differentiability on $(a, b)$ , and a Riemann-integrable extension of the interior derivative: for a continuous function on a closed interval whose interior derivative has a Riemann-integrable extension, the integral of that extension is the endpoint increment.

[F5]

The natural numbers $N$ (von Neumann): $0 \in N$ , so the order-zero base case is included.

[F6]

The principle of mathematical induction: a property true at $0$ and preserved by successor holds for every natural number.

Proof

1.1givenF1F5

For $q = 0$ , which is allowed by [F5], the integral convention gives $f (t_{0}) = f [t_{0}]$ and the zero-dimensional simplex has volume one.

1.2F1F2F3F4F6inductionalgebra

By [F6], use induction on $q \in N$ . Suppose the formula is established in order $q - 1$ , where $q \geq 1$ . In the recursive rule [F1], express the two order- $(q - 1)$ divided differences by the induction hypothesis. The bottom face $u_{q} = 0$ has argument $t_{0} + \sum_{j = 1}^{q - 1} u_{j} (t_{j} - t_{0})$ and gives $f [t_{0}, \dots, t_{q - 1}]$ . On the top face $u_{q} = 1 - \sum_{j = 1}^{q - 1} u_{j}$ , the same argument is $t_{q} + \sum_{j = 1}^{q - 1} u_{j} (t_{j} - t_{q})$ and gives $f [t_{1}, \dots, t_{q}]$ by symmetry [F2]. For fixed $u_{1}, \dots, u_{q - 1}$ with sum less than $1$ , the derivative with respect to $u_{q}$ of $f^{(q - 1)} ⁣ (t_{0} + \sum_{j = 1}^{q} u_{j} (t_{j} - t_{0}))$ is $(t_{q} - t_{0}) f^{(q)} (\sum_{j = 0}^{q} u_{j} t_{j})$ . The derivative is continuous by [F3], hence Riemann integrable. Applying [F4] on the $u_{q}$ interval, followed by integration over the other coordinates, shows that the difference of the two face integrals is $(t_{q} - t_{0})$ times the $q$ -dimensional integral in the statement. When the outer coordinates sum to $1$ , the inner $u_{q}$ interval is $[0, 0]$ and its integral is zero; for $q = 1$ there are no outer coordinates and the interval is $[0, 1]$ . Division by $t_{q} - t_{0} \neq 0$ gives the order- $q$ formula.

2.1step 1.1F4F6inductionalgebra∎

By [F6], prove the volume formula by induction, with its $q = 0$ case from step 1.1. For $q \geq 1$ , fix the outer coordinate $u_{1} \in [0, 1]$ . The remaining coordinates form a $(q - 1)$ -simplex scaled by $1 - u_{1}$ . If $q = 1$ , its zero-dimensional volume is $1$ ; if $q \geq 2$ and $u_{1} < 1$ , the dilation to $Δ_{q - 1}$ has Jacobian $(1 - u_{1})^{q - 1}$ , while the section at $u_{1} = 1$ has $(q - 1)$ -volume zero. Thus, by induction, the section volume is $(1 - u_{1})^{q - 1} / (q - 1)!$ . The function $H (u) = - (1 - u)^{q} / q!$ has derivative $(1 - u)^{q - 1} / (q - 1)!$ ; applying [F4] to $H$ on $[0, 1]$ gives $H (1) - H (0) = 1 / q!$ . $vol (Δ_{q}) = \int_{0}^{1} \frac{(1 - u_{1})^{q - 1}}{(q - 1)!} d u_{1} = \frac{1}{q!} .$ The induction and the divided-difference identity prove the statement.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Barycentric rational quotients are defined and interpolate at support nodes

Statement

For the data and nonzero weights in Barycentric rational quotient with support values and reduced poles, $Q$ is not the zero polynomial, the rational function $P / Q$ is well-defined, and $r (z_{j}) = f_{j}$ for each support node $z_{j}$ .

Facts & Assumptions

Given: Distinct complex nodes $z_{1}, \dots, z_{k}$ , complex data $f_{j}$ , and nonzero complex weights $w_{j}$ .

[F1]

Barycentric rational quotient with support values and reduced poles: $P (z) = \sum_{j} w_{j} f_{j} \prod_{ℓ \neq j} (z - z_{ℓ})$ and $Q (z) = \sum_{j} w_{j} \prod_{ℓ \neq j} (z - z_{ℓ})$ .

[F2]

A barycentric denominator polynomial with nonzero weights is nonzero: for distinct nodes and nonzero weights, $Q (z_{j}) = w_{j} \prod_{ℓ \neq j} (z_{j} - z_{ℓ}) \neq 0$ at each support node, so $Q$ is not the zero polynomial.

Proof

1.1givenF1F2

By [F2], $Q$ is not the zero polynomial. The definition [F1] therefore forms a rational function $P / Q$ for these data.

2.1F1F2step 1.1algebra∎

At $z_{j}$ , all summands of $P$ and $Q$ except the $j$ th vanish. Thus $P (z_{j}) = w_{j} f_{j} \prod_{ℓ \neq j} (z_{j} - z_{ℓ}) = f_{j} Q (z_{j})$ , and [F2] gives $Q (z_{j}) \neq 0$ . Hence $r (z_{j}) = f_{j}$ . Canceling a common polynomial factor cannot introduce a zero into the reduced denominator at $z_{j}$ , since the original $Q (z_{j})$ is nonzero. This proves the support convention.

AlgorithmAlgorithm: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Finite-sample AAA rational approximation with exact SVD

Definition

Input a nonempty ordered list of distinct sample points $Z = (z_{1}, \dots, z_{N}) \subset C$ , data $F = (f_{1}, \dots, f_{N}) \in C^{N}$ , an absolute tolerance $τ \geq 0$ , an integer support budget $B$ with $0 \leq B \leq ⌊ N / 2 ⌋$ , and a finite integer bidiagonal-QR sweep budget $S \geq 0$ . The procedure is evaluated in exact arithmetic; its finite-precision implementation makes no optimality claim. For $v \in C^{k}$ , use the standard Hermitian inner product and Euclidean norm $∥ v ∥_{2} = (\sum_{j = 1}^{k} ∣ v_{j} ∣^{2})^{1 / 2}$ . Initialize the accepted support-record list $S_{a c c} = \emptyset$ ; each record stores an original index, its sample point, and its data value. Set the remaining original-index list $R = (1, \dots, N)$ . Every return has form $(r_{a c c}, S_{a c c}, status, R)$ , where $R$ is the residual record maintained below.

