Bernstein polynomial

In the mathematical subfield of numerical analysis, a Bernstein polynomial, named after Sergei Natanovich Bernstein, is a polynomial in the Bernstein form, that is a linear combination of Bernstein basis polynomials.

Polynomials in Bernstein form were first used by Bernstein in a constructive proof for the Stone-Weierstrass approximation theorem. With the advent of computer graphics Bernstein polynomials, restricted to the interval [0,1], became important in the form of Bézier curves.

Polynomials in Bernstein form can be efficiently evaluated using De Casteljau's algorithm.

Definition

The n Bernstein basis polynomials of degree n are defined as

<math>b_{\nu,n}(x) = {n \choose \nu} x^{\nu} (1-x)^{n-\nu}, \qquad \nu=0,\ldots,n.<math>

The Bernstein basis polynomials of degree n form a basis for the vector space <math>\Pi_n<math> of polynomials of degree n.

A linear combination of Bernstein basis polynomials

<math>B(x) = \sum_{\nu=0}^{n} \beta_{\nu} b_{\nu,n}(x)<math>

is called Bernstein polynomial or polynomial in Bernstein form of degree n. The coefficients β_ν are called Bernstein coefficients or Bézier coefficients.

Notes

The Bernstein basis polynomials have the following properties

• b_ν,n(x) has a root with multiplicity ν at point x = 0
• b_ν,n(x) has a root with multiplicity n − ν at point x = 1
• b_ν,n(x) ≥ 0 if x in [0,1]
• b_ν,n(x) has a global maximum at x = ν/n

The Bernstein basis polynomials of degree n form a partition of unity

<math>\sum_{\nu=0}^n b_{\nu,n}(x) = \sum_{\nu=0}^n {n \choose \nu} x^{\nu}(1-x)^{n-\nu} = (x+(1-x))^n = 1.<math>

Example

The first few Bernstein basis polynomials are

<math>b_{0,0}(x) = 1<math>

<math>b_{0,1}(x) = 1-x \mbox{ , } b_{1,1}(x) = x<math>

<math>b_{0,2}(x) = (1-x)^2 \mbox{ , } b_{1,2}(x) = 2x(1-x) \mbox{ , } b_{0,2}(x) = x^2<math>

A theorem

It can be shown that

<math>\lim_{n\rightarrow\infty} B_n(f,x)=f(x)<math>

uniformly on the interval [0, 1]. This is a stronger statement than the proposition that the limit holds for each value of x separately; that would be pointwise convergence rather than uniform convergence. specifically, the word uniformly signifies that

<math>\lim_{n\rightarrow\infty}\sup\{\,\left|f(x)-B_n(f,x)\right|:0\leq x\leq 1\,\}=0.<math>

Bernstein polynomials thus afford one way to prove the Stone-Weierstrass approximation theorem that every continuous function on a closed bounded interval can be uniformly approximated by polynomial functions.

Proof

Suppose K is a random variable distributed as the number of successes in n independent Bernoulli trials with probability x of success on each trial; in other words, K has a binomial distribution with parameters n and x. Then we have the expected value E(K/n) = x.

Then the weak law of large numbers of probability theory tells us that

<math>\lim_{n\rightarrow\infty}P(\left|(K/n)-x\right|>\delta)=0.<math>

Because f, being continuous on a closed bounded interval, must be uniformly continuous on that interval, we can infer a statement of the form

<math>\lim_{n\rightarrow\infty} P(\left|f(K/n)-f(x)\right|>\varepsilon)=0.<math>

Consequently

<math>\lim_{n\rightarrow\infty}

P(\left|f(K/n)-E(f(K/n))\right|+\left|E(f(K/n))-f(x)\right|>\varepsilon)=0.<math>

<math>\lim_{n\rightarrow\infty}

P(\left|f(K/n)-E(f(K/n))\right|>\varepsilon/2)+P( \left|E(f(K/n))-f(x)\right|>\varepsilon/2)=0.<math>

And so the second probability above approaches 0 as n grows. But the second probability is either 0 or 1, since the only thing that is random is K, and that appears within the scope of the expectation operator E. Finally, observe that E(f(K/n)) is just the Bernstein polynomial B_n(f,x).

See also

• Bézier curve
• Polynomial interpolation
• Newton form
• Lagrange form