ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

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.

A small least-squares problem makes the gradient and Hessian formulas explicit

Example

Let

$A = (\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}), b = (\begin{array}{c} 1 \\ 0 \end{array}) .$

Then

$f (x_{1}, x_{2}) = \frac{1}{2} ((x_{1} + x_{2} - 1)^{2} + x_{2}^{2}) .$

Hence

$\nabla f (x) = (\begin{array}{c} x_{1} + x_{2} - 1 \\ x_{1} + 2 x_{2} - 1 \end{array}), \nabla^{2} f = A^{*} A = (\begin{array}{cc} 1 & 1 \\ 1 & 2 \end{array}) .$

Facts & Assumptions

Given: The displayed matrix $A$ and vector $b$ .

[L1]

For $f (x) = \frac{1}{2} ∥ A x - b ∥_{2}^{2}$ , one has $\nabla f (x) = A^{*} (A x - b)$ and $\nabla^{2} f = A^{*} A$ (The Frobenius least-squares objective has gradient $A^{*} (A x - b)$ and Hessian $A^{*} A$ in the vector variable).

Verification

technique · direct

1.1

Expanding the square gives the displayed scalar function. Differentiating entrywise yields $\partial_{x_{1}} f = x_{1} + x_{2} - 1$ and $\partial_{x_{2}} f = x_{1} + 2 x_{2} - 1$ , so the gradient is as shown.

algebra

2.1

Also $A^{*} (A x - b) = (\begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array}) (\begin{array}{c} x_{1} + x_{2} - 1 \\ x_{2} \end{array}) = (\begin{array}{c} x_{1} + x_{2} - 1 \\ x_{1} + 2 x_{2} - 1 \end{array})$ , and $A^{*} A = (\begin{array}{cc} 1 & 1 \\ 1 & 2 \end{array})$ . This matches [L1].

L1step 1.1algebra∎

Depends on

The Frobenius least-squares objective has gradient $A^*(Ax-b)$ and Hessian $A^*A$ in the vector variable

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 results within two dependency steps 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

Alan Edelman and Steven G. Johnson, Matrix Calculus for Machine Learning and Beyond (standard reference, not scraped)