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

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Quadratic interpolation at three nodes is exact

Example

For $f (x) = x^{2}$ at the three nodes $- 1, 0, 1$ , the cardinal interpolant is exactly $f$ .

Facts & Assumptions

Given: The ordered nodes $x_{0} = - 1, x_{1} = 0, x_{2} = 1$ and the function $f (x) = x^{2}$ .

[F1]

Cardinal-polynomial interpolation: $I_{2} f$ is the sum of the nodal values times the cardinal functions.

[F2]

Formal real polynomials, evaluation, degree, leading coefficient, and monic polynomials: $x^{2}$ is the evaluation of a formal polynomial of degree $2$ .

[F3]

Existence and uniqueness of the cardinal-polynomial interpolant: the degree-at-most- $2$ polynomial matching $f$ at these three distinct nodes is uniquely $I_{2} f$ .

Proof

1.1givenalgebra

The nodal values are $f (- 1) = 1$ , $f (0) = 0$ , and $f (1) = 1$ .

2.1givenF1step 1.1F2F3algebra∎

The polynomial $q (X) = X^{2}$ has degree $2$ by [F2] and takes these three values. By the definition [F1] and uniqueness [F3], $I_{2} f = q = f$ .

Depends on

Used by

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Dependency tree · two levels

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Sources

Mark Embree, Lecture Notes on Numerical Analysis, lecture 1 (standard reference, not scraped)
Jean-Paul Berrut and Lloyd N. Trefethen, Barycentric Lagrange Interpolation (standard reference, not scraped)