Alphabeta Math
LemmaStatement: AI-adaptedProof: 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.

A computed unitary radix-two butterfly has error at most (10u + εtw) times its input norm

Statement

Let u≤1/16, let ∣w∣=1, and let ∣w^−w∣≤ϵ≤1/4. For a,b∈C, perform the successful butterfly of Computed unitary radix-two FFT with certified twiddle phases using w^. If each exact nonzero real intermediate is normal and every operation succeeds under the operationwise model, then its computed pair (y^+,y^−) satisfies

∥(y^+,y^−)−2−1/2(a+wb,a−wb)∥2≤(10u+ϵ)∥(a,b)∥2.

Facts & Assumptions

Given: a,b,w,w^,u,ϵ with the stated bounds, and a successful rounded butterfly.

[F5]

For a pair (z1,z2)∈C2, ∥(z1,z2)∥2=(∣z1∣2+∣z2∣2)1/2 (The counting inner product on CZ/NZ).

[F1]

Six-operation complex multiplication obeys ∣fl⁡C(zw)−zw∣≤3u∣z∣∣w∣ when its six operations succeed (Six-operation complex multiplication has a normwise roundoff bound).

[F3]

Each real rounded addition, multiplication, or positive in-range square root has relative error at most u; complex addition and real scaling round their two components separately (The standard relative floating-point model and unit roundoff, Operationwise real and complex floating point arithmetic with rounded square roots and fused multiply-add).

Proof

1.1givenF1F2algebra

Let t=fl⁡C(w^b). By [F1] and [F2], ∣t−wb∣≤3u∣w^∣∣b∣+ϵ∣b∣≤(15u/4+ϵ)∣b∣. Thus ∣t∣≤(1+15u/4+ϵ)∣b∣≤95∣b∣/64.

1.2givenF3F4algebra

For q±=fl⁡C(a±t), [F3] gives ∣q±−(a±t)∣≤u(∣a∣+∣t∣) by [F4]. If c=1/2 and c^=fl⁡(1/2), then correct rounding gives ∣c^−c∣≤uc. The componentwise rounded scaling therefore satisfies ∣fl⁡C(c^q±)−cq±∣≤(2u+u2)c∣q±∣, and ∣q±∣≤(1+u)(∣a∣+∣t∣).

2.1step 1.1step 1.2algebra

Combining steps 1.1–1.2 and using 2c=1, the two-output error norm is at most (15u/4+ϵ)∣b∣+α(∣a∣+95∣b∣/64), where α=u+(2u+u2)(1+u)=3u+3u2+u3≤817u/256. Since 15/4+(817/256)(95/64)<9, this is at most (817u/256)∣a∣+(9u+ϵ)∣b∣.

3.1step 2.1F4F5algebra∎

By [F4], (817/256)2+92<10, so (817u/256)∣a∣+9u∣b∣≤10u∣a∣2+∣b∣2; [F5] identifies the square-root factor as ∥(a,b)∥2. Also ϵ∣b∣≤ϵ∣a∣2+∣b∣2. Adding the bounds proves the claim.

Depends on

Used by

Dependency tree · two levels

43 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