Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-02
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The Bartle-Sherbert bounds 2.828 < pi < 3.185

Example

Let γ=π/2. Then 2<γ<6−23,2.828<π<3.185.

Facts & Assumptions

Verification

technique · direct
1.1

At x=2, the first two cosine terms cancel and the alternating tail beginning with x4/4! is positive, so cos⁡(2)>0.

L2L3
1.2

Put a=6−23. Then 1−a2/2+a4/24=0, and the remaining alternating cosine tail begins negative with decreasing absolute terms, so cos⁡a<0.

L2L3algebra
2.1

Strict decrease and [L1] give 2<γ<a.

step 1.1step 1.2L1
3.1

Since (707/500)2<2, 22>707/250. Since (433/250)2<3 and (637/400)2>6−2(433/250), one has a<637/400, hence 2a<637/200.

step 2.1L3algebra
4.1

Doubling the bounds of step 2.1 proves the displayed decimal bounds for π.

step 2.1step 3.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

24 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