Bernstein polynomial

In the mathematical field of numerical analysis, a Bernstein polynomial is a polynomial expressed as a linear combination of Bernstein basis polynomials. The idea is named after mathematician Sergei Natanovich Bernstein.

Polynomials in Bernstein form were first used by Bernstein in a constructive proof for the Weierstrass approximation theorem. With the advent of computer graphics, Bernstein polynomials, restricted to the interval [0, 1], became important in the form of Bézier curves.

A numerically stable way to evaluate polynomials in Bernstein form is de Casteljau's algorithm.



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


 * $$b_{\nu,n}(x) \mathrel{:}\mathrel{=} \binom{n}{\nu} x^{\nu} \left( 1 - x \right)^{n - \nu}, \quad \nu = 0, \ldots, n,$$

where $$\tbinom{n}{\nu}$$ is a binomial coefficient.

So, for example, $$b_{2,5}(x) = \tbinom{5}{2}x^2(1-x)^3 = 10x^2(1-x)^3.$$

The first few Bernstein basis polynomials for blending 1, 2, 3 or 4 values together are:

\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 \end{align} $$

The Bernstein basis polynomials of degree n form a basis for the vector space $$\Pi_n$$ of polynomials of degree at most n with real coefficients.

Bernstein polynomials
A linear combination of Bernstein basis polynomials


 * $$B_n(x) \mathrel{:}\mathrel{=} \sum_{\nu=0}^{n} \beta_{\nu} b_{\nu,n}(x)$$

is called a Bernstein polynomial or polynomial in Bernstein form of degree n. The coefficients $$\beta_\nu$$ are called Bernstein coefficients or Bézier coefficients.

The first few Bernstein basis polynomials from above in monomial form are:

\begin{align} b_{0,0}(x) & = 1, \\ b_{0,1}(x) & = 1 - 1x, & b_{1,1}(x) & = 0 + 1x \\ b_{0,2}(x) & = 1 - 2x + 1x^2, & b_{1,2}(x) & = 0 + 2x - 2x^2, & b_{2,2}(x) & = 0 + 0x + 1x^2 \\ b_{0,3}(x) & = 1 - 3x + 3x^2 - 1x^3, & b_{1,3}(x) & = 0 + 3x - 6x^2 + 3x^3, & b_{2,3}(x) & = 0 + 0x + 3x^2 - 3x^3, & b_{3,3}(x) & = 0 + 0x + 0x^2 + 1x^3 \end{align} $$

