Advantage of Bernstein polynomial basis

Question

The well-known "Bernstein polynomials" on the interval [0,1] are defined as
$$ B_{N,i}(x)=\binom{n}{i}x^{i}(1-x)^{n-i}, \ \ i=0,...,N. $$ My question is about advantage of these polynomials in approximation theory:

If fact, all of $B_{N,i}(x),\ \ i=0,...,n$ are polynomials exactly degree $n$, i.e. $$ B_{N,0}(x)=(1-x)^{n},\ ...., B_{N,N}(x)=x^n, $$
but in other polynomials such as Legendre polynomials the n-th polynomial is of degree $n$. Is it an advantage of Bernestion polynomials that every of them is exactly degree n?

Answer 1

I assume your question is about the advantage of Bernstein polynomials compared with Legendre polynomials.

For numerical applications, there is no advantage. The convergence speed of Bernstein polynomials for continuous function approximation is very bad!

Usually, Bernstein polynomials are introduced in approximation theory lectures to prove the theorem of Weierstraß:

For each $f \in C[a, b]$ and each $\varepsilon > 0$ there is a polynomial $p$ so that $\|f − p\|_{\infty} < \varepsilon$.

The proof uses the fact that the Bernstein polynomials form a Korovkin sequence.

For practical applications, the Chebyshev polynomials are much more important than the Bernstein or Legendre polynomials.