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.

A symmetric Gaussian alternating sum vanishes in the Serre range

Statement

Let m≥1, and let d∈Z satisfy 1−m≤d≤m−1 and d≡m−1(mod2). With the symmetric Gaussian binomial Cm,n=(mn)ν from Quantum integers, factorials, Gaussian binomials and divided powers at qi, ∑n=0m(−1)nν−dnCm,n=0 for every real ν>1.

Facts & Assumptions

Given: The integer range and parity above, and the symmetric Gaussian coefficient convention.

[F1]

The Gauss product formula is a Laurent-polynomial identity and remains valid after substituting ν−1 for its parameter (The quantum Pascal recurrences, the Gauss product formula and Gaussian integrality).

Proof

technique · direct
1.1givenF1algebra

Substituting ν−1 for the parameter in [F1] gives ∏j=0m−1(1+ν−2jz)=∑n=0mν−n(m−1)Cm,nzn. This identity is valid at the real number ν>1 because both sides are Laurent polynomials. Set z=−νm−1−d; the right side becomes exactly ∑n=0m(−1)nν−dnCm,n.

2.1givenstep 1.1algebra∎

The parity assumption makes j=(m−1−d)/2 an integer, and the bounds on d give 0≤j≤m−1. The factor indexed by this j on the left side of step 1.1 is 1−ν−2jνm−1−d=0. Therefore the right side, which is the sum in the Statement, is zero.

Depends on

Used by

Dependency tree · two levels

6 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