Set the initial rational candidate $r_{0}$ to the constant mean $\overset{ˉ}{f} = N^{- 1} \sum_{i} f_{i}$ . Compute $η_{0} = \max_{i} ∣ f_{i} - \overset{ˉ}{f} ∣$ , set the last accepted candidate $r_{a c c} = r_{0}$ , and set $R = ((0, η_{0}))$ . If $η_{0} \leq τ$ , return $(r_{a c c}, S_{a c c}, tolerance-met, R)$ .
For $m = 1, \dots, B$ , choose from $R$ the least original index $s$ that maximizes the absolute residual from $r_{a c c}$ . Let $S_{p r o v}$ be $S_{a c c}$ with $(s, z_{s}, f_{s})$ appended; remove $s$ from $R$ .
Form the $(N - m) \times m$ Loewner matrix $L_{i j} = (f_{i} - f_{s_{j}}) / (z_{i} - z_{s_{j}})$ , where $i$ ranges over the indices in $R$ and $s_{j}$ are the original indices in $S_{p r o v}$ . The denominator is nonzero because all sample points are distinct.
Compute an exact full singular value decomposition (SVD) of $L$ by applying Householder bidiagonalization and its finite precision status followed by Finite budget implicit shifted QR for bidiagonal singular values with zero deflation tolerance and budget $S$ . If either call returns failure or the QR budget is exhausted, return $(r_{a c c}, S_{a c c}, inconclusive, R)$ .
Let $E$ be the right singular subspace for the smallest singular value of $L$ , and let $P_{E}$ be its orthogonal projection. Choose the least coordinate index $j$ for which $P_{E} e_{j} \neq 0$ and set $w = P_{E} e_{j} / ∥ P_{E} e_{j} ∥_{2}$ . This rule applies at every multiplicity, including dimension one. Multiply by the unique unit scalar that makes the first nonzero coordinate of $w$ positive real. If any coordinate of $w$ is zero, reject this provisional support and return $(r_{a c c}, S_{a c c}, zero-support-weight, R)$ .
Form the barycentric rational quotient using the nodes and data in $S_{p r o v}$ and $w$ as its weights. If this rational function has a genuine pole at any remaining sample point, reject this provisional support and return $(r_{a c c}, S_{a c c}, pole-status, R)$ . Otherwise accept the candidate by setting $r_{a c c}$ to it and $S_{a c c} = S_{p r o v}$ , and append $(m, η_{m}, σ_{m})$ , where $η_{m}$ is its maximum sampled residual and $σ_{m} = ∥ L w ∥_{2}$ , to $R$ . If $η_{m} \leq τ$ , return $(r_{a c c}, S_{a c c}, tolerance-met, R)$ .
If all $B$ iterations finish without another return, return $(r_{a c c}, S_{a c c}, support-budget-exhausted, R)$ . In particular $B = 0$ returns the initial constant and empty support list with this status when step 1 does not return.

The support tie rule and the projection rule make the exact procedure deterministic, including when the smallest singular value has multiplicity greater than one. Rank deficiency is permitted: if the smallest singular value is zero and all selected weights are nonzero and the quotient is finite at the remaining sample points, the candidate is accepted and its exact sample residual is evaluated. The algorithm asserts no uniform off-grid error bound and no minimax optimality. The Loewner residual $σ_{m}$ is a linear least-squares residual, not the nonlinear sampled residual $η_{m}$ .

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Local mesh ratio for degree-zero Floater–Hormann interpolation

Definition

For ordered nodes $x_{0} < \dots < x_{n}$ , $n \geq 1$ , set $β : = 0$ if $n \leq 2$ . If $n \geq 3$ , define the local mesh ratio $β = \max_{1 \leq α \leq n - 2} \min {\frac{x_{α + 1} - x_{α}}{x_{α} - x_{α - 1}}, \frac{x_{α + 1} - x_{α}}{x_{α + 2} - x_{α + 1}}} .$ All denominators are positive. This parameter is used only for the degree-zero error bound.

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Local polynomial in a Floater–Hormann window

Definition

For the nodes, data, and degree parameter of Floater–Hormann barycentric interpolant, let $0 \leq i \leq n - d$ . The local window polynomial is $p_{i} (t) = \sum_{k = i}^{i + d} f_{k} \prod_{\begin{array}{c} i \leq j \leq i + d \\ j \neq k \end{array}} \frac{t - x_{j}}{x_{k} - x_{j}} .$ All denominators are nonzero because the nodes are distinct. Empty products equal one, in particular when $d = 0$ .

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Bounded variation controls adjacent Hermite–Genocchi windows

Statement

Let $n \geq 1$ , let $0 \leq d < n$ , let $x_{0} < \dots < x_{n}$ be real nodes, and fix a real query point $x \in [x_{0}, x_{n}]$ distinct from the nodes. Let $f \in C^{d + 1} ([x_{0}, x_{n}])$ , put $g = f^{(d + 1)}$ , and suppose $g$ has finite total variation $V = {Var}_{[x_{0}, x_{n}]} (g)$ as defined in Bounded variation and total variation on an interval. For $0 \leq i \leq n - d$ , put $G_{i} = f [x_{i}, \dots, x_{i + d}, x]$ . Then $\sum_{i = 0}^{n - d - 1} ∣ G_{i + 1} - G_{i} ∣ \leq \frac{V}{(d + 1)!} .$

Facts & Assumptions

Given: $n \geq 1$ , $0 \leq d < n$ , ordered real nodes $x_{0} < \dots < x_{n}$ , a distinct query point $x$ in their interval, $f \in C^{d + 1}$ , and $Var (f^{(d + 1)}) = V < \infty$ .

[F1]

Hermite–Genocchi simplex formula for divided differences: an order- $(d + 1)$ divided difference is the integral of $f^{(d + 1)}$ over the standard simplex, whose volume is $1 / (d + 1)!$ .

[F2]

Bounded variation and total variation on an interval: $V$ is the supremum, over finite partitions of the interval, of the sums of absolute successive increments of $g$ .

[F3]

Higher derivatives and the classes $C^{k}$ and $C^{\infty}$ : $C^{d + 1}$ means that the derivatives through order $d + 1$ exist and are continuous on the closed interval.

[F4]

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

Proof

1.1givenF1F3F4construct

By [F4], $d + 1 \in N$ , so apply [F1] to each $G_{i}$ , using the same simplex coordinates for each window and using the final coordinate for the fixed point $x$ . Thus $G_{i} = \int_{Δ_{d + 1}} g (y_{i} (u)) d u, y_{i} (u) = u_{x} x + \sum_{k = 0}^{d} u_{k} x_{i + k},$ where all coefficients are nonnegative and sum to one. By [F3], $g$ is continuous on the closed interval. Therefore every $y_{i} (u)$ lies in $[x_{0}, x_{n}]$ .

2.1givenF1F2step 1.1algebra∎

For each fixed $u$ , the sequence $y_{0} (u), y_{1} (u), \dots, y_{n - d} (u)$ is nondecreasing: each summand $x_{i + k}$ is nondecreasing in $i$ and has coefficient $u_{k} \geq 0$ . Remove repeated values from this finite sequence and insert the interval endpoints; the definition [F2] then gives $\sum_{i = 0}^{n - d - 1} ∣ g (y_{i + 1} (u)) - g (y_{i} (u)) ∣ \leq V .$ Using [F1], the triangle inequality for integrals, and this pointwise bound, $\sum_{i} ∣ G_{i + 1} - G_{i} ∣ \leq \int_{Δ_{d + 1}} \sum_{i} ∣ g (y_{i + 1} (u)) - g (y_{i} (u)) ∣ d u \leq V vol (Δ_{d + 1}) = \frac{V}{(d + 1)!} .$ This is the claimed total variation bound.