Properties
The Bernstein basis polynomials have the following properties: 0 &\text{if } i \neq j,  \\ 1 &\text{if } i=j. \end{cases}$$
 * $$b_{\nu, n}(x) = 0$$, if $$\nu < 0$$ or $$\nu > n.$$
 * $$b_{\nu, n}(x) \ge 0$$ for $$x \in [0,\ 1].$$
 * $$b_{\nu, n}\left( 1 - x \right) = b_{n - \nu, n}(x).$$
 * $$b_{\nu, n}(0) = \delta_{\nu, 0}$$ and $$b_{\nu, n}(1) = \delta_{\nu, n}$$ where $$\delta$$ is the Kronecker delta function: $$\delta_{ij} = \begin{cases}
 * $$b_{\nu, n}(x)$$ has a root with multiplicity $$\nu$$ at point $$x = 0$$ (note: if $$\nu = 0$$, there is no root at 0).
 * $$b_{\nu, n}(x)$$ has a root with multiplicity $$\left( n - \nu \right)$$ at point $$x = 1$$ (note: if $$\nu = n$$, there is no root at 1).
 * The derivative can be written as a combination of two polynomials of lower degree: $$b'_{\nu, n}(x) = n \left( b_{\nu - 1, n - 1}(x) - b_{\nu, n - 1}(x) \right).$$
 * The k-th derivative at 0: $$b_{\nu, n}^{(k)}(0) = \frac{n!}{(n - k)!} \binom{k}{\nu} (-1)^{\nu + k}.$$
 * The k-th derivative at 1: $$b_{\nu, n}^{(k)}(1) = (-1)^k b_{n - \nu, n}^{(k)}(0).$$
 * The transformation of the Bernstein polynomial to monomials is $$b_{\nu,n}(x) = \binom{n}{\nu}\sum_{k=0}^{n-\nu} \binom{n-\nu}{k}(-1)^{n-\nu-k} x^{\nu+k} = \sum_{\ell=\nu}^n \binom{n}{\ell}\binom{\ell}{\nu}(-1)^{\ell-\nu}x^\ell,$$ and by the inverse binomial transformation, the reverse transformation is $$x^k = \sum_{i=0}^{n-k} \binom{n-k}{i} \frac{1}{\binom{n}{i}} b_{n-i,n}(x) = \frac{1}{\binom{n}{k}} \sum_{j=k}^n \binom{j}{k}b_{j,n}(x).$$
 * The indefinite integral is given by $$\int b_{\nu, n}(x) \, dx = \frac{1}{n+1} \sum_{j=\nu+1}^{n+1} b_{j, n+1}(x).$$
 * The definite integral is constant for a given n: $$\int_0^1 b_{\nu, n}(x) \, dx = \frac{1}{n+1} \quad\ \, \text{for all } \nu = 0,1, \dots, n.$$
 * If $$n \ne 0$$, then $$b_{\nu, n}(x)$$ has a unique local maximum on the interval $$[0,\, 1]$$ at $$x = \frac{\nu}{n}$$. This maximum takes the value $$\nu^\nu n^{-n} \left( n - \nu \right)^{n - \nu} {n \choose \nu}.$$
 * The Bernstein basis polynomials of degree $$n$$ form a partition of unity: $$\sum_{\nu = 0}^n b_{\nu, n}(x) = \sum_{\nu = 0}^n {n \choose \nu} x^\nu \left(1 - x\right)^{n - \nu} = \left(x + \left( 1 - x \right) \right)^n = 1.$$
 * By taking the first $$x$$-derivative of $$(x+y)^n$$, treating $$y$$ as constant, then substituting the value $$y = 1-x$$, it can be shown that $$\sum_{\nu=0}^{n} \nu b_{\nu, n}(x) = nx.$$
 * Similarly the second $$x$$-derivative of $$(x+y)^n$$, with $$y$$ again then substituted $$y = 1-x$$, shows that $$\sum_{\nu=1}^{n}\nu(\nu-1) b_{\nu, n}(x) = n(n-1)x^2.$$
 * A Bernstein polynomial can always be written as a linear combination of polynomials of higher degree: $$b_{\nu, n - 1}(x) = \frac{n - \nu}{n} b_{\nu, n}(x) + \frac{\nu + 1}{n} b_{\nu + 1, n}(x).$$
 * The expansion of the Chebyshev Polynomials of the First Kind into the Bernstein basis is $$T_n(u) = (2n-1)!! \sum_{k=0}^n \frac{(-1)^{n-k}}{(2k-1)!!(2n-2k-1)!!} b_{k,n}(u).$$

Approximating continuous functions
Let &fnof; 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
 * $$\lim_{n \to \infty}{ B_n(f) } = f $$

uniformly on the interval [0, 1].

Bernstein polynomials thus provide one way to prove the Weierstrass approximation theorem that every real-valued continuous function on a real interval [a, b] can be uniformly approximated by polynomial functions over $$\mathbb R$$.

A more general statement for a function with continuous kth derivative is
 * $${\left\| B_n(f)^{(k)} \right\|}_\infty \le \frac{ (n)_k }{ n^k } \left\| f^{(k)} \right\|_\infty \quad\ \text{and} \quad\ \left\| f^{(k)}- B_n(f)^{(k)} \right\|_\infty \to 0,$$

where additionally
 * $$\frac{ (n)_k }{ n^k } = \left( 1 - \frac{0}{n} \right) \left( 1 - \frac{1}{n} \right) \cdots \left( 1 - \frac{k - 1}{n} \right)$$

is an eigenvalue of Bn; the corresponding eigenfunction is a polynomial of degree k.

Probabilistic proof
This proof follows Bernstein's original proof of 1912. See also Feller (1966) or Koralov & Sinai (2007).

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 $$\operatorname{\mathcal E}\left[\frac{K}{n}\right] = x\ $$ and
 * $$p(K) = {n \choose K} x^{K} \left( 1 - x \right)^{n - K} = b_{K,n}(x)$$

By the weak law of large numbers of probability theory,
 * $$\lim_{n \to \infty}{ P\left( \left| \frac{K}{n} - x \right|>\delta \right) } = 0$$

for every &delta; > 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 $1/n$ K, equal to $1/n$ x(1&minus;x), is bounded from above by $1/(4n)$ irrespective of x.

