LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02

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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.

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

Statement

For $n \geq 1$ , $0 \leq x \leq 1$ , and $b_{n, k} (x) = ι (\binom{n}{k}) x^{k} (1 - x)^{n - k}$ ,

$\sum_{k = 0}^{n} b_{n, k} = 1, \sum_{k = 0}^{n} \frac{k}{n} b_{n, k} = x, \sum_{k = 0}^{n} {(\frac{k}{n} - x)}^{2} b_{n, k} = \frac{x (1 - x)}{n} .$

Facts & Assumptions

Given: $n \geq 1$ and $x \in [0, 1]$ .

[L1]

The binomial theorem gives $(u + v)^{n} = \sum_{k = 0}^{n} ι (\binom{n}{k}) u^{k} v^{n - k}$ (The binomial theorem in $R$ : $(x + y)^{n} = \sum_{k < n + 1} ι ⁣ (\binom{n}{k}) x^{k} y^{n - k}$ ).

[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 $u = x$ and $v = 1 - x$ to obtain the zeroth identity.

L1algebra

1.2

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

L1L2algebra

2.1

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

step 1.2L1L2algebra

3.1

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

step 2.1algebra∎

Depends on

Used by

Dependency tree · two levels

23 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

Bernstein polynomial (Encyclopedia of Mathematics) (standard reference, not scraped)