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A global mesh ratio bounds the local degree-zero ratio

Statement

Let $x_{0} < \dots < x_{n}$ , $n \geq 1$ , put $h = \max_{0 \leq j < n} (x_{j + 1} - x_{j})$ and $h_{\min} = \min_{0 \leq j < n} (x_{j + 1} - x_{j})$ , and let $ρ = h / h_{\min}$ . The local ratio $β$ in Local mesh ratio for degree-zero Floater–Hormann interpolation satisfies $0 \leq β \leq ρ$ .

Facts & Assumptions

Given: Ordered distinct nodes, their largest and smallest adjacent gaps $h, h_{\min}$ , and the local ratio $β$ .

[F1]

Local mesh ratio for degree-zero Floater–Hormann interpolation: for $n \geq 3$ , $β$ is the maximum over interior indices of the smaller of two adjacent-gap ratios; for $n \leq 2$ , $β = 0$ .

Proof

1.1givenF1algebra

Since every adjacent gap is at most $h$ and at least $h_{\min} > 0$ , each ratio appearing in [F1] is positive and at most $h / h_{\min} = ρ$ .

2.1givenF1step 1.1algebra∎

By step 1.1, their minimum is in $[0, ρ]$ , and the maximum of these minima also lies in $[0, ρ]$ . When $n \leq 2$ , $β = 0 \leq ρ$ .

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Floater–Hormann window polynomials interpolate their local data

Statement

For each window $i$ , the polynomial $p_{i}$ in Local polynomial in a Floater–Hormann window has degree at most $d$ and satisfies $p_{i} (x_{k}) = f_{k}$ for every $i \leq k \leq i + d$ . It is the unique polynomial of degree at most $d$ with those values. In particular, when all local data equal $1$ , the local Lagrange basis functions sum to the constant polynomial $1$ .

Facts & Assumptions

Given: Distinct real nodes and data as in the Floater–Hormann definition, an integer $0 \leq d < n$ , and a window index $0 \leq i \leq n - d$ .

[F1]

Local polynomial in a Floater–Hormann window: $p_{i}$ is the sum of the local Lagrange terms $f_{k} \prod_{j \neq k} (t - x_{j}) / (x_{k} - x_{j})$ .

[F2]

Cardinal-polynomial interpolation and Existence and uniqueness of the cardinal-polynomial interpolant: the displayed local Lagrange sum is the unique polynomial of degree at most $d$ taking the data on the $d + 1$ window nodes.

[F3]

The natural numbers $N$ (von Neumann): every nonnegative integer $d$ belongs to $N$ .

Proof

1.1givenF1algebra

Each summand in [F1] is a polynomial of degree at most $d$ , so $p_{i}$ has degree at most $d$ . At a local node $x_{ℓ}$ , every summand with $k \neq ℓ$ contains the zero factor $x_{ℓ} - x_{ℓ}$ , while the $k = ℓ$ product equals one. Therefore $p_{i} (x_{ℓ}) = f_{ℓ}$ .

2.1givenF2F3step 1.1algebra∎

By [F3], $d \in N$ , so [F2] applies and $p_{i}$ is the unique polynomial of degree at most $d$ with these window values. The constant polynomial $1$ has the value $1$ at each local node; uniqueness makes its Lagrange interpolant equal to $1$ .

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AAA weights minimize the Loewner residual and interpolate support data

Statement

At an AAA iteration whose exact SVD call completes successfully, with Loewner matrix $L \in C^{(N - m) \times m}$ , the selected unit weight vector $w$ minimizes $∥ L v ∥_{2}$ among all $v \in C^{m}$ with $∥ v ∥_{2} = 1$ ; its residual is the smallest singular value of $L$ . If every selected weight is nonzero and the iteration does not return pole status, its barycentric candidate interpolates all selected support data. The finite support budget makes the procedure terminate after at most $B$ iterations. These claims concern finite sampled data and do not assert minimax optimality or uniform off-grid convergence.

Facts & Assumptions

Given: A successful exact-SVD iteration of Finite-sample AAA rational approximation with exact SVD, with Loewner matrix $L$ and its selected right singular vector.

[F1]

Finite-sample AAA rational approximation with exact SVD: the algorithm selects a unit vector in the smallest right singular subspace of $L$ , then forms the barycentric quotient if all weights are nonzero and returns it only after checking its reduced denominator at remaining samples.

[F2]

Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition: a finite-dimensional linear map has orthonormal left and right singular-vector bases and nonnegative singular values, with zero singular values in its kernel directions.

[F3]

Barycentric rational quotients are defined and interpolate at support nodes: nonzero support weights give a well-defined barycentric quotient whose value at each support node is its assigned datum.

Proof

1.1givenF1F2algebra

Let $v_{1}, \dots, v_{m}$ be a full orthonormal right singular-vector basis for $L$ , with singular values $σ_{1} \geq \dots \geq σ_{m} \geq 0$ and $σ_{m} = σ_{\min} (L)$ . For any unit vector $v = \sum_{j} c_{j} v_{j}$ , orthonormality gives $\sum_{j} ∣ c_{j} ∣^{2} = 1$ , and the SVD gives $∥ L v ∥_{2}^{2} = \sum_{j = 1}^{m} σ_{j}^{2} ∣ c_{j} ∣^{2} \geq σ_{m}^{2} .$ The selected $w$ lies in the right singular subspace for $σ_{m}$ , so equality holds. Hence $∥ L w ∥_{2} = σ_{\min} (L)$ and no unit vector gives a smaller Loewner residual.

1.2givenF1F2algebra

The right singular subspace $E$ is nonzero. If every projection $P_{E} e_{j}$ of a coordinate basis vector vanished, then every vector of $E$ would be orthogonal to every coordinate basis vector and hence would be zero, a contradiction. Thus the least index with nonzero projection exists; the normalization and phase convention in [F1] choose a unique unit vector in $E$ , also when the smallest singular value is repeated or zero.

1.3givenF1F3

On the successful-weight branch, every support weight is nonzero. By [F3], the barycentric quotient is defined and its value at each selected support node is the corresponding datum. The candidate is returned only after the algorithm's reduced-denominator pole check succeeds at all remaining sample points, so its finite residual there is defined.

2.1givenF1step 1.1∎

Each iteration removes one previously remaining point and the budget satisfies $B \leq ⌊ N / 2 ⌋$ . Every branch either returns immediately or proceeds to the next integer $m \leq B$ , so the algorithm makes at most $B$ iterations and terminates. The minimized quantity in step 1.1 is $∥ L w ∥_{2}$ ; the procedure separately records $η_{m}$ and makes no minimax or uniform-error claim.

