TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generated

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An accepted damped Newton step strictly decreases the residual merit

Statement

Use the inputs and residual merit $Φ (x) = \frac{1}{2} ∥ F (x) ∥_{2}^{2}$ of Finite-budget residual-merit damped Newton. At any iterate where the procedure accepts a step $t > 0$ with $F (x) \neq 0$ , the accepted point $x_{+} = x + t s$ satisfies $Φ (x_{+}) \leq Φ (x) - α t ∥ F (x) ∥_{2}^{2} < Φ (x) .$ This statement concerns accepted steps only; it does not assert that a chosen finite backtrack budget will accept a trial.

Facts & Assumptions

Given: An admissible execution of Finite-budget residual-merit damped Newton, with its parameter $α \in (0, 1)$ , a nonzero residual $F (x)$ , and an accepted step size $t = β^{j}$ for some tested integer $j \geq 0$ .

[F1]

Finite-budget residual-merit damped Newton: a tested step is accepted only if $Φ (x + t s) \leq Φ (x) - α t ∥ F (x) ∥_{2}^{2}$ .

Proof

Given: The hypotheses in the Statement.

1.1givenalgebra

Since $0 < β < 1$ and $j \geq 0$ , the tested size $t = β^{j}$ is positive. Because the Euclidean norm is defined by the square root of the sum of coordinate squares, $F (x) \neq 0$ gives a nonzero coordinate and hence $∥ F (x) ∥_{2}^{2} > 0$ ; with $α > 0$ , this gives $α t ∥ F (x) ∥_{2}^{2} > 0$ .

2.1step 1.1F1algebra∎

The acceptance condition in [F1] therefore gives $Φ (x_{+}) \leq Φ (x) - α t ∥ F (x) ∥_{2}^{2} < Φ (x)$ , as claimed.

Depends on

Finite-budget residual-merit damped Newton

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

E. Hairer and G. Wanner, Solving Ordinary Differential Equations II, Chapter VI: Iterative Methods—Nonlinear Equations (standard reference, not scraped)
Tobin A. Driscoll and Richard J. Braun, Fundamentals of Numerical Computation, version 1.0 (standard reference, not scraped)