Bernstein–Bézier form

Remarks

On $[0, 1]$ the basis consists of the Bernstein basis polynomials $\binom{n}{k} x^k (1-x)^{n-k}$. Since the basis functions are nonnegative on $[a, b]$ and sum to $1$, the values of $p$ on $[a, b]$ lie between $\min_k c_k$ and $\max_k c_k$, and $p(a) = c_0$, $p(b) = c_n$. Example: on $[0, 1]$ with $n = 2$, $x = \tfrac{1}{2} \cdot 2x(1-x) + 1 \cdot x^2$, so $x$ has Bézier coefficients $0, 1/2, 1$. Non-example: $p(x) = x^3$ has no Bernstein–Bézier form of degree $2$, since it is not a polynomial of degree at most $2$; one must take $n \ge 3$. In several variables one uses barycentric coordinates $\lambda_0, \dots, \lambda_d$ of a simplex and the basis $\frac{n!}{\alpha_0! \cdots \alpha_d!} \lambda_0^{\alpha_0} \cdots \lambda_d^{\alpha_d}$ with $\abs{\alpha} = n$.