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How statement and proof provenance work

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A second-order response surface with a saddle point

Example

On $R = [- 1, 1]^{2}$ , consider $m (x_{1}, x_{2}) = x_{1}^{2} - x_{2}^{2}$ . Its curvature matrix is $B = (\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array})$ , so the stationary point at the origin is a saddle in the fitted polynomial.

Facts & Assumptions

Given: The region and polynomial in the Example.

[F1]

Local response-surface models: a second-order response-surface model has $m (x) = β_{0} + x^{T} β + x^{T} B x$ with symmetric $B$ and is interpreted only on its declared region.

[F2]

Stationary point and classification of a nonsingular quadratic response surface: a nonsingular symmetric curvature matrix with eigenvalues of both signs gives a saddle at its stationary point; an interior stationary point has that local classification in the fitted region.

Verification

1.1givenF1F2algebra

The matrix $B$ is symmetric and nonsingular, and its eigenvalues are $1$ and $- 1$ . The stationary point formula in [F2] gives $x_{0} = 0$ , which lies in the interior of $R$ .

2.1givenF2choosealgebra∎

Directly, $m (h, 0) - m (0, 0) = h^{2} > 0$ and $m (0, h) - m (0, 0) = - h^{2} < 0$ for every $0 < ∣ h ∣ < 1$ . Thus arbitrarily close points in the region give increments of both signs, verifying the saddle classification in [F2].

Depends on

Used by

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Dependency tree · two levels

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Sources

Gary W. Oehlert, A First Course in Design and Analysis of Experiments, Chapter 19, §§19.5–19.7 (standard reference, not scraped)