Editing Bernstein polynomial (section)

====Motivation====

We will first give intuition for Bernstein's original proof. A continuous function on a compact interval must be uniformly continuous. Thus, the value of any continuous function can be uniformly approximated by its value on some finite net of points in the interval. This consideration renders the approximation theorem intuitive, given that polynomials should be flexible enough to match (or nearly match) a finite number of pairs <math>(x, f(x))</math>. To do so, we might (1) construct a function close to <math>f</math> on a lattice, and then (2) smooth out the function outside the lattice to make a polynomial. 

The probabilistic proof below simply provides a constructive method to create a polynomial which is approximately equal to <math>f</math> on such a point lattice, given that "smoothing out" a function is not always trivial. Taking the expectation of a random variable with a simple distribution is a common way to smooth. Here, we take advantage of the fact that Bernstein polynomials look like Binomial expectations. We split the interval into a lattice of ''n'' discrete values.  Then, to evaluate any ''f(x)'', we evaluate ''f'' at one of the ''n'' lattice points close to ''x'', randomly chosen by the Binomial distribution.  The expectation of this approximation technique is polynomial, as it is the expectation of a function of a binomial RV. The proof below illustrates that this achieves a uniform approximation of ''f''. The crux of the proof is to (1) justify replacing an arbitrary point with a binomially chosen lattice point by concentration properties of a Binomial distribution, and (2) justify the inference from <math>x \approx X</math> to <math>f(x) \approx f(X)</math> by uniform continuity.