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Floater–Hormann weights equal the local-polynomial blend weights

Statement

For $n \geq 1$ and $0 \leq d < n$ , define $λ_{i} (x) = \frac{(- 1)^{i}}{\prod_{k = i}^{i + d} (x - x_{k})}, μ_{i} (x) = \prod_{j < i} (x - x_{j}) \prod_{k > i + d} (x_{k} - x), 0 \leq i \leq n - d,$ where empty products equal $1$ . Put $s (x) = \sum_{i} μ_{i} (x)$ and let $p_{i}$ be the window polynomials in Local polynomial in a Floater–Hormann window. For $x \notin {x_{0}, \dots, x_{n}}$ , the Floater–Hormann quotient is $r_{d} (x) = \frac{\sum_{i} λ_{i} (x) p_{i} (x)}{\sum_{i} λ_{i} (x)} = \frac{\sum_{i} μ_{i} (x) p_{i} (x)}{s (x)} .$ Moreover, clearing the node factors in the barycentric denominator gives $\prod_{j = 0}^{n} (x - x_{j}) \sum_{i} λ_{i} (x) = (- 1)^{n - d} s (x)$ . After clearing and canceling the displayed factors, this is a polynomial identity valid at every real $x$ , including the nodes. Expanding this expression into a support-node barycentric quotient gives exactly the weights in Floater–Hormann barycentric interpolant. Its continuous values at the nodes are the assigned data.

Facts & Assumptions

Given: $n \geq 1$ , ordered distinct real nodes $x_{0} < \dots < x_{n}$ , real data $f_{j}$ , $0 \leq d < n$ , and $m = n - d$ .

[F1]

Floater–Hormann barycentric interpolant: the support weights are $w_{j} = (- 1)^{j - d} \sum_{i = \max (0, j - d)}^{\min (j, n - d)} \prod_{i \leq k \leq i + d, k \neq j} ∣ x_{j} - x_{k} ∣^{- 1}$ .

[F2]

Local polynomial in a Floater–Hormann window and Floater–Hormann window polynomials interpolate their local data: each window polynomial is its displayed Lagrange sum, has degree at most $d$ , interpolates the window data, and the same Lagrange basis sums to $1$ for constant data.

[F3]

Barycentric rational quotient with support values and reduced poles: the support quotient uses weights $w_{j}$ in both its numerator and denominator and takes the data values at its support nodes.

Proof

1.1givenconstructalgebra

Put $P_{0} (x) = \prod_{j = 0}^{n} (x - x_{j})$ and $m = n - d$ . Counting the $m$ factors to the right of a window gives $μ_{i} (x) = (- 1)^{m} P_{0} (x) λ_{i} (x) .$ Indeed, after cancellation each right factor $x_{k} - x$ contributes one minus sign when rewritten as $x - x_{k}$ , and the remaining denominator factors are exactly the window nodes. Thus the cleared products agree as polynomials, including at nodes; summing gives the stated cleared-denominator identity. Away from nodes it also shows that the blend quotient with $λ_{i}$ equals the quotient with $μ_{i}$ and $s$ .

1.2givenF1F2algebra

Expand each $p_{i}$ using [F2], and interchange the two finite sums. The coefficient of $f_{j} / (x - x_{j})$ in $\sum_{i} λ_{i} p_{i}$ is ${\tilde{w}}_{j} = \sum_{i = \max (0, j - d)}^{\min (j, n - d)} (- 1)^{i} \prod_{\begin{array}{c} i \leq k \leq i + d \\ k \neq j \end{array}} \frac{1}{x_{j} - x_{k}} .$ Within a window exactly $i + d - j$ factors $x_{j} - x_{k}$ are negative, so each summand has sign $(- 1)^{2 i + d - j} = (- 1)^{j - d}$ and absolute value $\prod_{k \neq j} ∣ x_{j} - x_{k} ∣^{- 1}$ . Thus ${\tilde{w}}_{j} = w_{j}$ as in [F1].

2.1F1F2F3step 1.2algebra

The same expansion for constant data $f_{k} = 1$ has each window polynomial equal to $1$ by [F2]. Therefore the coefficient of $1 / (x - x_{j})$ in $\sum_{i} λ_{i}$ is also ${\tilde{w}}_{j} = w_{j}$ . Dividing the expanded numerator by the expanded denominator proves that the blend equals the barycentric quotient in [F3] away from support nodes.

3.1F2F3step 1.1algebra∎

At a node $x_{α}$ , $μ_{i} (x_{α})$ is zero unless $\max (0, α - d) \leq i \leq \min (α, m)$ ; for these indices all its factors are positive. This index set is nonempty, and every corresponding window contains $x_{α}$ . By [F2], $p_{i} (x_{α}) = f_{α}$ , so the blend has numerator $f_{α} s (x_{α})$ and $s (x_{α}) > 0$ . Its value is therefore $f_{α}$ , agreeing with the support convention in [F3].

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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.

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Floater–Hormann blend partial sums are positive in each node gap

Statement

Let $x_{0} < \dots < x_{n}$ and $0 \leq d < n$ , and use the $μ_{i}$ and $s = \sum_{i = 0}^{n - d} μ_{i}$ of Floater–Hormann weights equal the local-polynomial blend weights. For $x_{α} < x < x_{α + 1}$ , partition the indices into $I_{1} = {i \leq α - d}, I_{2} = {α - d + 1 \leq i \leq α}, I_{3} = {i \geq α + 1},$ intersecting each set with ${0, \dots, n - d}$ . Each nonempty partial sum $\sum_{i \in I_{k}} μ_{i} (x)$ is positive. More precisely, if $I_{3}$ is nonempty, its terms alternate starting positive with strictly decreasing magnitudes; if $I_{1}$ is nonempty, its terms have the same property when read backward from $i = α - d$ . If $d \geq 1$ , every term in $I_{2}$ is positive.

Facts & Assumptions

Given: Ordered distinct nodes $x_{0} < \dots < x_{n}$ , $0 \leq d < n$ , and a query point in one open node gap.

[F1]

Floater–Hormann weights equal the local-polynomial blend weights: $μ_{i} (x) = \prod_{j < i} (x - x_{j}) \prod_{k > i + d} (x_{k} - x)$ .

Proof

1.1givenF1algebra

If $i \in I_{2}$ , every factor in $μ_{i} (x)$ is positive: left indices lie strictly below $x_{α} < x$ , and right indices lie strictly above $x < x_{α + 1}$ . Thus all terms in a nonempty $I_{2}$ are positive. If $d = 0$ , $I_{2}$ is empty.

1.2givenF1algebra

If $I_{3}$ is nonempty, its first index is $α + 1$ and $μ_{α + 1} (x) > 0$ . For $α + 1 \leq i < n - d$ , $\frac{μ_{i + 1} (x)}{μ_{i} (x)} = \frac{x - x_{i}}{x_{i + d + 1} - x} < 0, ∣ \frac{μ_{i + 1} (x)}{μ_{i} (x)} ∣ < 1.$ The numerator is negative and its magnitude is smaller than the positive denominator because $x_{i} < x_{i + d + 1}$ . Consecutive terms therefore alternate with strictly decreasing magnitudes. Pairing from the first positive term gives a positive sum, including when one positive term remains.