Because &fnof;, being continuous on a closed bounded interval, must be uniformly continuous on that interval, one infers a statement of the form
 * $$\lim_{n \to \infty}{ P\left( \left| f\left( \frac{K}{n} \right) - f\left( x \right) \right| > \varepsilon \right) } = 0$$

uniformly in x. Taking into account that ƒ is bounded (on the given interval) one gets for the expectation
 * $$\lim_{n \to \infty}{ \operatorname{\mathcal E}\left( \left| f\left( \frac{K}{n} \right) - f\left( x \right) \right| \right) } = 0$$

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 2M, where M is an upper bound for |&fnof;(x)|; this part cannot contribute more than 2M 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
 * $$\operatorname{\mathcal E}\left[f\left(\frac{K}{n}\right)\right] = \sum_{K=0}^n f\left(\frac{K}{n}\right) p(K) = \sum_{K=0}^n f\left(\frac{K}{n}\right) b_{K,n}(x) = B_n(f)(x)$$

Elementary proof
The probabilistic proof can also be rephrased in an elementary way, using the underlying probabilistic ideas but proceeding by direct verification:

The following identities can be verified:


 * 1) $$ \sum_k {n \choose k} x^k (1-x)^{n-k} = 1$$ ("probability")
 * 2) $$ \sum_k {k\over n} {n \choose k} x^k (1-x)^{n-k} = x$$ ("mean")
 * 3) $$ \sum_k \left( x -{k\over n}\right)^2 {n \choose k} x^k (1-x)^{n-k} = {x(1-x)\over n}. $$ ("variance")

In fact, by the binomial theorem

$$(1+t)^n = \sum_k {n \choose k} t^k,$$

and this equation can be applied twice to $$t\frac{d}{dt}$$. The identities (1), (2), and (3) follow easily using the substitution $$t = x/ (1 - x)$$.

Within these three identities, use the above basis polynomial notation


 * $$ b_{k,n}(x) = {n\choose k} x^k (1-x)^{n-k},$$

and let


 * $$ f_n(x) = \sum_k f(k/n)\, b_{k,n}(x).$$

Thus, by identity (1)


 * $$f_n(x) - f(x) = \sum_k [f(k/n) - f(x)] \,b_{k,n}(x), $$

so that


 * $$|f_n(x) - f(x)| \le  \sum_k |f(k/n) - f(x)| \, b_{k,n}(x).$$

Since f is uniformly continuous, given $$\varepsilon > 0$$, there is a $$\delta > 0$$ such that $$|f(a) - f(b)| < \varepsilon$$ whenever $$|a-b| < \delta$$. Moreover, by continuity, $$M= \sup |f| < \infty$$. But then


 * $$ |f_n(x) - f(x)| \le \sum_{|x -{k\over n}|< \delta} |f(k/n) - f(x)|\, b_{k,n}(x) + \sum_{|x -{k\over n}|\ge \delta} |f(k/n) - f(x)|\, b_{k,n}(x) .$$

The first sum is less than ε. On the other hand, by identity (3) above, and since $$|x - k/n| \ge \delta$$, the second sum is bounded by $$2M$$ times


 * $$\sum_{|x - k/n|\ge \delta} b_{k,n}(x) \le \sum_k \delta^{-2} \left(x -{k\over n}\right)^2 b_{k,n}(x)  = \delta^{-2} {x(1-x)\over n} <  {1\over4} \delta^{-2} n^{-1}.$$


 * (Chebyshev's inequality)

It follows that the polynomials fn tend to f uniformly.

Generalizations to higher dimension
Bernstein polynomials can be generalized to $k$ dimensions – the resulting polynomials have the form $B_{i_{1}}(x_{1}) B_{i_{2}}(x_{2}) ... B_{i_{k}}(x_{k})|undefined$. In the simplest case only products of the unit interval $[0,1]$ are considered; but, using affine transformations of the line, Bernstein polynomials can also be defined for products $[a_{1}, b_{1}] × [a_{2}, b_{2}] × ... × [a_{k}, b_{k}]$. For a continuous function $f$ on the $k$-fold product of the unit interval, the proof that $f(x_{1}, x_{2}, ..., x_{k})$ can be uniformly approximated by


 * $$\sum_{i_1} \sum_{i_2} \cdots \sum_{i_k} {n_1\choose i_1} {n_2\choose i_2} \cdots {n_k\choose i_k}

f\left({i_1\over n_1}, {i_2\over n_2}, \dots, {i_k\over n_k}\right) x_1^{i_1} (1-x_1)^{n_1-i_1} x_2^{i_2} (1-x_2)^{n_2-i_2} \cdots x_k^{i_k} (1-x_k)^{n_k - i_k} $$ is a straightforward extension of Bernstein's proof in one dimension.