ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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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.

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Natural cubic interpolation with two and three knots

Example

With two knots, the natural cubic interpolant is the line joining the two values. With knots $0, 1, 2$ and values $0, 1, 0$ , the middle second derivative is $- 3$ , and the interpolant is $s (x) = {\begin{cases} \frac{3}{2} x - \frac{1}{2} x^{3}, & 0 \leq x \leq 1, \\ \frac{3}{2} (2 - x) - \frac{1}{2} (2 - x)^{3}, & 1 \leq x \leq 2. \end{cases}$

Facts & Assumptions

Given: The natural cubic convention and distinct-knot interpolation theorem.

[F1]

Natural cubic splines are $C^{2}$ , piecewise cubic, have zero second derivative at the extreme knots, and have linear tails (Natural cubic splines).

[F2]

At two or more distinct ordered knots, arbitrary interpolation values determine a unique natural cubic spline via the tridiagonal second-derivative system (Natural cubic interpolation at distinct knots).

[A1]

The first data set has two distinct knots; the second has knots $0, 1, 2$ and values $0, 1, 0$ .

Verification

Given: The data sets $(0, y_{1}), (1, y_{2})$ and $(0, 0), (1, 1), (2, 0)$ .

1.1A1F1F2algebra

For two knots, there are no interior second derivatives in the system of [F2]. The endpoint second derivatives are both zero, so the cubic segment has affine second derivative vanishing at both ends, hence second derivative identically zero. The segment is the unique line through the two values, and its linear extension on any data-free tail is the same line.

1.2A1F2algebra

For the three-knot data, $h_{1} = h_{2} = 1$ and the secant slopes are $d_{1} = 1$ , $d_{2} = - 1$ . The one interior system equation is $4 M_{2} = 6 (d_{2} - d_{1}) = - 12$ , so $M_{2} = - 3$ while $M_{1} = M_{3} = 0$ .

2.1A1F1F2step 1.2algebra∎

Substitution in the interval formula gives the displayed pieces. They interpolate $0, 1, 0$ ; their first derivatives at $1$ are both $0$ , their second derivatives at $1$ are both $- 3$ , and the second derivatives at $0, 2$ are $0$ . Thus they join to a natural cubic spline, which is unique by [F2].

Depends on

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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)