Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)
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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.

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Constant BBHT stage success beyond the threshold scale

Statement

Let 0<a=t/N<3/4, θ=arcsin⁡a, and let J be uniform on {1,…,M}. If the positive integer M satisfies M≥4/a, then E[sin⁡2((2J+1)θ)]≥38.

Facts & Assumptions

Given: 0<a<3/4, sin⁡θ=a, θ∈(0,π/2), and an integer M≥4/a.

[F1]

Exact success probability after an integer number of Grover iterates: after j integer iterates, the marked probability is sin⁡2((2j+1)θ).

[F3]

The addition formulas for sine and cosine: sine and cosine obey their addition formulas.

[F4]

Double-angle and quadratic power-reduction identities: sin⁡(2θ)=2sin⁡θcos⁡θ and sin⁡2x=(1−cos⁡2x)/2.

[F5]

Principal inverse sine and inverse cosine and Signs, monotonicity intervals, and ranges of sine and cosine: the principal angle is in [−π/2,π/2], sine is strictly increasing there, and cosine is positive on (0,π/2).

[F6]

Parity and the Pythagorean identity for sine and cosine: sin⁡2θ+cos⁡2θ=1.

[F9]
[F10]

Squaring is monotone on the nonnegatives: square roots preserve order on nonnegative reals.

Proof

1.1givenF4F5F6F7F8F9F10algebra

Since 0<a<3/4, [F7] and [F10] give 0<a<1. From [F4], sin⁡θ=a and θ∈[−π/2,π/2]. By [F8], sin⁡0=0, so strict increase [F5] gives θ>0. Since sin⁡(π/2)=1 by [F9], strict increase and sin⁡θ<1 give θ<π/2; then cos⁡θ>0 by [F5]. The Pythagorean identity [F6] gives cos⁡θ=1−a.

1.2givenF1F4algebra

By [F1] and [F4], the average equals 12−12M∑j=1Mcos⁡((4j+2)θ).

2.1step 1.2F3algebra

By [F3], for every j, 2sin⁡(2θ)cos⁡((4j+2)θ)=sin⁡((4j+4)θ)−sin⁡(4jθ). Summing from j=1 to M and using sin⁡u−sin⁡v=2sin⁡((u−v)/2)cos⁡((u+v)/2) gives ∑j=1Mcos⁡((4j+2)θ)=sin⁡(2Mθ)cos⁡((2M+4)θ)/sin⁡(2θ). Thus the average in step 1.2 is 12−sin⁡(2Mθ)cos⁡((2M+4)θ)2Msin⁡(2θ).

3.1givenstep 1.1step 2.1F4F5F6F7F10algebra∎

Since a<3/4, 1−a>1/4, so [F10] gives 1−a>1/2. Steps 1.1 and [F4]–[F7] then give sin⁡(2θ)=2a1−a>a. Therefore Msin⁡(2θ)>4 and the absolute value of the correction in step 2.1 is less than 1/8. The average is greater than 3/8, hence at least 3/8.

Depends on

Used by

Dependency tree · two levels

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Sources