2.1givenF1algebra∎

If $I_{1}$ is nonempty, its last index is $α - d$ and $μ_{α - d} (x) > 0$ . Read backward. For $i \leq α - d$ with $i > 0$ , $\frac{μ_{i - 1} (x)}{μ_{i} (x)} = \frac{x_{i + d} - x}{x - x_{i - 1}} < 0, ∣ \frac{μ_{i - 1} (x)}{μ_{i} (x)} ∣ < 1.$ Here $x_{i + d} \leq x_{α} < x$ and $x_{i - 1} < x_{i} \leq x_{i + d}$ , so the numerator is negative and its magnitude is smaller than the positive denominator. Pairing backward from the first positive term proves the $I_{1}$ sum is positive.

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Floater–Hormann barycentric denominator lower bound

Statement

Let $n \geq 1$ , $0 \leq d < n$ , $x_{0} < \dots < x_{n}$ , and $h = \max_{j} (x_{j + 1} - x_{j})$ . Let $D_{d} (x) = \sum_{j = 0}^{n} w_{j} / (x - x_{j})$ be the Floater–Hormann denominator from Floater–Hormann barycentric interpolant. For every open gap $x_{α} < x < x_{α + 1}$ , $∣ D_{d} (x) ∣ \geq {\begin{cases} \frac{1}{d! h^{d + 1}}, & d \geq 1, \\ \frac{1}{h (1 + β)}, & d = 0, \end{cases}$ where $β$ is the local ratio of Local mesh ratio for degree-zero Floater–Hormann interpolation.

Facts & Assumptions

Given: Ordered distinct real nodes, $n \geq 1$ , $0 \leq d < n$ , an open node gap, and the associated Floater–Hormann denominator.

[F1]

Floater–Hormann weights equal the local-polynomial blend weights: $D_{d} (x) = \sum_{i} λ_{i} (x)$ and $\sum_{i} μ_{i} (x) = (- 1)^{n - d} (\prod_{j} (x - x_{j})) D_{d} (x)$ away from nodes.

[F2]

Floater–Hormann blend partial sums are positive in each node gap: each nonempty block sum over $I_{1}, I_{2}, I_{3}$ is positive; for $d = 0$ , the first two terms of $I_{3}$ and the last two terms of $I_{1}$ form positive pairs.

[F3]

Local mesh ratio for degree-zero Floater–Hormann interpolation: for $d = 0$ , $β$ is the maximum of the interior minima of the two adjacent-gap ratios, and is zero when $n \leq 2$ .

Proof

1.1givenF1F2choosealgebra

Suppose $d \geq 1$ . The middle block $I_{2}$ contains at least one index: $j = \max (0, α - d + 1)$ satisfies $j \leq α$ , $j + d \geq α + 1$ , and $j \leq n - d$ . Thus its window contains both endpoints of the gap and $μ_{j} (x) > 0$ . By [F2], $s (x) = s_{1} (x) + s_{2} (x) + s_{3} (x) \geq s_{2} (x) \geq μ_{j} (x)$ , with empty block sums interpreted as zero. From [F1], $∣ D_{d} (x) ∣ = \frac{s (x)}{\prod_{k = 0}^{n} ∣ x - x_{k} ∣} \geq \frac{μ_{j} (x)}{\prod_{k = 0}^{n} ∣ x - x_{k} ∣} = ∣ λ_{j} (x) ∣ = \frac{1}{\prod_{k = j}^{j + d} ∣ x - x_{k} ∣} .$ This reduces the denominator estimate to one window product.

1.2givenF1F2algebra

Now let $d = 0$ . On the first gap, $I_{1} = {0}$ and the other nonempty block sums are positive by [F2], hence $s \geq μ_{0}$ and $∣ D_{0} (x) ∣ \geq ∣ λ_{0} (x) ∣ = 1 / (x - x_{0}) \geq 1 / h$ . On the last gap it gives $I_{3} = {n}$ and the other nonempty block sums are positive, hence $s \geq μ_{n}$ and $∣ D_{0} (x) ∣ \geq 1 / (x_{n} - x) \geq 1 / h$ . These bounds imply the claimed inequality because $β \geq 0$ .

2.1givenstep 1.1algebra

Put $u = α - j + 1$ and $v = j + d - α$ , so $u, v \geq 1$ and $u + v = d + 1$ . For $j \leq k \leq α$ , $∣ x - x_{k} ∣ \leq x_{α + 1} - x_{k} \leq (α + 1 - k) h$ ; for $α + 1 \leq k \leq j + d$ , $∣ x - x_{k} ∣ \leq x_{k} - x_{α} \leq (k - α) h$ . Multiplying these $d + 1$ bounds gives $\prod_{k = j}^{j + d} ∣ x - x_{k} ∣ \leq u! v! h^{d + 1} \leq d! h^{d + 1} .$ The last inequality follows from $(\binom{d + 1}{u}) = (d + 1)! / (u! v!) \geq d + 1$ . Substitution in step 1.1 proves the $d \geq 1$ bound.

3.1givenF1F2F3step 1.2algebra∎

If $1 \leq α \leq n - 2$ , both $I_{1}$ and $I_{3}$ are nonempty. The first two terms from $I_{3}$ and first two read backward from $I_{1}$ form positive pairs. With $Δ = x_{α + 1} - x_{α}$ , $δ_{R} = x_{α + 2} - x_{α + 1}$ and $δ_{L} = x_{α} - x_{α - 1}$ , [F1]–[F2] yield $∣ D_{0} (x) ∣ \geq \max {\frac{1}{x_{α + 1} - x} - \frac{1}{x_{α + 2} - x}, \frac{1}{x - x_{α}} - \frac{1}{x - x_{α - 1}}} .$ Writing $t = x_{α + 1} - x \leq Δ$ in the first difference gives $δ_{R} / [t (t + δ_{R})] \geq 1 / [h (1 + Δ / δ_{R})]$ . Writing $u = x - x_{α} \leq Δ$ in the second gives $δ_{L} / [u (u + δ_{L})] \geq 1 / [h (1 + Δ / δ_{L})]$ . Hence $∣ D_{0} (x) ∣ \geq \frac{1}{h (1 + \min {Δ / δ_{L}, Δ / δ_{R}})} \geq \frac{1}{h (1 + β)}$ by [F3]. For $n \leq 2$ there is no such interior index, so steps 1.2 cover all gaps.

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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.

