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{{for|the Bernstein polynomial in [[D-module]] theory|Bernstein–Sato polynomial}}
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[[Image:Bernstein Approximation.gif|thumb|right|Bernstein polynomials approximating a curve]]
In the [[mathematics|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 [[numerical stability|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 theorem|Stone–Weierstrass approximation theorem]]. With the advent of computer graphics, Bernstein polynomials, restricted to the interval ''x''&nbsp;∈&nbsp;[0,&nbsp;1], became important in the form of [[Bézier curve]]s.
 
==Definition==
The ''n''&nbsp;+&nbsp;1 '''Bernstein basis polynomials''' of degree ''n'' are defined as
 
: <math>b_{\nu,n}(x) = {n \choose \nu} x^{\nu} \left( 1 - x \right)^{n - \nu}, \quad \nu = 0, \ldots, n.</math>
 
where <math>{n \choose \nu}</math> is a [[binomial coefficient]].
 
The Bernstein basis polynomials of degree ''n'' form a [[basis (linear algebra)|basis]] for the [[vector space]] Π<sub>''n''</sub> of polynomials of degree at most&nbsp;''n''.  
 
A linear combination of Bernstein basis polynomials
 
:<math>B_n(x) = \sum_{\nu=0}^{n} \beta_{\nu} b_{\nu,n}(x)</math>
 
is called a '''Bernstein polynomial''' or '''polynomial in Bernstein form''' of degree&nbsp;''n''. The coefficients <math>\beta_\nu</math> are called '''Bernstein coefficients''' or '''Bézier coefficients'''.
 
==Example==
The first few Bernstein basis polynomials are:
 
: <math>
\begin{align}
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 \\
b_{0,3}(x) & = (1 - x)^3, & b_{1,3}(x) & = 3x(1 - x)^2, & b_{2,3}(x) & = 3x^2(1 - x), & b_{3,3}(x) & = x^3  \\
b_{0,4}(x) & = (1 - x)^4, & b_{1,4}(x) & = 4x(1 - x)^3, & b_{2,4}(x) & = 6x^2(1 - x)^2, & b_{3,4}(x) & = 4x^3(1 - x), & b_{4,4}(x) & = x^4
\end{align}
</math>
 
==Properties==
The Bernstein basis polynomials have the following properties:
* <math>b_{\nu, n}(x) = 0</math>, if <math>\nu < 0</math> or <math>\nu > n</math>.
* <math>b_{\nu, n}(0) = \delta_{\nu, 0}</math> and <math>b_{\nu, n}(1) = \delta_{\nu, n}</math> where <math>\delta</math> is the [[Kronecker delta]] function.
* <math>b_{\nu, n}(x)</math> has a root with multiplicity <math>\nu</math> at point <math>x = 0</math> (note: if <math>\nu = 0</math>, there is no root at 0).
* <math>b_{\nu, n}(x)</math> has a root with multiplicity <math>\left( n - \nu \right)</math> at point <math>x = 1</math> (note: if <math>\nu = n</math>, there is no root at 1).
* <math>b_{\nu, n}(x) \ge 0</math> for <math>x \in [0,\ 1]</math>.
* <math>b_{\nu, n}\left( 1 - x \right) = b_{n - \nu, n}(x)</math>.
 
* The [[derivative]] can be written as a combination of two polynomials of lower degree:
*: <math>b'_{\nu, n}(x) = n \left( b_{\nu - 1, n - 1}(x) - b_{\nu, n - 1}(x) \right).</math>
 
* The [[integral]] is constant for a given <math>n</math>
*: <math>\int_{0}^{1}b_{\nu, n}(x)dx = \frac{1}{n+1}  \forall \nu = 0,1 \dots n</math>
 
* If <math>n \ne 0</math>, then <math>b_{\nu, n}(x)</math> has a unique local maximum on the interval <math>[0,\ 1]</math> at <math>x = \frac{\nu}{n}</math>. This maximum takes the value:
*: <math>\nu^\nu n^{-n} \left( n - \nu \right)^{n - \nu} {n \choose \nu}.</math>
 
* The Bernstein basis polynomials of degree <math>n</math> form a [[partition of unity]]:
*: <math>\sum_{\nu = 0}^n b_{\nu, n}(x) = \sum_{\nu = 0}^n {n \choose \nu} x^\nu \left( 1 - x \right)^{n - \nu} = \left(x + \left( 1 - x \right) \right)^n = 1.</math>
 
* By taking the first derivative of <math>(x+y)^n</math> where <math>y = 1-x</math>, it can be shown that
*: <math>\sum_{\nu=0}^{n}\nu b_{\nu, n}(x) = nx</math>
 
* The second derivative of <math>(x+y)^n</math> where <math>y = 1-x</math> can be used to show
*: <math>\sum_{\nu=1}^{n}\nu(\nu-1) b_{\nu, n}(x) = n(n-1)x^2</math>
 
* A Bernstein polynomial can always be written as a linear combination of polynomials of higher degree:
*: <math>b_{\nu, n - 1}(x) = \frac{n - \nu}{n} b_{\nu, n}(x) + \frac{\nu + 1}{n} b_{\nu + 1, n}(x).</math>
 
==Approximating continuous functions==
Let ''&fnof;'' be a [[continuous function]] on the interval [0,&nbsp;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 \to \infty}{ B_n(f)(x) } = f(x) \,</math>
 
[[uniform convergence|uniformly]] on the interval&nbsp;[0,&nbsp;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 \to \infty} \sup \left\{\, \left| f(x) - B_n(f)(x) \right| \,:\, 0 \leq x \leq 1 \,\right\} = 0.</math>
 
Bernstein polynomials thus afford one way to prove the [[Stone&ndash;Weierstrass theorem#Weierstrass_approximation_theorem|Weierstrass approximation theorem]] that every real-valued continuous function on a real interval [''a'',&nbsp;''b''] can be uniformly approximated by polynomial functions over&nbsp;'''R'''.
 
