TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Double-node Hermite interpolation has the stated pointwise remainder

Statement

Let $m \geq 1$ , let $a < b$ , let $x_{1} < \dots < x_{m}$ lie in $(a, b)$ , and let $f \in C^{2 m} ([a, b], R)$ . Let $ℓ_{i}$ be the Lagrange cardinal polynomials for these nodes as in Cardinal-polynomial interpolation. Define $H (t) : = \sum_{i = 1}^{m} [f (x_{i}) (1 - 2 ℓ_{i}^{'} (x_{i}) (t - x_{i})) + f^{'} (x_{i}) (t - x_{i})] ℓ_{i} (t)^{2} .$ Then $H$ has degree at most $2 m - 1$ , and $H (x_{i}) = f (x_{i})$ and $H^{'} (x_{i}) = f^{'} (x_{i})$ for every $i$ . If $x \in [a, b]$ is not a node, there is $ξ_{x} \in (a, b)$ such that $f (x) - H (x) = \frac{f^{(2 m)} (ξ_{x})}{(2 m)!_{R}} \prod_{i = 1}^{m} (x - x_{i})^{2},$ where $(2 m)!_{R} : = \prod_{j = 1}^{2 m} j$ . At a node the interpolation error and the displayed node product are both zero.

Facts & Assumptions

Given: Distinct interior nodes $x_{1} < \dots < x_{m}$ , $f \in C^{2 m} ([a, b])$ , and a query $x \in [a, b]$ .

[F1]

Cardinal-polynomial interpolation and Cardinal interpolation functions are polynomials of the stated degrees: $ℓ_{i}$ is a polynomial of degree at most $m - 1$ , $ℓ_{i} (x_{j}) = 1$ when $i = j$ and $0$ otherwise, and the node product $ω (t) = \prod_{i} (t - x_{i})$ is nonzero off the nodes.

[F2]

Higher derivatives and the classes $C^{k}$ and $C^{\infty}$ : all derivatives of $f$ through order $2 m$ exist on $[a, b]$ and are continuous there.

[F3]

Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ : finite sums, scalar multiples and products obey the derivative rules on the interval.

[F4]

Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function: polynomials and their finite sums and products are continuous.

[F5]

Higher-order Rolle theorem: a function with $n + 1$ distinct ordered zeros, continuous on their closed span and $n$ -times differentiable in its interior, has an $n$ th-derivative zero between the extreme zeros.

[F6]

A finite product of monic linear factors is monic: the product of $2 m$ monic linear factors is monic of degree $2 m$ .

[F7]

The highest derivative of a polynomial: a polynomial of degree at most $2 m$ has constant $2 m$ th derivative equal to $(2 m)!_{R}$ times its coefficient of $X^{2 m}$ .

[F8]

Degree inequalities for sums and products over a commutative ring and The reals form a field: sums and products of the displayed polynomial factors have the stated degree bounds and coefficient arithmetic.

[F9]

The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ and Laws of finite sums and products in $N$ , and $ι (\sum_{k < n} a_{k}) = \sum_{k < n} ι (a_{k})$ : $(2 m)!_{R} = \prod_{j = 1}^{2 m} j \neq 0$ .

Proof

1.1F1F3algebra

For each $i$ , set $h_{i} (t) = (1 - 2 ℓ_{i}^{'} (x_{i}) (t - x_{i})) ℓ_{i} (t)^{2}, k_{i} (t) = (t - x_{i}) ℓ_{i} (t)^{2} .$ At $x_{i}$ , $ℓ_{i} (x_{i}) = 1$ , so $h_{i} (x_{i}) = 1$ , $h_{i}^{'} (x_{i}) = 0$ , $k_{i} (x_{i}) = 0$ , and $k_{i}^{'} (x_{i}) = 1$ by the product rule [F3]. At $x_{j}$ with $j \neq i$ , $ℓ_{i} (x_{j}) = 0$ , so both $ℓ_{i}^{2}$ and its derivative vanish there; hence $h_{i} (x_{j}) = h_{i}^{'} (x_{j}) = k_{i} (x_{j}) = k_{i}^{'} (x_{j}) = 0$ . Therefore the displayed sum defining $H$ satisfies every value and first-derivative condition.

1.2F1F8construct

Each $ℓ_{i}$ has degree at most $m - 1$ by [F1]. Thus $ℓ_{i}^{2}$ has degree at most $2 m - 2$ , and multiplication by an affine polynomial gives degree at most $2 m - 1$ by [F8]. The finite sum $H$ also has degree at most $2 m - 1$ .

2.1F1F3step 1.1constructalgebra

Suppose $x$ is not a node. By [F1], $ω (x)^{2} \neq 0$ , so put $c = (f (x) - H (x)) / ω (x)^{2}$ and define $g (t) = f (t) - H (t) - c ω (t)^{2}$ . At every node, step 1.1 and $ω (x_{i}) = 0$ give $g (x_{i}) = 0$ and $g^{'} (x_{i}) = 0$ , since $(ω^{2})^{'} = 2 ω ω^{'}$ by [F3]; by the choice of $c$ , also $g (x) = 0$ .

3.1F2F3F4step 2.1choose

Sort the $m$ nodes and $x$ as $z_{0} < \dots < z_{m}$ . Rolle's theorem on each consecutive interval gives a zero of $g^{'}$ strictly between its endpoints; these $m$ zeros are distinct and are not nodes. Along with $g^{'} (x_{i}) = 0$ at the $m$ nodes from step 2.1, this gives $2 m$ distinct ordered zeros of $g^{'}$ .

4.1F2F3F4F5step 3.1

The function $g^{'}$ is continuous on the closed span of those zeros and is $(2 m - 1)$ -times differentiable on its interior by [F2]–[F4]. Apply [F5] to those $2 m$ zeros. There is $ξ_{x} \in (a, b)$ with $g^{(2 m)} (ξ_{x}) = 0$ .

5.1F6F7step 1.2step 2.1step 4.1algebra

By step 1.2, $H^{(2 m)} = 0$ . By [F6], $ω^{2}$ is monic of degree $2 m$ , so [F7] gives $(ω^{2})^{(2 m)} = (2 m)!_{R}$ . Differentiating the definition of $g$ in step 2.1 at $ξ_{x}$ therefore gives $0 = f^{(2 m)} (ξ_{x}) - c (2 m)!_{R}$ .

6.1F1F9step 1.1step 2.1step 5.1algebra∎

By [F9], $(2 m)!_{R} \neq 0$ , so step 5.1 gives $c = f^{(2 m)} (ξ_{x}) / (2 m)!_{R}$ . Substituting this into the definition of $c$ in step 2.1 proves the remainder formula. At a node, step 1.1 gives zero interpolation error and $ω = 0$ .

Depends on

Used by

Gaussian quadrature has the Hermite mean-value remainder Theorem

Dependency tree · two levels

76 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

John A. Gubner, Gaussian Quadrature and the Eigenvalue Problem, §§2–5, 2020 (standard reference, not scraped)
MATH 373 Lecture Notes, Lecture 14: Gaussian Quadrature, §§10.2–10.5 (standard reference, not scraped)
Walter Gautschi, The Interplay Between Classical Analysis and (Numerical) Linear Algebra, §2.1 (standard reference, not scraped)