TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge 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.

Natural cubic interpolation at distinct knots

Statement

Let $r \geq 2$ , $a \leq x_{1} < \dots < x_{r} \leq b$ , and $y_{1}, \dots, y_{r} \in R$ . There is exactly one natural cubic spline $s$ on $[a, b]$ with $s (x_{i}) = y_{i}$ for each $i$ . On each interval $[x_{i}, x_{i + 1}]$ it is the cubic with endpoint values $y_{i}, y_{i + 1}$ and endpoint second derivatives $M_{i}, M_{i + 1}$ , where $M_{1} = M_{r} = 0$ and the interior values solve the positive-definite tridiagonal system $h_{i - 1} M_{i - 1} + 2 (h_{i - 1} + h_{i}) M_{i} + h_{i} M_{i + 1} = 6 (\frac{y_{i + 1} - y_{i}}{h_{i}} - \frac{y_{i} - y_{i - 1}}{h_{i - 1}}), 2 \leq i \leq r - 1,$ with $h_{i} = x_{i + 1} - x_{i} > 0$ . The interpolant is linear outside $[x_{1}, x_{r}]$ .

Facts & Assumptions

Given: $r \geq 2$ , strictly increasing knots $x_{1} < \dots < x_{r}$ in $[a, b]$ , and arbitrary values $y_{i} \in R$ .

[F1]

A natural cubic spline is $C^{2}$ , piecewise cubic between its distinct knots, has second derivative zero at the extreme knots, and is linear on data-free end intervals (Natural cubic splines).

[A1]

$r \geq 2$ , $x_{1} < \dots < x_{r}$ in $[a, b]$ , and $y_{1}, \dots, y_{r}$ are arbitrary real values.

Proof

technique · direct

1.1A1algebra

Put $h_{i} = x_{i + 1} - x_{i}$ and $d_{i} = (y_{i + 1} - y_{i}) / h_{i}$ . For $r = 2$ there are no interior second derivatives to solve. For $r \geq 3$ , let $A$ be the $(r - 2) \times (r - 2)$ symmetric tridiagonal matrix with diagonal entry $2 (h_{i - 1} + h_{i})$ in row $i$ and adjacent off-diagonal entries $h_{i}$ . For a nonzero vector $v = (v_{2}, \dots, v_{r - 1})$ , using $2 v_{i} v_{i + 1} \geq - v_{i}^{2} - v_{i + 1}^{2}$ gives $v^{T} A v \geq \sum_{i = 2}^{r - 1} (h_{i - 1} + h_{i}) v_{i}^{2} > 0.$ Thus $A$ is positive definite and invertible, so the displayed system has exactly one solution $M_{2}, \dots, M_{r - 1}$ .

2.1A1F1step 1.1algebraconstruct

On $[x_{i}, x_{i + 1}]$ , define $s_{i} (x) = \frac{M_{i} (x_{i + 1} - x)^{3} + M_{i + 1} (x - x_{i})^{3}}{6 h_{i}} + (y_{i} - \frac{M_{i} h_{i}^{2}}{6}) \frac{x_{i + 1} - x}{h_{i}} + (y_{i + 1} - \frac{M_{i + 1} h_{i}^{2}}{6}) \frac{x - x_{i}}{h_{i}} .$ Direct substitution gives $s_{i} (x_{i}) = y_{i}$ , $s_{i} (x_{i + 1}) = y_{i + 1}$ , and $s_{i}^{''} (x_{i}) = M_{i}$ , $s_{i}^{''} (x_{i + 1}) = M_{i + 1}$ . Its left and right endpoint slopes are respectively $s_{i}^{'} (x_{i}) = d_{i} - M_{i} h_{i} / 3 - M_{i + 1} h_{i} / 6$ and $s_{i}^{'} (x_{i + 1}) = d_{i} + M_{i} h_{i} / 6 + M_{i + 1} h_{i} / 3$ . Equating the right slope from $s_{i - 1}$ and the left slope from $s_{i}$ gives the tridiagonal equation at $x_{i}$ . Hence the pieces join $C^{1}$ ; their second derivatives are linear with the same nodal values, so they join $C^{2}$ . At the extreme knots $M_{1} = M_{r} = 0$ , so adjoining the tangent line gives the required linear tails and preserves $C^{2}$ . For $r = 2$ , the formula is the straight line between the two values and the same endpoint argument applies.

3.1A1F1step 1.1step 2.1algebra∎

Any natural cubic interpolant has nodal second derivatives with $M_{1} = M_{r} = 0$ . Equality of its adjacent endpoint slopes forces the displayed tridiagonal equations, so by step 1.1 its nodal second derivatives equal the constructed ones. On each interval, the difference from $s_{i}$ is a cubic with zero second derivative at both endpoints; its second derivative is affine and therefore zero throughout, so the difference is affine and vanishes at both endpoint values. It is identically zero. Thus the interpolant is unique.

Depends on

Natural cubic splines

Used by

Dependency tree · one level

1 result within one dependency step 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

P. J. Green and B. W. Silverman, Nonparametric Regression and Generalized Linear Models: A Roughness Penalty Approach, §2.2.1, Theorem 2.2, printed p. 15 (standard reference, not scraped)