Distance of a Bézier curve and control points

Question

Derive a formula (independent of $n$) for $ H(P_n, B_n)$ if $B_n$ is the Bézier curve which is determined by the control points $ p_0 := (−4, 0), p_1 = p_2 = . . . = p_{n−1} := (0, 4) $ and $ p_n := (4, 0) $, and $P_n = \{p_0,\ldots,p_n\}$. Here $ \quad H(A, B) \quad is \quad sup_{a∈A} (inf_{b∈B} d(a, b))$ for two non-empty subsets $A$, $B$ of $\mathbb{R}^2 $ also $d(a, b)$ denotes the Euclidean distance of $ a, b ∈ \mathbb{R}^2 $.

Answer 1

The Bézier curve looks like this:

It's obvious by symmetry that the maximum distance between the curve and its control points occurs at the mid-point $q = B_n(\tfrac12)$ of the curve. If you don't believe that it's obvious, you can reason as follows: the control points are symmetric about the $y$-axis, so the curve will also be symmetric. So, at the point where it crosses the $y$-axis, the curve tangent must be horizontal. This means that the line from $p_1, \ldots, p_{n-1}$ to $q$ is normal to the curve at $q$, so $q$ is the point on the curve that is closest to $p_1, \ldots, p_{n-1}$.

Let $\phi_0, \ldots, \phi_n$ be the Bernstein polynomials of degree $n$, then the point $q$ is given by $$ q = \phi_0(\tfrac12)p_0 + \phi_1(\tfrac12)p_1 + \cdots + \phi_n(\tfrac12)p_n $$ So, using the coordinates of $p_0, \ldots, p_n$, the $y$-coordinate of the point $q$ is \begin{align} &(0)\phi_0(\tfrac12) + (4)\phi_1(\tfrac12) + (4)\phi_2(\tfrac12) + \cdots + (4)\phi_{n-1}(\tfrac12) + (0)\phi_n(\tfrac12) \\ = \;&4\big[\phi_0(\tfrac12) + \phi_1(\tfrac12) + \cdots + \phi_n(\tfrac12)\big] - 4\phi_0(\tfrac12) - 4\phi_n(\tfrac12) \\ = \;&4 - 4(1-\tfrac12)^n - 4(\tfrac12)^n \\ = \;&4 - (\tfrac12)^{n-3} \\ \end{align} So the desired distance is $(\tfrac12)^{n-3}$. Note that this is not independent of $n$; in fact, it decreases as $n$ increases. This should make sense -- putting more control points at the location $(0,4)$ will pull the curve more strongly towards this point.