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Uniform meshes refute the C^(d+1)-seminorm-only Floater–Hormann rate

Statement refuted

The refuted claim is that, for every integer $d \geq 0$ and $ρ \geq 1$ , there exists a constant $C (d, ρ)$ such that for every $n > d$ , every ordered distinct real mesh $x_{0} < \dots < x_{n}$ in a compact interval $[a, b]$ , and every $f \in C^{d + 1} ([a, b])$ , the exact Floater–Hormann interpolant $r_{d}$ satisfies the following bound whenever $h / h_{\min} \leq ρ$ , where $h = \max_{j} (x_{j + 1} - x_{j})$ and $h_{\min} = \min_{j} (x_{j + 1} - x_{j})$ : $∥ f - r_{d} ∥_{\infty} \leq C (d, ρ) h^{d + 1} ∥ f^{(d + 1)} ∥_{\infty}$ Here $∥ q ∥_{\infty} : = \sup_{t \in [a, b]} ∣ q (t) ∣$ for continuous $q$ on $[a, b]$ .

Facts & Assumptions

Given: A fixed integer $d \geq 0$ , the exact Floater–Hormann weights on uniform nodes, and the construction below.

[F1]

Floater–Hormann barycentric interpolant: the weights are $w_{j} = (- 1)^{j - d} \sum_{i = \max (0, j - d)}^{\min (j, n - d)} \prod_{k = i, k \neq j}^{i + d} ∣ x_{j} - x_{k} ∣^{- 1}$ .

[F2]

Floater–Hormann interpolants have no real poles: the exact denominator is nonzero at every real query point.

[F3]

Higher derivatives and the classes $C^{k}$ and $C^{\infty}$ : $C^{d + 1}$ means derivatives through order $d + 1$ exist and are continuous on the closed interval.

[F4]

Barycentric rational quotient with support values and reduced poles: away from support nodes, the rational quotient is the barycentric numerator divided by its denominator; common nonzero scaling of all weights cancels.

[F5]

Extreme value theorem: a continuous real function on a nonempty compact subset of $R$ attains a greatest and a least value: a continuous real function on the nonempty compact interval has a finite attained maximum and minimum.

Counterexample

1.1givenF1constructalgebra

Fix $d \geq 0$ . For every integer $n \geq 2 d + 2$ , take $h = 1 / n$ , $x_{j} = j h$ on $[0, 1]$ , and query $x = h / 2$ . This family has $h / h_{\min} = 1$ . From [F1], after multiplying every weight by the same positive factor $h^{d} d!$ , its weights are ${\tilde{w}}_{j} = (- 1)^{j - d} δ_{j}, δ_{j} = \sum_{i = \max (0, j - d)}^{\min (j, n - d)} (\binom{d}{j - i}) .$ Indeed, on a uniform mesh each reciprocal-distance product in [F1] equals $h^{- d} / ((j - i)! (i + d - j)!)$ . Thus $1 \leq δ_{j} \leq 2^{d}$ , and $δ_{j} = 2^{d}$ for $d \leq j \leq n - d$ .

1.2givenconstructalgebra

Define $ϕ (t) = {\begin{cases} (1 - 16 t^{2})^{d + 2}, & ∣ t ∣ \leq 1 / 4, \\ 0, & ∣ t ∣ > 1 / 4. \end{cases}$ At both support endpoints the polynomial has a zero of multiplicity $d + 2$ , so its derivatives through order $d + 1$ vanish there and agree with the zero extension; hence $ϕ \in C^{d + 1} (R)$ and $ϕ (0) = 1$ . Expanding the polynomial and differentiating each term gives $∥ ϕ^{(d + 1)} ∥_{\infty} \leq M_{d} : = 4^{d + 1} \sum_{k = ⌈ (d + 1) / 2 ⌉}^{d + 2} (\binom{d + 2}{k}) \frac{(2 k)!}{(2 k - d - 1)!} < \infty .$ This is a finite bound depending only on $d$ , and its $k = d + 2$ term shows $M_{d} > 0$ .

2.1givenF1F2step 1.1algebra

Let $D (x) = \sum_{j = 0}^{n} {\tilde{w}}_{j} / (x - x_{j})$ be the denominator at the query. If $d = 0$ , its $j = 0$ term has magnitude $2 / h$ , and the remaining terms form an alternating decreasing sequence beginning with magnitude $2 / h$ , so $∣ D (x) ∣ \leq 4 / h$ . If $d \geq 1$ , the endpoint blocks $j = 0, \dots, d - 1$ and $j = n - d + 1, \dots, n$ each contain $d$ terms of magnitude at most $2^{d + 1} / h$ . The middle block $j = d, \dots, n - d$ has constant $δ_{j} = 2^{d}$ and alternating decreasing magnitudes, so its sum has magnitude at most $2^{d + 1} / h$ . Hence in both cases $0 < ∣ D (x) ∣ \leq \frac{2^{d} (4 d + 4)}{h},$ The strict lower inequality follows from [F2].

2.2givenF3step 1.2constructalgebra

The real number ${\tilde{w}}_{j} / (x - x_{j})$ is nonzero. Define $σ_{j} = 1$ if it is positive and $σ_{j} = - 1$ if it is negative, and set $f_{n} (t) = h^{d + 1} \sum_{j = 0}^{n} σ_{j} ϕ ⁣ (\frac{t - x_{j}}{h}), 0 \leq t \leq 1.$ The supports have radius $h / 4$ about distinct nodes, so they are disjoint; the query $x = h / 2$ is outside every support. Therefore $f_{n} (x) = 0$ , $f_{n} (x_{j}) = h^{d + 1} σ_{j}$ , $f_{n} \in C^{d + 1} ([0, 1])$ , and $∥ f_{n}^{(d + 1)} ∥_{\infty} \leq M_{d}$ by [F3] and the derivative estimate in step 1.2.

3.1givenF1F2F4step 2.1step 2.2algebra

By the barycentric quotient formula [F4] and the signs $σ_{j}$ , the numerator at $x$ is $h^{d + 1} \sum_{j} ∣ {\tilde{w}}_{j} / (x - x_{j}) ∣$ . Restricting to $j = d + 1, \dots, n - d$ and using $δ_{j} = 2^{d}$ and $∣ x - x_{j} ∣ = h (j - \frac{1}{2})$ , step 2.1 gives $∣ f_{n} (x) - r_{d} (x) ∣ \geq \frac{h^{d + 1}}{4 d + 4} \sum_{j = d + 1}^{n - d} \frac{1}{j - \frac{1}{2}} .$ Since $t \mapsto 1 / (t - \frac{1}{2})$ is decreasing, $\sum_{j = d + 1}^{n - d} \frac{1}{j - \frac{1}{2}} \geq \int_{d + 1}^{n - d + 1} \frac{d t}{t - \frac{1}{2}} = \log \frac{n - d + \frac{1}{2}}{d + \frac{1}{2}} .$ This lower bound grows without limit as $n$ increases.

4.1F5step 2.2step 3.1algebra∎

By [F2], $f_{n} - r_{d}$ is continuous on $[0, 1]$ , so [F5] makes its uniform norm finite. It follows that $\frac{∥ f_{n} - r_{d} ∥_{\infty}}{h^{d + 1} M_{d}} \geq \frac{\log ((n - d + \frac{1}{2}) / (d + \frac{1}{2}))}{(4 d + 4) M_{d}} ⟶ \infty (n \to \infty) .$ Every mesh in the family has ratio one, but no fixed $C (d, 1)$ can satisfy the refuted inequality. Thus it fails even at $ρ = 1$ for each fixed $d$ .

