Talk:Bernstein polynomial

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how is the pointwise convergence works for this polynomial? — Preceding unsigned comment added by 130.113.226.6 (talk) 19:33, 22 April 2004 (UTC)

Bernstein polynomial stability

Where does this notion that deCastlejau's is the stable way to evaluate Bernstein polynomials come from? I don't see it in the references for this article, and as discussed in http://mae.engr.ucdavis.edu/~farouki/bernstein.pdf (one of the references on the Bezier Curves Wikipedia page): "it was shown that, in fact, the Bernstein basis is “optimally stable”". As far as I know, no sane person actually writes code that evaluates Bezier curves using de Castlejau's algorithm, unless they are doing it to teach about the properties of Bezier curves, or if the only thing they know about Bezier curves is what they read on Wikipedia (which currently seems to suggest that if you want numerical stability you have to use the inefficient de Castlejau's algo). Most code in the real world evaluates Beziers using Bernstein polynomials. — Preceding unsigned comment added by 2620:0:1009:0:baac:6fff:fe98:e716 (talk) 00:40, 26 April 2013 (UTC)

Merging Bezier curve & Bernstein polynomial

I want to merge Bézier curve and Bernstein polynomial. I do not care what the resulting article is called. The two article are talking about the same subject and so there is a lot of duplicated material and inconsistent notation. Any comments ?MathMartin 12:53, 19 Sep 2004 (UTC)

Wrong statement

b_ν,n(x) = n · (b_{ν − 1,n − 1}(x) − b_{ν,n − 1}(x))???

is it just me or is this equation wrong?!

Looks wrong to me:

b2,4(3) = 216
b1,3(3) = 36
b2,3(3) = -54
4 * (36 - -54) = 360

Last time I checked, 360 and 216 were different numbers.

I'm going to pull that statement from the article.

shouldn't that equation be for the derivative?

${\tfrac {d}{dx}}b_{\nu ,n}(x)=n\cdot (b_{\nu -1,n-1}(x)-b_{\nu ,n-1}(x))$ —MoA)gnome (talk) 17:29, 7 February 2008 (UTC)

Yes, second that. Also just proved this on paper.Methossant (talk) 16:14, 29 May 2009 (UTC)

Proof problem?

Could you please refer to this proof? Johnnyaug (talk) 01:07, 21 January 2010 (UTC)

I am not sure what the "proof" is proving, but it seems the logic of the proof is wrong to me. The triangle inequality is used backwards after the word "Consequently". Also

$\lim _{n\rightarrow \infty }P(|f({\frac {K}{n}})-E(f({\frac {K}{n}}))|>\epsilon /2)=0$

can't be true since $E(f( \frac{K}{n} )$ is a constant and $f( \frac{K}{n} )$ most likely is not. --MarcMehlman (talk) 19:01, 21 September 2008 (UTC)

"Can't be true"? You're saying a sequence of random variables cannot converge in probability to a constant. If you're right about that, then the weak law of large numbers is not true. Michael Hardy (talk) 21:57, 21 January 2010 (UTC)

Second on that. That's rather old and I'm also not sure the proof proves anything. Can someone clarify that please? Johnnyaug (talk) 20:44, 20 January 2010 (UTC)

Re triangle inequality: Start with |a|+|b|≥|a+b| the inequality as I remember it from school. Then |a+b|>ε implies |a|+|b|>ε. So if a and b are random variables, P(|a+b|>ε)≤P(|a|+|b|>ε). Therefore P(|a|+|b|>ε)=0 implies P(|a|+|b|>ε)=0 is the form of the triangle inequality being used here which seems to be the opposite of what is used in the proof. Someone else should check this, I tend to get a bit dyslexic around inequalities.--RDBury (talk) 14:01, 21 January 2010 (UTC)

The proof is attempting the show that B_n converges uniformly to f on [0,1]. The use of probability is throwing me off a bit and it's not clear to me that probability =1 implies certainty here. Does anyone have a proof that uses more standard techniques?--RDBury (talk) 14:32, 21 January 2010 (UTC)

As far as I've heard the most standard proof does use probability, I just didn't hear enough to actually know how to prove it. Johnnyaug (talk) 15:45, 21 January 2010 (UTC)

First of all, such a proof is well-known and contained in many textbooks; here is one (and I can give more): 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". Another source: Durrett, Richard "Probability: theory and examples" (second edition); see Example 5.1 on page 37. See also Probabilistic proofs of non-probabilistic theorems. On the other hand, of course, analytic proofs are also well-known (and appeared earlier); here is the idea of one: consider the convolution of the given function and the polynomial $(1-x^{2})^{n}.$

Second, it may happen that the proof is distorted here; I'll look closely. Boris Tsirelson (talk) 19:48, 21 January 2010 (UTC)

Yes, after the word "Consequently" the proof is distorted. The point should be like this: we see that f(K/n) converges to f(x) in probability (analysts would say, in measure); also, f(K/n) are uniformly bounded; convergence of expectations (analysts would say, integrals) follows. Boris Tsirelson (talk) 20:34, 21 January 2010 (UTC). In other words: the rescaled binomial distribution (of K/n) converges weakly (analysts would say, weak^*) to the degenerate distribution (of x), the atom at the point x. And the weak convergence of measures implies convergence of integrals of the bounded continuous function f. However, one should bother to get uniform convergence...Boris Tsirelson (talk) 06:41, 22 January 2010 (UTC)

Well, the proof is now correct. Boris Tsirelson (talk) 20:49, 24 January 2010 (UTC)

Go above it?

Specifically, the word uniformly signifies that the approximation will meet the original function, but not go above it…

I don't know what "go above it" means here, but if it means that the approximation is bounded above by the true function, then it is not only wrong but contradicted by the graphic at the top of the page. — Preceding unsigned comment added by 207.10.142.163 (talk) 01:47, 6 October 2011 (UTC)

Yes, you are right. The wrong phrase is now deleted. Thank you. Boris Tsirelson (talk) 13:20, 6 October 2011 (UTC)

Plot is Incorrect

Correct me if I'm wrong, but I can't think of a Bernstein polynomial that would look like the one depicted in the plot.

Specifically, at the value x=1, the Bernstein polynomial always has value zero, since it has a factor of (1-x) in it. The only case in which it doesn't have this zero value is when v=x, in which case, the Bernstein polynomial just becomes x^n, which certainly doesn't look like the plot shown.

Am I missing something?

Monsterman222 (talk) 00:40, 2 December 2013 (UTC)

Probably you miss the distinction between "Bernstein polynomial" and "Bernstein basis polynomial". Boris Tsirelson (talk) 06:14, 2 December 2013 (UTC)

Reference to planet math does not work

It seems something is wrong in the attribution page for planet math, but I am not insightful enough to correct it. — Preceding unsigned comment added by Dansynek (talk • contribs) 05:47, 3 January 2014 (UTC)