Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generated
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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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Rational bounds for pi used in finite schedules

Statement

The circle constant satisfies 3<π<22/7.

Facts & Assumptions

Given: The principal inverse tangent and the usual positive constant π.

[F1]

arctan⁡x=∫0x(1+t2)−1 dt and π/4=∑n=0∞(−1)n/(2n+1) (Principal arctangent: derivative, integral, power series, and the Gregory–Leibniz series). The positive terms 1/(2n+1) decrease to zero; the alternating-series test places the sum between consecutive partial sums and bounds the error by the first omitted term (The alternating series test: if (bk) is nonincreasing with bk→0 then ∑k(−1)kbk converges, the sum lies between any two consecutive partial sums, and the error after n terms is at most bn, Every complete ordered field is Archimedean).

[F2]

The definition identifies π/2 as the first positive zero of cosine and fixes the same circle constant used in the arctangent identity (Pi as twice the smallest positive zero of cosine).

Proof

1.1givenF1F2algebra

Group the first eight terms of the alternating series in [F1] into four positive pairs. Their sum is 2/3+2/35+2/99+2/195=33976/45045>3/4, since 4⋅33976=135904>135135=3⋅45045. The alternating-series remainder after an even partial sum is positive, so π/4>3/4 and π>3.

2.1givenF1F2algebra∎

For 0<x<1, the integrand x4(1−x)4/(1+x2) is positive. Polynomial division gives x4(1−x)4/(1+x2)=x6−4x5+5x4−4x2+4−4/(1+x2). Its integral over [0,1] is 1/7−2/3+1−4/3+4−4arctan⁡(1)=22/7−π by [F1]. The integral is strictly positive, hence π<22/7.

Depends on

Used by

Dependency tree · two levels

35 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