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Floater–Hormann error bounds for a bounded-variation highest derivative

Statement

Let $n \geq 1$ , $0 \leq d < n$ , and $x_{0} < \dots < x_{n}$ be real nodes in $[a, b] = [x_{0}, x_{n}]$ . Put $h = \max_{0 \leq j < n} (x_{j + 1} - x_{j})$ , $m = n - d$ , and $ϵ = 1$ if $m$ is even and $ϵ = 0$ if $m$ is odd. Let $f \in C^{d + 1} ([a, b])$ , set $g = f^{(d + 1)}$ , and suppose $g$ has finite total variation $V = {Var}_{[a, b]} (g)$ as defined in Bounded variation and total variation on an interval. For a continuous $q$ on $[a, b]$ , write $∥ q ∥_{\infty} = \sup_{x \in [a, b]} ∣ q (x) ∣$ . If $d \geq 1$ , then $∥ f - r_{d} ∥_{\infty} \leq \frac{h^{d + 1}}{d + 1} (V + ϵ ∥ g ∥_{\infty}),$ If $d = 0$ , let $β$ be the local mesh ratio in Local mesh ratio for degree-zero Floater–Hormann interpolation. Then $∥ f - r_{0} ∥_{\infty} \leq h (1 + β) (V + ϵ ∥ f^{'} ∥_{\infty}) .$ No mesh-ratio restriction is needed when $d \geq 1$ . For $d = 0$ , a fixed global mesh-ratio bound $ρ = h / h_{\min}$ implies $β \leq ρ$ by A global mesh ratio bounds the local degree-zero ratio.

Facts & Assumptions

Given: Ordered distinct nodes $x_{0}, \dots, x_{n}$ , $n \geq 1$ , $0 \leq d < n$ , $f \in C^{d + 1} ([x_{0}, x_{n}])$ , and $g = f^{(d + 1)}$ of finite total variation $V$ .

[F1]

Floater–Hormann weights equal the local-polynomial blend weights and Newton remainder for a Floater–Hormann window polynomial: off the nodes, $f (x) - r_{d} (x) = (\sum_{i = 0}^{n - d} (- 1)^{i} G_{i}) / D (x)$ with $G_{i} = f [x_{i}, \dots, x_{i + d}, x]$ and $D (x) = \sum_{i} λ_{i} (x)$ .

[F2]

Bounded variation controls adjacent Hermite–Genocchi windows: the adjacent divided differences satisfy $\sum_{i = 0}^{n - d - 1} ∣ G_{i + 1} - G_{i} ∣ \leq V / (d + 1)!$ .

[F3]

Hermite–Genocchi simplex formula for divided differences: an order- $(d + 1)$ divided difference is the simplex integral of $g$ ; the simplex volume is $1 / (d + 1)!$ .

[F4]

Floater–Hormann barycentric denominator lower bound: on an open gap, $∣ D (x) ∣ \geq 1 / (d! h^{d + 1})$ if $d \geq 1$ , and $∣ D (x) ∣ \geq 1 / [h (1 + β)]$ if $d = 0$ .

[F5]

Floater–Hormann interpolants have no real poles: the quotient is continuous on $[a, b]$ and interpolates the data at every node.

[F6]

Bounded variation and total variation on an interval: $V$ is the supremum of the sums of absolute successive increments over all finite partitions of $[a, b]$ .

[F7]

Higher derivatives and the classes $C^{k}$ and $C^{\infty}$ : $C^{d + 1}$ means derivatives through order $d + 1$ exist and are continuous on the closed interval; under the additional condition $f \in C^{d + 2}$ , the recursive definition gives $g^{'} = f^{(d + 2)}$ .

[F8]

Mean value theorem on a closed real interval: on every partition subinterval, a function continuous on the closed subinterval and differentiable on its interior has an increment equal to its derivative at an interior point times the interval length.

[F9]

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

[F10]

Extreme value theorem: a continuous real function on a nonempty compact subset of $R$ attains a greatest and a least value: a continuous real function on a nonempty compact interval has a finite attained maximum and minimum.

Proof

1.1givenF1F2F9F10algebra

By [F9], $d + 1 \in N$ . Fix $x$ in an open node gap. By [F7], the derivatives in [F1] exist and are continuous. By [F1], $f (x) - r_{d} (x) = N (x) / D (x)$ with $N (x) = \sum_{i = 0}^{m} (- 1)^{i} G_{i}$ . [F10] makes the derivative sup norm in the statement finite. Pair successive terms as $(G_{0} - G_{1}) + (G_{2} - G_{3}) + \dots$ . The sum of the absolute values of these pair differences is bounded by the full adjacent-increment sum in [F2], hence by $V / (d + 1)!$ . If $m$ is even there is one unpaired term $G_{m}$ ; otherwise there is none.

2.1givenF3F10step 1.1algebra

The integral formula [F3] and its simplex volume give $∣ G_{m} ∣ \leq ∥ g ∥_{\infty} / (d + 1)!$ . Thus in both parity cases $∣ N (x) ∣ \leq \frac{V + ϵ ∥ g ∥_{\infty}}{(d + 1)!} .$ This bounds the numerator in either parity case.

3.1givenF1F4step 2.1algebra

If $d \geq 1$ , divide step 2.1 by the lower bound $∣ D (x) ∣ \geq 1 / (d! h^{d + 1})$ from [F4] to obtain $∣ f (x) - r_{d} (x) ∣ \leq \frac{h^{d + 1}}{d + 1} (V + ϵ ∥ g ∥_{\infty}) .$ This is the required open-gap bound for $d \geq 1$ .

3.2givenF1F4step 2.1algebra

If $d = 0$ , $g = f^{'}$ and [F4] instead gives $∣ f (x) - r_{0} (x) ∣ \leq h (1 + β) (V + ϵ ∥ f^{'} ∥_{\infty}) .$ This is the required open-gap bound for $d = 0$ .

4.1givenF5F6F7F8F10step 3.1step 3.2algebra∎

At every node the error is zero by [F5]. Since $r_{d}$ has no real poles, $f - r_{d}$ is continuous on the compact interval and has a finite supremum by [F10]. The estimates on the open gaps therefore imply the stated supremum bounds. If $f \in C^{d + 2}$ , then for any partition [F8] gives $∣ g (t_{i + 1}) - g (t_{i}) ∣ \leq ∥ f^{(d + 2)} ∥_{\infty} (t_{i + 1} - t_{i})$ ; summing and taking the supremum yields $V \leq (b - a) ∥ f^{(d + 2)} ∥_{\infty}$ .

