Diegodp

For the Bernstein polynomial in D-module theory, see Bernstein-Sato polynomial.

Bernstein polynomials approximating a curve

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 x ∈ [0, 1], became important in the form of Bézier curves.

Definizione

I n + 1 Polinomi di base di Bernstein di grado n sono definiti come

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

dove

${n \choose \nu }$

è un coefficiente binomiale.

I Polinomi di base di Bernstein di grado n formano una base per lo [spazio vettoriale |[vector space]] $\Pi _{n}$ dei polimomi di grado n.

Una combinazione lineare di Polinomi di base di Bernstein

$B(x)=\sum _{\nu =0}^{n}\beta _{\nu }b_{\nu ,n}(x)$

è detta un polinomio di Bernstein di grado n. I coefficienti β_ν sono detti coefficienti di Bernstein.

Osservazioni

I Polinomi di base di Bernstein hanno le seguenti proprietà:

• $b_{\nu ,n}(x)=0$, se ν < 0 o ν > n
• $b_{\nu ,n}(0)=\delta _{\nu ,0}$ e $b_{\nu ,n}(1)=\delta _{\nu ,n}$ ove $\delta$ è la funzione delta di Kronecker.
• $b_{\nu ,n}(x)$ ha una radice con molteplicità ν in x = 0 (oss: se ν è 0 non ci sono radici in 0)
• $b_{\nu ,n}(x)$ ha una radice con moltiplicità n − ν nel punto x = 1 (oss: se ν = n non ci sono radici in 1)
• $b_{\nu ,n}(x)$ ≥ 0 per x in [0,1]
• $b_{\nu ,n}(1-x)=b_{n-\nu ,n}(x)$
• SE ν ≠ 0, allora $b_{\nu ,n}(x)$ ha un solo massimo locale in [0,1] in x = ν/n. Il suo valore è

$\nu ^{\nu }n^{-n}(n-\nu )^{n-\nu }{n \choose \nu }.$

• I Polinomi di base di Bernstein di grado n formano una [partition of unity|[partizione dell' unità]]:

$\sum _{\nu =0}^{n}b_{\nu ,n}(x)=\sum _{\nu =0}^{n}{n \choose \nu }x^{\nu }(1-x)^{n-\nu }=(x+(1-x))^{n}=1.$

Esempi

The first few Bernstein basis polynomials are

$b_{0,0}(x)=1\,$

$b_{0,1}(x)=1-x{\mbox{ , }}b_{1,1}(x)=x\,$

$b_{0,2}(x)=(1-x)^{2}{\mbox{ , }}b_{1,2}(x)=2x(1-x){\mbox{ , }}b_{2,2}(x)=x^{2}.\,$

Approssimazione di funzioni continue

Let f(x) be a continuous function on the interval [0, 1]. Consider the Bernstein polynomial

$B_{n}(f)(x)=\sum _{\nu =0}^{n}f\left({\frac {\nu }{n}}\right)b_{\nu ,n}(x).$

It can be shown that

$\lim _{n\rightarrow \infty }B_{n}(f)(x)=f(x)$

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

$\lim _{n\rightarrow \infty }\sup\{\,\left|f(x)-B_{n}(f)(x)\right|:0\leq x\leq 1\,\}=0.$

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

$\|B_{n}(f)^{(k)}\|_{\infty }\leq {(n)_{k} \over n^{k}}\|f^{(k)}\|_{\infty }$ and $\|f^{(k)}-B_{n}(f)^{(k)}\|_{\infty }\rightarrow 0,$

where additionally ${(n)_{k} \over n^{k}}=\left(1-{0 \over n}\right)\left(1-{1 \over n}\right)\cdots \left(1-{k-1 \over n}\right)$ is an eigenvalue of $B_{n}$; the corresponding eigenfunction is a polynomial of degree k.

Dimostrazione

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

$\lim _{n\to \infty }P\left(\left|{\frac {K}{n}}-x\right|>\delta \right)=0$

for every $\delta >0$.

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

$\lim _{n\rightarrow \infty }P\left(\left|f(K/n)-f(x)\right|>\varepsilon \right)=0.$

Consequently

$\lim _{n\rightarrow \infty }P\left(\left|f(K/n)-E(f(K/n))\right|+\left|E(f(K/n))-f(x)\right|>\varepsilon \right)=0.$

$\lim _{n\rightarrow \infty }P\left(\left|f\left({\frac {K}{n}}\right)-E\left(f\left({\frac {K}{n}}\right)\right)\right|>\varepsilon /2\right)+P\left(\left|E\left(f\left({\frac {K}{n}}\right)\right)-f(x)\right|>\varepsilon /2\right)=0.$

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).

Vedi anche

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

Riferimenti

• Weisstein, Eric W. "Bernstein Polynomial". MathWorld.