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A mismatched terminal adjoint condition changes the discrete gradient

Example

Take $x_{0} (θ) = 0$ , one step $x_{1} (θ) = θ$ , and terminal objective $J (θ) = \frac{1}{2} x_{1} (θ)^{2}$ . The correct terminal adjoint seed is $p_{1} = D ϕ (x_{1}) = x_{1}$ .

Facts & Assumptions

Given: The one-step scalar graph, terminal loss, and exact reverse recursion.

[F1]

The terminal seed and reverse recursion are the exact derivative rule for a finite state-step objective (The reverse recursion is the exact gradient of a finite state-step objective).

Verification

1.1F1givenalgebra

Directly, $J (θ) = \frac{1}{2} θ^{2}$ , so $J^{'} (0) = 0$ . The correct seed is $p_{1} = x_{1} = 0$ at $θ = 0$ and the transition derivative is $D_{θ} x_{1} = 1$ , so the exact adjoint gradient is also zero.

2.1F1step 1.1algebra∎

If the terminal seed is incorrectly replaced by $p_{1} = 1$ , reverse accumulation returns $p_{1} D_{θ} x_{1} = 1$ at $θ = 0$ , which differs from $J^{'} (0)$ . Thus the terminal condition is load-bearing for the discrete gradient.

Depends on

The reverse recursion is the exact gradient of a finite state-step objective

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 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

William W. Hager, Runge-Kutta methods in optimal control and the transformed adjoint system, §3 (standard reference, not scraped)
Michael B. Giles and Niles A. Pierce, An Introduction to the Adjoint Approach to Design, §2.4 (standard reference, not scraped)