Bernstein polynomial

Bernstein polynomial

In the mathematical field 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.

Definition

The "n" + 1 Bernstein basis polynomials of degree "n" are defined as

:b_{ u,n}(x) = {n choose u} x^{ u} (1-x)^{n- u}, qquad u=0,ldots,n.

where {n choose u} is a binomial coefficient.

The Bernstein basis polynomials of degree "n" form a basis for the vector space Pi_n of polynomials of degree "n".

A linear combination of Bernstein basis polynomials

:B(x) = sum_{ u=0}^{n} eta_{ u} b_{ u,n}(x)

is called a Bernstein polynomial or polynomial in Bernstein form of degree "n". The coefficients βν are called Bernstein coefficients or Bézier coefficients.

Example

The first few Bernstein basis polynomials are:b_{0,0}(x) = 1 ,

:b_{0,1}(x) = 1-x ,: b_{1,1}(x) = x ,

:b_{0,2}(x) = (1-x)^2 ,:b_{1,2}(x) = 2x(1-x) ,: b_{2,2}(x) = x^2 .

Properties

The Bernstein basis polynomials have the following properties:
*b_{ u,n}(x) = 0, if &nu; < 0 or &nu; > "n"
*b_{ u,n}(0) = delta_{ u,0} and b_{ u,n}(1) = delta_{ u,n} where delta is the Kronecker delta function.
*b_{ u,n}(x) has a root with multiplicity &nu; at point "x" = 0 (note if &nu; is 0 there is no root at 0)
*b_{ u,n}(x) has a root with multiplicity "n" − &nu; at point "x" = 1 (note if &nu; = "n" there is no root at 1)
*b_{ u,n}(x) &ge; 0 for "x" in [0,1]
*b_{ u,n}(1 - x) = b_{n- u,n}(x)
*The derivative can be written as a combination of two polynomials of lower degree:::b'_{ u,n}(x) = nleft(b_{ u-1,n-1}(x) - b_{ u,n-1}(x) ight)

*If &nu; &ne; 0, then b_{ u,n}(x) has a unique local maximum on the interval [0,1] at "x" = &nu;/"n". This maximum takes the value:: u^{ u}n^{-n}(n- u)^{n- u}{nchoose u}.

*The Bernstein basis polynomials of degree "n" form a partition of unity:

::sum_{ u=0}^n b_{ u,n}(x) = sum_{ u=0}^n {n choose u} x^{ u}(1-x)^{n- u} = (x+(1-x))^n = 1.

*A Bernstein polynomial can always be written as a linear combination of polynomials of higher degree.

::b_{ u,n-1}(x)=frac{n- u}{n}b_{ u,n}(x)+frac{ u+1}{n}b_{ u+1,n}(x).

Approximating continuous functions

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

:B_n(f)(x) = sum_{ u=0}^{n} fleft(frac{ u}{n} ight) b_{ u,n}(x).

It can be shown that :lim_{n ightarrowinfty} 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 ightarrowinfty} sup{,left|f(x)-B_n(f)(x) ight|:0leq xleq 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 ightarrow 0,

where additionally {(n)_k over n^k}= left(1-{0 over n} ight)left(1-{1 over n} ight) cdots left(1-{k-1 over n} ight) is an eigenvalue of B_n; 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

:lim_{n oinfty}Pleft(left|frac{K}{n}-x ight|>delta ight)=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 ightarrowinfty} Pleft(left|f(K/n)-f(x) ight|>varepsilon ight)=0.

Consequently

:lim_{n ightarrowinfty}Pleft(left|f(K/n)-E(f(K/n)) ight|+left|E(f(K/n))-f(x) ight|>varepsilon ight)=0.

:lim_{n ightarrowinfty}Pleft(left|fleft(frac{K}{n} ight)-Eleft(fleft(frac{K}{n} ight) ight) ight|>varepsilon/2 ight)+Pleft(left|Eleft(fleft(frac{K}{n} ight) ight)-f(x) ight|>varepsilon/2 ight)=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").

ee also

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

References

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