Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-27
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The dyadic simple approximants for x2 on [0,2] at k=1 and k=2

Example

For f(x)=x2 on [0,2], the explicit dyadic truncations are:

s1(x)={0,0x<1/2,1/2,1/2x<1,1,1x2,

and

s2(x)={0,0x<1/2,1/4,1/2x<2/2,1/2,2/2x<3/2,3/4,3/2x<1,1,1x<5/2,5/4,5/2x<6/2,3/2,6/2x<7/2,7/4,7/2x<2,2,2x2.

Facts & Assumptions

Given: The nonnegative measurable function f(x)=x2 on [0,2] and the explicit dyadic truncations from the simple-approximation theorem.

[L1]

For a nonnegative measurable function, the approximants are

sk=j<k2kj2k1{j2kf<(j+1)2k}+k1{fk}.

Verification

technique · direct
1.1

For k=1, the two dyadic cells are [L1, algebra] 0x2<1/2 and 1/2x2<1, while the truncation cell is x21. On [0,2] these become exactly the three intervals displayed in the first formula.

L1algebra
2.1

For k=2, the eight nontrivial dyadic cells are [step 1.1, L1, algebra] j/4x2<(j+1)/4 for j=0,,7, together with the truncation cell x22. On [0,2] these become [0,1/2), [1/2,2/2), [2/2,3/2), [3/2,1), [1,5/2), [5/2,6/2), [6/2,7/2), [7/2,2), and [2,2], exactly as displayed.

step 1.1L1algebra

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