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The Bernstein polynomials of $f (x) = x^{2}$ equal $x^{2} + x (1 - x) / n$

Statement

For $f (x) = x^{2}$ and $n \geq 1$ , $B_{n} (f) (x) = x^{2} + x (1 - x) / n$ . Hence $∥ B_{n} (f) - f ∥_{\infty} = 1 / (4 n)$ . For $n = 0$ , $B_{0} (f) = 0$ .

Facts & Assumptions

Given: $f (x) = x^{2}$ on $[0, 1]$ .

[L1]

The first two Bernstein binomial moments are those of The zeroth, first, and second centred moments of the Bernstein basis.

[L2]

The Bernstein polynomial is defined in The Bernstein polynomial $B_{n} (f)$ on $[0, 1]$ .

Proof

technique · calculation

1.1

For $n \geq 1$ , substitute $f (k / n) = (k / n)^{2}$ in the definition and use [L1] to obtain $B_{n} (f) = x^{2} + x (1 - x) / n$ .

L1L2algebra

2.1

Since $0 \leq x (1 - x) \leq 1 / 4$ with equality at $x = 1 / 2$ , the stated supremum error is $1 / (4 n)$ .

step 1.1algebra

3.1

The separate definition at degree zero gives $B_{0} (f) = f (0) = 0$ .

L2algebra∎

Depends on

Used by

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Sources

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

The Bernstein polynomials of f(x)=x2f(x)=x2 equal x2+x(1−x)/nx2+x(1−x)/n