Bernstein polynomial

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

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 Weierstrass approximation theorem. With the advent of computer graphics, Bernstein polynomials, restricted to the interval , 1 became important in the form of Bézier curves.

Definition

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

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

where $\tbinom$ is a binomial coefficient.

So, for example, $b_(x) = \tbinomx^2(1-x)^3 = 10x^2(1-x)^3.$

The first few Bernstein basis polynomials for blending 1, 2, 3 or 4 values together are:
: $\begin b_(x) & = 1, \\ b_(x) & = 1 - x, & b_(x) & = x \\ b_(x) & = (1 - x)^2, & b_(x) & = 2x(1 - x), & b_(x) & = x^2 \\ b_(x) & = (1 - x)^3, & b_(x) & = 3x(1 - x)^2, & b_(x) & = 3x^2(1 - x), & b_(x) & = x^3 \end$
:

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. A linear combination of Bernstein basis polynomials

:$B_n(x) = \sum_^ \beta_ b_(x)$

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

Properties

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

Approximating continuous functions

Let ''ƒ'' be a continuous function on the interval , 1 Consider the Bernstein polynomial
:$B_n(f)(x) = \sum_^n f\left( \frac \right) b_(x).$

It can be shown that
:$\lim_ = f$

uniformly on the interval  , 1Natanson (1964) p. 6

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 $\mathbb R$.Natanson (1964) p. 3

A more general statement for a function with continuous ''k''^th derivative is
:$_\infty \le \frac \left\, f^ \right\, _\infty \quad\ \text \quad\ \left\, f^- B_n(f)^ \right\, _\infty \to 0,$
where additionally
:$\frac = \left( 1 - \frac \right) \left( 1 - \frac \right) \cdots \left( 1 - \frac \right)$
is an eigenvalue of ''B''_''n''; 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\left frac\right= x\ and
:$p(K) = x^ \left( 1 - x \right)^ = b_(x)$

By the weak law of large numbers of probability theory,
:$\lim_ = 0$

for every ''δ'' > 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'', equal to  ''x''(1−''x''), is bounded from above by irrespective of ''x''.

Because ''ƒ'', being continuous on a closed bounded interval, must be uniformly continuous on that interval, one infers a statement of the form
:$\lim_ = 0$

uniformly in ''x''. Taking into account that ''ƒ'' is bounded (on the given interval) one gets for the expectation
: $\lim_ = 0$
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 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
:\operatorname\left \left(\frac\right)\right= \sum_^n f\left(\frac\right) p(K) = \sum_^n f\left(\frac\right) b_(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:

# $\sum_k x^k (1-x)^ = 1$ ("probability")
# $\sum_k x^k (1-x)^ = x$ ("mean")
# $\sum_k \left( x -\right)^2 x^k (1-x)^ = .$ ("variance")

In fact, by the binomial theorem

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

and this equation can be applied twice to $t\frac$. 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

:$b_(x) = x^k (1-x)^,$

and let

:$f_n(x) = \sum_k f(k/n)\, b_(x).$

Thus, by identity (1)

:f_n(x) - f(x) = \sum_k (k/n) - f(x)\,b_(x),

so that

:$, f_n(x) - f(x), \le \sum_k , f(k/n) - f(x), \, b_(x).$

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

:$, f_n(x) - f(x), \le \sum_ , f(k/n) - f(x), \, b_(x) + \sum_ , f(k/n) - f(x), \, b_(x) .$

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 2''M'' times

:$\sum_ b_(x) \le \sum_k \delta^ \left(x -\right)^2 b_(x) = \delta^ < \delta^ n^.$

:( Chebyshev's inequality)

It follows that the polynomials ''f''_''n'' tend to ''f'' uniformly.

Generalizations to higher dimension

Bernstein polynomials can be generalized to dimensions – the resulting polynomials have the form . In the simplest case only products of the unit interval are considered; but, using affine transformations of the line, Bernstein polynomials can also be defined for products . For a continuous function on the -fold product of the unit interval, the proof that can be uniformly approximated by

:$\sum_ \sum_ \cdots \sum_ \cdots f\left(, , \dots, \right) x_1^ (1-x_1)^ x_2^ (1-x_2)^ \cdots x_k^ (1-x_k)^$

is a straightforward extension of Bernstein's proof in one dimension.

See also

*Polynomial interpolation
*Newton form
*Lagrange form
* Binomial QMF (also known as Daubechies wavelet)

Notes

References

*, English translation
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*, Russian edition first published in 1940
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External links

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* from University of California, Davis. Note the error in the summation limits in the first formula on page 9.
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* Feature Column from American Mathematical Society
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