A more general statement for a function with continuous ''k''<sup>th</sup> derivative is
 
: <math>{\left\| B_n(f)^{(k)} \right\|}_\infty \le \frac{ (n)_k }{ n^k } \left\| f^{(k)} \right\|_\infty \text{ and } \left\| f^{(k)}- B_n(f)^{(k)} \right\|_\infty \to 0</math>
 
where additionally
 
: <math>\frac{ (n)_k }{ n^k } = \left( 1 - \frac{0}{n} \right) \left( 1 - \frac{1}{n} \right) \cdots \left( 1 - \frac{k - 1}{n} \right)</math>
 
is an [[eigenvalue]] of ''B''<sub>''n''</sub>; the corresponding eigenfunction is a polynomial of degree&nbsp;''k''.
 
===Proof===
 
Suppose ''K'' is a [[random variable]] distributed as the number of successes in ''n'' independent [[Bernoulli trial]]s with probability ''x'' of success on each trial; in other words, ''K'' has a [[binomial distribution]] with parameters ''n'' and&nbsp;''x''. Then we have the [[expected value]] E(''K''/''n'')&nbsp;=&nbsp;''x''.
 
By the [[law of large numbers|weak law of large numbers]] of [[probability theory]],
: <math>\lim_{n \to \infty}{ P\left( \left| \frac{K}{n} - x \right|>\delta \right) } = 0</math>
for every ''&delta;''&nbsp;>&nbsp;0. Moreover, this relation holds uniformly in ''x'', which can be seen from its proof via [[Chebyshev's inequality]], taking into account that the variance of ''K''/''n'', equal to ''x''(1-''x'')/''n'', is bounded from above by 1/(4''n'') irrespective of ''x''.
 
Because ''&fnof;'', being continuous on a closed bounded interval, must be [[uniform continuity|uniformly continuous]] on that interval, one infers a statement of the form
: <math>\lim_{n \to \infty}{ P\left( \left| f\left( \frac{K}{n} \right) - f\left( x \right) \right| > \varepsilon \right) } = 0</math>
uniformly in ''x''. Taking into account that ''ƒ'' is bounded (on the given interval) one gets for the expectation
: <math>\lim_{n \to \infty}{ E\left( \left| f\left( \frac{K}{n} \right) - f\left( x \right) \right| \right) } = 0</math>
uniformly in ''x''. To this end one splits the sum for the expectation in two parts. On one part the difference does not exceed ε; this part cannot contribute more than ε.
On the other part the difference exceeds ε, but does not exceed 2''M'', where ''M'' is an upper bound for |''ƒ''(x)|; this part cannot contribute more than 2''M'' times the small probability that the difference exceeds ε.
 
Finally, one observes that the absolute value of the difference between expectations never exceeds the expectation of the absolute value of the difference, and that E(''&fnof;''(''K''/''n'')) is just the Bernstein polynomial&nbsp;''B''<sub>''n''</sub>(''ƒ'',&nbsp;''x'').
 
See for instance.<ref>L. Koralov and Y. Sinai, "Theory of probability and random processes" (second edition), Springer 2007; see page 29, Section "Probabilistic proof of the Weierstrass theorem".</ref>
 
==See also==
*[[Bézier curve]]
*[[Polynomial interpolation]]
*[[Newton polynomial|Newton form]]
*[[Lagrange polynomial|Lagrange form]]
 
==Notes==
<references />
 
==References==
* [http://www.idav.ucdavis.edu/education/CAGDNotes/Bernstein-Polynomials.pdf BERNSTEIN POLYNOMIALS by Kenneth I. Joy ]
* H. Caglar and A. N. Akansu, [http://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=224242&userType=inst "A Generalized Parametric PR-QMF Design Technique Based on Bernstein Polynomial Approximation,"] IEEE Transactions on Signal Processing, vol. 41, no. 7, pp.&nbsp;2314–2321, July 1993.
* [http://www.ams.org/featurecolumn/archive/bezier.html From Bézier to Bernstein]
* {{springer|title=Bernstein polynomials|id=B/b015730|last=Korovkin|first=P.P.}}
* {{mathworld|urlname=BernsteinPolynomial|title=Bernstein Polynomial}}
* {{PlanetMath attribution|id=9775|title=properties of Bernstein polynomial}}
 
{{DEFAULTSORT:Bernstein Polynomial}}
[[Category:Numerical analysis]]
[[Category:Polynomials]]
[[Category:Articles containing proofs]]

Revision as of 14:06, 28 February 2014

Many individuals aren't positive how to begin trying to find a great legal professional. Legal difficulties can be produced very much worse when a particular person carries a poor attorney. As a result, you should carefully read the below post as a way to learn how to find the best legal representative to help relieve your anxieties.



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Come up with a spending budget with regards to paying for legal costs. If you contact an lawyer plus they are above your price range, try to look for somebody else. As there is no problem with going just a little around your financial allowance, you may not wish to go with a legal representative that you may have issues trying to pay out.

A great tip to bear in mind when getting a legal professional will be extremely wary of any lawyer who seems more interested in acquiring paid out than profitable your situation. There are many dishonest legal professionals around which will try to help you to pay for a contingency charge, as well as enable you to get to mortgage loan your home.

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