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How statement and proof provenance work

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Branchwise AD for absolute value and the nondifferentiable switch

Example

The primitive $f (x) = ∣ x ∣$ has two strict branches: $f (x) = x$ for $x > 0$ and $f (x) = - x$ for $x < 0$ . Branchwise AD returns derivative $1$ on the positive branch and $- 1$ on the negative branch. At the switch $x = 0$ it returns NONDIFFERENTIABLE.

Facts & Assumptions

Given: The absolute-value primitive with the two strict branch formulas.

[F1]

Finite straight-line computational programs and real-arithmetic tapes: a piecewise primitive uses the derivative of its unique strict active branch and returns NONDIFFERENTIABLE at a branch boundary.

Proof

technique · one-sided difference quotients

1.1F1givenalgebra

For $x > 0$ , the active formula is $f (x) = x$ , whose derivative is $1$ ; for $x < 0$ , the active formula is $f (x) = - x$ , whose derivative is $- 1$ . These are the two derivatives used by the strict-branch rule in [F1].

1.2givenalgebra

At zero, the right difference quotient is $(∣ h ∣ - 0) / h = 1$ for $h > 0$ , while the left quotient is $(∣ h ∣ - 0) / h = - 1$ for $h < 0$ . Since the one-sided limits differ, the derivative does not exist at zero.

2.1F1step 1.2∎

By [F1], the branch boundary therefore returns NONDIFFERENTIABLE and no derivative; selecting a generalized derivative would require a separately supplied rule.

Depends on

Finite straight-line computational programs and real-arithmetic tapes

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 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

MIT OpenCourseWare 18.S096, Matrix Calculus for Machine Learning and Beyond, complete lecture notes, Chapter 8 §§8.3–8.4 (standard reference, not scraped)