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Floater–Hormann C^(d+2) parity error bounds

Statement

Let $n \geq 1$ , $0 \leq d < n$ , and $x_{0} < \dots < x_{n}$ be real nodes in $[a, b] = [x_{0}, x_{n}]$ . Put $h = \max_{0 \leq j < n} (x_{j + 1} - x_{j})$ , $m = n - d$ , and $ϵ = 1$ if $m$ is even and $ϵ = 0$ if $m$ is odd. Let $f \in C^{d + 2} ([a, b])$ and let $r_{d}$ be the exact Floater–Hormann interpolant. For a continuous $q$ on $[a, b]$ , write $∥ q ∥_{\infty} = \sup_{x \in [a, b]} ∣ q (x) ∣$ . If $d \geq 1$ , then $∥ f - r_{d} ∥_{\infty} \leq h^{d + 1} (\frac{(b - a) ∥ f^{(d + 2)} ∥_{\infty}}{d + 2} + ϵ \frac{∥ f^{(d + 1)} ∥_{\infty}}{d + 1}) .$ If $d = 0$ , let $β$ be the local mesh ratio in Local mesh ratio for degree-zero Floater–Hormann interpolation. Then $∥ f - r_{0} ∥_{\infty} \leq h (1 + β) (\frac{(b - a) ∥ f^{''} ∥_{\infty}}{2} + ϵ ∥ f^{'} ∥_{\infty}) .$ No quasi-uniformity hypothesis is needed when $d \geq 1$ . If $d = 0$ , a bound $β \leq ρ$ follows from a global mesh-ratio bound $ρ$ by A global mesh ratio bounds the local degree-zero ratio.

Facts & Assumptions

Given: Ordered distinct nodes $x_{0}, \dots, x_{n}$ , an integer $0 \leq d < n$ , $f \in C^{d + 2} ([x_{0}, x_{n}])$ , and the exact Floater–Hormann interpolant.

[F1]

Floater–Hormann weights equal the local-polynomial blend weights and Newton remainder for a Floater–Hormann window polynomial: off the nodes, $f (x) - r_{d} (x) = (\sum_{i = 0}^{n - d} (- 1)^{i} f [x_{i}, \dots, x_{i + d}, x]) / (\sum_{i} λ_{i} (x))$ .

[F2]

Adjacent-window divided-difference recurrence: adjacent terms satisfy $G_{i} - G_{i + 1} = - (x_{i + d + 1} - x_{i}) f [x_{i}, \dots, x_{i + d + 1}, x]$ .

[F3]

Rolle bound for a divided difference: for $f \in C^{q}$ , an order- $q$ divided difference has magnitude at most $∥ f^{(q)} ∥_{\infty} / q!$ .

[F4]

Floater–Hormann barycentric denominator lower bound: on each open gap the denominator has magnitude at least $1 / (d! h^{d + 1})$ for $d \geq 1$ , and at least $1 / [h (1 + β)]$ for $d = 0$ .

[F5]

Floater–Hormann interpolants have no real poles: the quotient has no real poles and interpolates the data at every node.

[F6]

Higher derivatives and the classes $C^{k}$ and $C^{\infty}$ : $C^{d + 2}$ means derivatives through order $d + 2$ exist and are continuous on the closed interval, with one-sided endpoint derivatives.

[F7]

The natural numbers $N$ (von Neumann): $N$ contains $0$ and is closed under successor, so $d + 1, d + 2 \in N$ .

[F8]

Proof

1.1givenF1F6F8algebra

Fix a point $x$ in an open node gap. By [F6], the stated smoothness hypothesis supplies the derivatives used below. By [F8], their sup norms are finite. By [F1], the error equals $N (x) / D (x)$ , where $N (x) = \sum_{i = 0}^{m} (- 1)^{i} G_{i}, G_{i} = f [x_{i}, \dots, x_{i + d}, x], D (x) = \sum_{i} λ_{i} (x) .$ This identity follows by writing $f - r_{d} = \sum_{i} λ_{i} (f - p_{i}) / D$ and substituting the window remainder; $λ_{i} \prod_{k = i}^{i + d} (x - x_{k}) = (- 1)^{i}$ .

2.1givenF2F3F7F8step 1.1algebra

Pair the terms of $N$ as $(G_{0} - G_{1}) + (G_{2} - G_{3}) + \dots$ . If $m$ is odd, these pairs exhaust the sum. If $m$ is even, the final term $G_{m}$ remains. By [F2] each pair is $- (x_{i + d + 1} - x_{i}) f [x_{i}, \dots, x_{i + d + 1}, x]$ . The order- $(d + 2)$ Rolle bound [F3] bounds its divided difference by $∥ f^{(d + 2)} ∥_{\infty} / (d + 2)!$ . Each interval length $x_{i + d + 1} - x_{i}$ is the sum of $d + 1$ adjacent mesh gaps, and each gap occurs in at most $d + 1$ windows. Hence the sum of lengths over the even indices in the pairing is at most $(d + 1) (b - a)$ .

3.1givenF3F7F8step 2.1algebra

If $m$ is odd, step 2.1 gives $∣ N ∣ \leq (d + 1) (b - a) ∥ f^{(d + 2)} ∥_{\infty} / (d + 2)!$ . If $m$ is even, it gives the same bound plus $∣ G_{m} ∣ \leq ∥ f^{(d + 1)} ∥_{\infty} / (d + 1)!$ by [F3] and [F7]. Thus both cases are $∣ N ∣ \leq \frac{(d + 1) (b - a) ∥ f^{(d + 2)} ∥_{\infty}}{(d + 2)!} + ϵ \frac{∥ f^{(d + 1)} ∥_{\infty}}{(d + 1)!} .$ This bounds the numerator in either parity case.

4.1givenF1F4step 3.1algebra

If $d \geq 1$ , divide the numerator bound in step 3.1 by $∣ D ∣ \geq 1 / (d! h^{d + 1})$ from [F4]. Simplifying factorials gives $∣ f (x) - r_{d} (x) ∣ \leq h^{d + 1} (\frac{(b - a) ∥ f^{(d + 2)} ∥_{\infty}}{d + 2} + ϵ \frac{∥ f^{(d + 1)} ∥_{\infty}}{d + 1}) .$ This is the required open-gap bound for $d \geq 1$ .

4.2givenF1F4step 3.1algebra

If $d = 0$ , the same numerator bound in step 3.1, with $∣ D ∣ \geq 1 / [h (1 + β)]$ from [F4], gives $∣ f (x) - r_{0} (x) ∣ \leq h (1 + β) (\frac{(b - a) ∥ f^{''} ∥_{\infty}}{2} + ϵ ∥ f^{'} ∥_{\infty}) .$ This is the required open-gap bound for $d = 0$ .

5.1givenF5F8step 4.1step 4.2∎

At every node the error is zero by [F5]. The no-real-pole property makes $r_{d}$ continuous on the compact interval, so the pointwise estimates on the open gaps extend to the supremum over all $[a, b]$ ; finiteness of that supremum follows from [F8]. The two cases prove the stated bounds.

5 · Examples, counterexamples and false statements

None yet.