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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 x &isin; [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 &beta;&nu; sono detti  coefficienti di Bernstein.

Osservazioni
I Polinomi di base di Bernstein hanno le seguenti proprietà:
 * $$b_{\nu,n}(x) = 0$$, se &nu; < 0 o &nu; > 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 &nu; in x = 0 (oss: se &nu; è 0 non ci sono radici in 0)
 * $$b_{\nu,n}(x)$$ ha una radice con moltiplicità n &minus; &nu; nel punto x = 1 (oss: se &nu; = n non ci sono radici in 1)
 * $$b_{\nu,n}(x)$$ &ge; 0 per x in [0,1]
 * $$b_{\nu,n}(1 - x) = b_{n-\nu,n}(x) $$
 * SE &nu; &ne; 0, allora $$b_{\nu,n}(x)$$ ha un solo massimo locale in [0,1] in x = &nu;/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


 * $$\| B_n(f)^{(k)} \|_\infty \le {(n)_k \over n^k} \| f^{(k)} \|_\infty$$ 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 Bn(f,x).

Vedi anche

 * Bézier curve
 * Polynomial interpolation
 * Newton form
 * Lagrange form

Riferimenti


Bernsteinpolynom Polynôme de Bernstein Polinomio di Bernstein Wielomiany Bernsteina Polinómios de Bernstein Bernsteinpolynom