Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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.

The zeroth, first, and second centred moments of the Bernstein basis

Statement

For n≥1n\ge1, 0≤x≤10\le x\le1, and bn,k(x)=ι(nk)xk(1−x)n−kb_{n,k}(x)=\iota\binom nkx^k(1-x)^{n-k},

∑k=0nbn,k=1,∑k=0nknbn,k=x,∑k=0n(kn−x)2bn,k=x(1−x)n.\sum_{k=0}^n b_{n,k}=1,\qquad \sum_{k=0}^n\frac{k}{n}b_{n,k}=x,\qquad \sum_{k=0}^n\left(\frac{k}{n}-x\right)^2b_{n,k}=\frac{x(1-x)}n.

Facts & Assumptions

Given: n≥1n\ge1 and x∈[0,1]x\in[0,1].

[L1]

The binomial theorem gives (u+v)n=∑k=0nι(nk)ukvn−k(u+v)^n=\sum_{k=0}^n\iota\binom nk u^kv^{n-k} (The binomial theorem in R\mathbb{R}: (x+y)n=∑k<n+1ι ⁣(nk) xky n−k(x+y)^{n} = \sum_{k<n+1} \iota\!\binom{n}{k}\, x^{k} y^{\,n-k}).

[L2]

Finite sums are additive, scale, and split as in Laws of finite sums and finite products.

Proof

technique · direct
1.1

Apply [L1] to u=xu=x and v=1−xv=1-x to obtain the zeroth identity.

L1algebra
1.2

Reindex the terms k(nk)=n(n−1k−1)k\binom nk=n\binom{n-1}{k-1} and apply [L1] at exponent n−1n-1 to obtain the first identity.

L1L2algebra
2.1

Apply the same reindexing to k(k−1)(nk)=n(n−1)(n−2k−2)k(k-1)\binom nk=n(n-1)\binom{n-2}{k-2} when n≥2n\ge2, combine it with step 1.2, and expand the square.

step 1.2L1L2algebra
3.1

For n=1n=1 the displayed centred identity is checked directly; together with step 2.1 this proves it for all n≥1n\ge1.

step 2.1algebra∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 62 results over 19 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources