Control Points of Bézier Curve?

Question

Could someone give me a reason/proof why the control points do not lie on the Bézier Curve? Perhaps involving Bernstein Polynomials, if possible?

Thanks!

Answer 1

I assume that by "control points" you mean the intermediate control points, because the curve interpolates between the first control point and the last one.

Your statement as phrased is not true in all cases. However, if I guess at your intention...

Consider that a point $P(t)$ on the curve defined by control points $P_i$ is $P(t) = \Sigma_{i=0}^n P_i b_{i,n}(t)$ (using Wikipedia's notation for the Bernstein basis polynomials). For the right hand side of that to be identically $P_j$ we require $b_{j,n}(t) = 1$ and $\forall k\ne j : b_{k,n}(t) = 0$. But the only cases for which this is possible are $j=0$, $t=0$ and $j=n$, $t=1$.

To prove that latter statement, take the formula for the Bernstein basis polynomials: $b_{k,n}(t) = {}^{n}C_{k} t^k (1-t)^{n-k}$. For this to evaluate to $0$ when $0 \le k \le n$ and $k, n \in \mathbb{N}$ we require $t=0$ or $t=1$. It is left as an exercise to demonstrate that these correspond to the first and last control points.

Answer 2

Note that the Bernstein basis polynomials are just the terms of the binomial expansion of $(t + (1-t))^n$.

Since this expression equals 1, they sum to 1. For $t$ between 0 and 1 inclusive, they are all positive. So each point on the Bezier curve is a simple weighted average of the control points.

If the control points all lie on their convex hull and are not collinear, any weighted average of them must lie within the convex hull.

(If the points do not all lie on their convex hull -- or they are collinear -- your statement is not necessarily true; the curve certainly might pass through a control point in the interior of the convex hull.)