ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generated

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.

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An exponential-decay kinetics fit has a full-rank exact solution

Example

Fit the one-rate exponential-decay model $y (t; a, k) = a e^{- k t}$ to observations at $t = 0$ and $t = 1$ with values $2$ and $2 e^{- 1}$ . Its residual is $r (a, k) = (a - 2, a e^{- k} - 2 e^{- 1})^{T}$ . The parameter $x_{*} = (2, 1)$ is an exact fit, and the Jacobian there has full column rank. Thus the exact full-rank Gauss-Newton step at $x_{*}$ is zero. This is a local exact-fit illustration; it does not claim that a nonlinear iteration from every initial parameter converges to $x_{*}$ .

Facts & Assumptions

Given: The model, observations, residual, and candidate parameters in the Example.

[F1]

Gauss-Newton linearized least-squares step with QR/SVD rank safeguards: with a supplied exact reduced QR factorization, the full-column-rank exact branch returns the unique minimizer of the linearized residual problem.

[F2]

The exponential function is smooth and $(\exp)^{'} = \exp$ : the derivative of $e^{u}$ is $e^{u}$ .

[F3]

The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ : $e^{- 1} > 0$ .

Verification

1.1givenF2algebra

At $x_{*} = (2, 1)$ , the residual components are $r_{1} = 2 - 2 = 0$ and $r_{2} = 2 e^{- 1} - 2 e^{- 1} = 0$ . By [F2], the Jacobian is $J (a, k) = [\begin{array}{cc} 1 & 0 \\ e^{- k} & - a e^{- k} \end{array}]$ , hence $J (x_{*}) = [\begin{array}{cc} 1 & 0 \\ e^{- 1} & - 2 e^{- 1} \end{array}]$ .

2.1step 1.1F1F3algebra∎

Put $c = e^{- 1} > 0$ and $s = \sqrt{1 + c^{2}}$ . Supply the exact reduced QR factorization $Q = [\begin{array}{cc} 1 / s & c / s \\ c / s & - 1 / s \end{array}]$ and $R = [\begin{array}{cc} s & - 2 c^{2} / s \\ 0 & 2 c / s \end{array}]$ , for which $Q^{T} Q = I$ and $Q R = J (x_{*})$ ; its diagonal entries are positive. The determinant of $J (x_{*})$ is $- 2 e^{- 1} \neq 0$ by [F3], so the parameters are locally identifiable by the linearized model. Since $r (x_{*}) = 0$ , the linearized least-squares objective is $∥ J (x_{*}) p ∥_{2}^{2} / 2$ , whose unique minimizer is $p = 0$ . By [F1], the supplied exact QR branch returns this step.

Depends on

Used by

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Sources

Jorge Nocedal and Stephen J. Wright, Numerical Optimization, second edition (standard reference, not scraped)
David Bindel, CS 4220: Numerical Analysis, Gauss-Newton and nonlinear least squares (standard reference, not scraped)