Bernstein Polynomials

Subjects: Special Polynomials
Links: Rectifiable Curves in Rn
A Bernstein polynomial is a polynomial of the form:

b_{ν, n} (x) = (\binom{n}{ν}) x^{ν} (1 - x)^{n - ν}

The set ${b_{ν, n}}_{ν = 0}^{n}$ is a basis for $P_{n} (R) = Π_{n}$ or the polynomials with real coefficients of at most degree $n$ , then:

p (x) = \sum_{ν = 0}^{n} β_{ν} b_{ν, n} (x)

is the Bernstein form of the polynomial $p (x)$ , the coefficietns are called Bernstein coefficients or Bézier coefficients.

Given a collection of points ${P_{k}}_{k = 0}^{n}$ , then the Bézier Curve with $P_{k}$ as its control points it’s:

B (t) = \sum_{k = 0}^{n} P_{k} b_{k, n} (t)

and $t \in [0, 1]$ .
In particular it has a neat property, that for any $n \in N$ :

\sum_{ν = 0}^{n} b_{ν, n} = 1

$b_{ν, n} (x) = 0$ if $ν < 0$ or $ν > n$

$b_{ν, n} (x) \geq 0$ , for $x \in [0, 1]$

$b_{ν, n} (1 - x) = b_{n - ν, n} (x)$

$b_{ν, n} (0) = δ_{ν, 0}$ and $b_{ν, n} (1) = δ_{ν, n}$ , where $δ_{i j}$ is the Kronecker delta

$b_{ν, n} (x)$ has a root with multiplicity $ν$ at $x = 0$ , if $ν = 0$ , then there’s no root

$b_{ν, n} (x)$ has a root with multiplicity $n - ν$ at $x = 1$ , if $ν = n$ , then there’s no root

$b_{ν, n}^{'} (x) = n (b_{ν - 1, n - 1} (x) - b_{ν, n - 1} (x))$

If $n \neq 0$ , then $b_{ν, n} (x)$ has a unique local maximum on $[0, 1]$ at $x = \frac{ν}{n}$ , with the value

ν^{ν} n^{- n} (n - ν)^{n - ν} (\binom{n}{ν})

We can show that

\sum_{ν = 0}^{n} ν b_{ν, n} (x) = n x \sum_{ν = 0}^{n} ν (ν - 1) b_{ν, n} (x) = n (n - 1) x^{2}

The antiderivative of $b_{ν, n}$ is given by

\int b_{ν, n} (x) d x = \frac{1}{n + 1} \sum_{j = ν + 1}^{n + 1} b_{j, n + 1} (x)

the integral over $[0, 1]$ of $b_{ν, n}$ is

\int_{0}^{1} b_{ν, n} (x) d x = \frac{1}{n + 1}

The transformations of Berstein Polynomials to monomials is

b_{ν, n} (x) = (\binom{n}{ν}) \sum_{k = 0}^{n - ν} (\binom{n - ν}{k}) (- 1)^{n - ν - k} x^{ν + k} = \sum_{k = ν}^{n} (\binom{n}{k}) (\binom{k}{ν}) (- 1)^{k - ν} x^{k}

using the inverse binomial transformation we get that

x^{k} = \sum_{i = 0}^{n - k} (\binom{n - k}{i}) \frac{1}{(\binom{n}{i})} b_{n - i, n} (x) = \frac{1}{(\binom{n}{k})} \sum_{j = k}^{n} (\binom{j}{k}) b_{j, n} (x)