Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04
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.

Fourier partial sums of the sawtooth

Example

Assume the Axiom of Countable Choice.

Let f be the one-periodic sawtooth given by f(0)=0 and f(x)=x12 for 0<x<1. Then f^(0)=0 and, for k0,

f^(k)=12πik.

Hence

SNf(x)=k=1Nsin(2πkx)πk.

At every noninteger x, SNf(x)f(x), while at every integer x, SNf(x)0.

Facts & Assumptions

Given: The Axiom of Countable Choice and the one-periodic sawtooth f(0)=0 and f(x)=x12 for 0<x<1.

[L1]

Fourier coefficients and partial sums are defined by the one-period formulas in Period-one Fourier coefficients, partial sums, and convolution on the torus.

[L2]

Assuming the Axiom of Countable Choice, a one-periodic bounded-variation function converges at each point to the midpoint of its one-sided limits under Fourier partial sums (Dirichlet-Jordan pointwise convergence).

Verification

technique · direct
1.1

Since 01(x12)dx=0, [L1] gives f^(0)=0. For k0, direct integration gives f^(k)=01(x12)e2πikxdx=12πik.

L1algebra
2.1

Insert the coefficients from step 1.1 into the partial-sum formula [L1]. Pairing the k and k terms gives SNf(x)=k=1Nsin(2πkx)πk.

L1step 1.1algebra
3.1

The sawtooth is piecewise C1, hence of bounded variation on one period. If xZ, choose the unique integer m with y:=xm(0,1). Because both f and its Fourier partial sums are one-periodic, one has SNf(x)=SNf(y) and f(x)=f(y)=y12. The one-sided limits at y therefore both equal f(x), so [L2] gives SNf(x)f(x). At an integer x, the one-sided limits are 1/2 and 1/2, whose midpoint is 0, so [L2] gives SNf(x)0.

L2step 2.1choosealgebra

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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