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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The standard Rosenbrock start accepts a full Newton step

Example

For $f (x, y) = 100 (y - x^{2})^{2} + (1 - x)^{2}$ at $(x_{0}, y_{0}) = (- 6 / 5, 1)$ , use an exact unshifted Newton solve, descent threshold $σ = 1 / 10$ , and strong Wolfe parameters $c_{1} = 1 / 10000$ , $c_{2} = 2 / 5$ . The initial unit trial satisfies both strong Wolfe inequalities, so the bracketing search returns $t = 1$ .

Use the deterministic Krylov procedure which, for a two-dimensional operator $A$ and right-hand side $b$ , forms $V = [b A b]$ when $M \geq 2$ . If $V$ and $V^{T} A V$ are invertible, it returns $p = V (V^{T} A V)^{- 1} V^{T} b$ with iteration count $m = 2$ ; otherwise it returns no-candidate. On other input dimensions or budgets below two it returns no-candidate. Its candidate lies in $K_{2} (A, b)$ by construction. Take $M \geq 2$ , $N \geq 1$ , $E, Z \geq 1$ , $J = 0$ , $η_{k} = \overset{ˉ}{η} = 0$ , $γ = 2$ , and $λ_{0} = λ_{\max} = 1$ .

Facts & Assumptions

Given: The Rosenbrock objective and parameters in the Example.

[F1]

The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case: the gradient is the vector of first partial derivatives in Euclidean coordinates.

[F2]

The Hessian matrix and critical points of a scalar field: the Hessian matrix consists of the second partial derivatives.

[F3]

Finite-budget inexact Newton–Krylov with a descent safeguard: with $λ = 0$ , the algorithm accepts a candidate only after it verifies the Newton residual bound and strict descent test.

[F4]

Finite-budget bracketing and zoom for a strong Wolfe step: the expansion phase tests $t = 1$ first and returns a step only after both strong Wolfe inequalities are checked.

[F5]

Wolfe and strong Wolfe line-search conditions: a strong Wolfe step satisfies Armijo decrease and $∣ ϕ^{'} (t) ∣ \leq c_{2} ∣ ϕ^{'} (0) ∣$ .

Verification

Given: The objective, initial point, and parameters of the Example.

1.1givenF1F2algebra

Put $r : = y - x^{2}$ . Direct differentiation using [F1]–[F2] gives $\nabla f = (- 400 x r + 2 (x - 1), 200 r)$ and $H_{f} = [\begin{array}{cc} 1200 x^{2} - 400 y + 2 & - 400 x \\ - 400 x & 200 \end{array}]$ . At $(- 6 / 5, 1)$ , $r = - 11 / 25$ , $g = (- 1078 / 5, - 88)$ , and $H = [\begin{array}{cc} 1330 & 480 \\ 480 & 200 \end{array}]$ .

1.2givenF3algebra

With $b = - g$ , direct calculation gives $\det [b H b] = - 14221856 / 5 \neq 0$ and $\det H = 35600 \neq 0$ . Thus $V$ and $V^{T} H V$ are invertible. The specified solver returns $p \in K_{2} (H, b)$ with $V^{T} (H p - b) = 0$ ; since $V$ is invertible, $H p = b$ . Solving $H p = - g$ gives $p = (11 / 445, 847 / 2225)$ ; substitution into the matrix product verifies the equation exactly. Hence the recomputed unshifted residual is zero and $g^{T} p = - 86394 / 2225 < - 38$ . Also $∥ g ∥_{2} < 234$ and $∥ p ∥_{2} < 41 / 100$ , so $g^{T} p < - 38 < - (1 / 10) (234) (41 / 100) < - σ ∥ g ∥_{2} ∥ p ∥_{2}$ . The descent safeguard accepts this direction.

2.1givenF1F5step 1.2algebra

The full trial is $(x_{1}, y_{1}) = (- 523 / 445, 3072 / 2225)$ , with $y_{1} - x_{1}^{2} = - 121 / 198025$ . Thus $f (x_{0}, y_{0}) = 121 / 5$ , while $f (x_{1}, y_{1}) = (968 / 445)^{2} + 100 (121 / 198025)^{2} < 5$ . Since $ϕ^{'} (0) = g^{T} p > - 39$ , the Armijo right side is greater than $121 / 5 - 39 / 10000 > 24$ ; therefore the full trial satisfies Armijo.

2.2givenF1F5step 1.2algebra

At the trial point, $∣ x_{1} ∣ < 6 / 5$ , $∣ x_{1} - 1 ∣ < 11 / 5$ , and $∣ y_{1} - x_{1}^{2} ∣ < 1 / 1600$ . The gradient formula of step 1.1 gives $∣ (\nabla f (x_{1}, y_{1}))_{1} ∣ < 47 / 10$ and $∣ (\nabla f (x_{1}, y_{1}))_{2} ∣ < 1 / 8$ . Since $∣ p_{1} ∣ < 1 / 40$ and $∣ p_{2} ∣ < 2 / 5$ , it follows that $∣ ϕ^{'} (1) ∣ = ∣ ⟨ \nabla f (x_{1}, y_{1}), p ⟩ ∣ < 1 / 5$ . Meanwhile $c_{2} ∣ ϕ^{'} (0) ∣ > (2 / 5) 38 > 15$ . The strong curvature inequality therefore holds.

3.1F4step 2.1step 2.2∎

Steps 2.1 and 2.2 verify both strong Wolfe tests at the first trial $t = 1$ . By [F4] the search returns this verified full step. This example is a single accepted update and gives no global convergence claim.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

29 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.