TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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A computed natural-cubic reconstruction has a forward-error certificate

Statement

Let $r \geq 2$ and use the stored knots and values and the algorithm of Natural-cubic tridiagonal Cholesky solve with a residual certificate. Let $s$ be the exact natural cubic interpolant of those data, supplied by Natural cubic interpolation at distinct knots. On a successful call, let $\hat{M}$ be the returned second-derivative vector, put $\hat{m} = ({\hat{M}}_{2}, \dots, {\hat{M}}_{r - 1})$ , and let $\hat{s}$ be the piecewise cubic reconstruction defined by the algorithm from $\hat{M}$ , evaluated exactly. For $r \geq 3$ , the success output is $success (\hat{M}, C, α, β, B)$ . The algorithm computes $H$ , $δ$ , and $Q = ∥ \hat{m} ∥_{\infty}$ internally when forming $B$ ; these are not separate returned fields. The returned certificate fields satisfy

∥ \hat{s} - s ∥_{L^{\infty} ([x_{1}, x_{r}])} \leq B = \frac{H^{2}}{8 δ} (C + α ∥ \hat{m} ∥_{\infty} + β) .

Here $α$ and $β$ account for the differences between the exact natural-cubic system assembled from the stored input data and the rounded matrix and right-hand side used by the factorization. If those systems are exactly the same, then $α = β = 0$ and $B = H^{2} C / (8 δ)$ . For $r = 2$ , both the natural interpolant and the algorithm-defined reconstruction are the straight line, and $B = 0$ .

The algorithm-defined reconstruction interpolates the stored data, but for $r \geq 3$ it need not be $C^{1}$ at its knots and is not asserted to be an exact natural spline. The bound excludes floating-point error from evaluating the cubic pieces and makes no unconditional arithmetic-success or general natural-spline fourth-order claim.

Facts & Assumptions

Given: Strictly increasing stored knots and finite stored values, $r \geq 2$ , and a successful algorithm call including its exact rational certificate.

[F1]

Natural-cubic tridiagonal Cholesky solve with a residual certificate defines the exact canonical system $A m = b$ , the stored matrix $T$ and right-hand side $f$ used by the rounded solve, and the discrepancies $α = \max_{k} \sum_{j} ∣ A_{k j} - T_{k j} ∣$ and $β = \max_{k} ∣ b_{k} - f_{k} ∣$ . On a successful $r \geq 3$ call it returns $\hat{M}, C, α, β, B$ and computes $H$ , $δ$ , and $Q = ∥ \hat{m} ∥_{\infty}$ internally to form $B = H^{2} (C + α Q + β) / (8 δ)$ .

[F2]

A successful natural-cubic tridiagonal Cholesky solve has a certified residual gives $∥ T \hat{m} - f ∥_{\infty} \leq C$ on a full successful call.

[F3]

Natural cubic interpolation at distinct knots: for $r \geq 3$ , the exact natural interpolant's interior second derivatives solve $A m = b$ ; for $r = 2$ it is the straight line.

[F4]

A natural-spline tridiagonal residual bounds the spline value error: if $ρ = A \hat{m} - b$ is the exact residual for the interior natural-cubic system and the endpoint second derivatives agree at zero, then the piecewise cubic reconstruction differs from the exact natural interpolant on the knot hull by at most $H^{2} ∥ ρ ∥_{\infty} / (8 δ)$ .

[A1]

The data knots are exact stored real numbers with $x_{1} < \dots < x_{r}$ ; for $r \geq 3$ , each $h_{i} > 0$ and $δ > 0$ .

Proof

1.1A1F1F2algebra

Suppose $r \geq 3$ . Let $\hat{M}$ be the returned vector and $\hat{m} = ({\hat{M}}_{2}, \dots, {\hat{M}}_{r - 1})$ its interior subvector. By [F2] and the exact identity $A \hat{m} - b = (T \hat{m} - f) + (A - T) \hat{m} + (f - b)$ , each row satisfies $∣ (A \hat{m} - b)_{k} ∣ \leq ∣ (T \hat{m} - f)_{k} ∣ + \sum_{j} ∣ A_{k j} - T_{k j} ∣ ∣ {\hat{m}}_{j} ∣ + ∣ f_{k} - b_{k} ∣ \leq C + α ∥ \hat{m} ∥_{\infty} + β$ . Taking the maximum proves $∥ A \hat{m} - b ∥_{\infty} \leq C + α ∥ \hat{m} ∥_{\infty} + β$ ; these exact discrepancies account for all rounded system assembly.

2.1A1F1F4step 1.1algebra

The returned endpoint second derivatives and those of $s$ are both zero. Apply [F4] to the exact residual from step 1.1 and the original widths. It gives $∥ \hat{s} - s ∥_{L^{\infty} ([x_{1}, x_{r}])} \leq (H^{2} / (8 δ)) ∥ A \hat{m} - b ∥_{\infty} \leq (H^{2} / (8 δ)) (C + α ∥ \hat{m} ∥_{\infty} + β) = B$ . The transfer theorem applies to any proposed interior derivatives with zero endpoint values; it does not require the algorithm-defined pieces to join with matching first derivatives.

3.1A1F1F3algebra∎

If $r = 2$ , the algorithm returns zero endpoint second derivatives and its associated reconstruction is the exact straight-line piece. By [F3] this is also $s$ , and the algorithm defines $B = 0$ , so the stated bound holds with equality.

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Sources

Carl de Boor, A Practical Guide to Splines, revised edition, Chapter IV, pp. 42–45 (standard reference, not scraped)
Brian LeMesurier, Introduction to Numerical Methods and Analysis with Python, Chapter 3.4 (standard reference, not scraped)