Bernstein polynomial
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- For the Bernstein polynomial in D-module theory, see Bernstein-Sato 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.
A numerically stable way to evaluate polynomials in Bernstein form is de Casteljau's algorithm.
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.
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Definition
The n + 1 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 a Bernstein polynomial or polynomial in Bernstein form of degree n. The coefficients βν are called Bernstein coefficients or Bézier coefficients.
Notes
The Bernstein basis polynomials have the following properties:
- bν,n(x) has a root with multiplicity ν at point x = 0
- bν,n(x) has a root with multiplicity n − ν at point x = 1
- bν,n(x) ≥ 0 if x ∈ [0,1]
- bν,n(x) has a maximum at x = ν/n on the interval [0,1]
- b’ν,n(x) = n [bν − 1, n − 1(x) − bν,n − 1(x)]
- bν,n(x) = 0, if ν < 0 or ν > n
- bν,n(0) = δν0 and bν,n(1) = δνn where δ is the Kronecker delta function
- bn−ν,n(x) = bν,n(1 − x)
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_{2,2}(x) = x^2.\,</math>
Approximating continuous functions
Let f(x) be a continuous function on the interval [0, 1]. Consider the Bernstein polynomial
- <math>B_n(f)(x) = \sum_{\nu=0}^{n} f\left(\frac{\nu}{n}\right) b_{\nu,n}(x).</math>
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 real-valued continuous function on a real interval [a,b] can be uniformly approximated by polynomial functions over R.
A more general statement for a function with continuous k-th derivative is
- <math>\| B_n(f)^{(k)} \|_\infty \le {(n)_k \over n^k} \| f^{(k)} \|_\infty</math> and <math>\|f^{(k)}- B_n(f)^{(k)} \|_\infty \rightarrow 0</math>,
where additionally <math>{(n)_k \over n^k}= \left(1-{0 \over n}\right)\left(1-{1 \over n}\right) \cdots \left(1-{k-1 \over n}\right)</math> is an eigenvalue of <math>B_n</math>; the corresponding eigenfunction is a polynomial of degree k.
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 Bn(f,x).