I am currently learning numerical method and I am somehow confused about the Bezier Curve, Bernstein polynomial and control polygon.
For example, if we have a curve $p(u)=-12u^2+12u+1$,and we want to find the Bezier representation of degree 3 in the interval $[1,4]$ and draw the control polygon.
- In many textbooks, they say that I should change the variable $u$ to $t=\frac{u-a}{b-a}$, if $u$ is defined on arbitrary interval $[a,b]$, but how should I exactly do it? Do I write it as $\displaystyle p(u)=\sum^{n}_{i=1} \beta_i B^n_i\left(\frac{u-a}{b-a}\right)$ and then find $\beta_i$ by solving a system of linear equations? Can anyone show my by this example?
- And my second question is about its geometric interpretation. Again, in textbook and also Wikipedia, they call $\beta_i$ the control points of control polygon. However, I really couldn't understand in our example that how we can draw a control polygon only by single value not by points? For example, what do $\beta_i$ mean exactly geometrically? Again, can you show me by our examples?
Part 1: Yes, you use the substitution $t=(u-a)/(b-a)$. Or, reversing this, $u = a + (b-a)t$. So, in your example $p(u)=-12u^2+12u+1 \quad (1 \le u \le 4)$, we put $u = 1 + 3t$, and we get $p(t) = -12(3t+1)^2 + 12(3t+1) + 1 \quad (0 \le t \le 1).$
Part 2: You are dealing with real-valued functions. The $\beta_i$ are just numbers, so calling them "control points" is not really appropriate. However, all the Bezier-Bernstein formulae you see will work regardless of whether the $\beta_i$ are in $\mathbb R$, $\mathbb R^2$, $\mathbb R^3$, or in any $\mathbb R^n$. In $\mathbb R^2$ and $\mathbb R^3$, the term "control point" certainly makes sense, but it's a little strange in $\mathbb R^1$. Some people would use the term "Bezier ordinates" for the $\beta_i$ in the case of $\mathbb R^1$.
Take a look at these notes, especially section 2.14.