Patches

Bernstein Polynomials

$B_{i}^{n} (t) = (\begin{matrix} n \\ i \end{matrix}) t^{i} (1 - t)^{n - i}$
$B_{i}^{n} (t) = \frac{n!}{(n - i)! i!} t^{i} (1 - t)^{n - i}$

These have been generalized for two dimensions in Bezier Triangles.

Bernstein polynomials form a basis, which I believe lets them be used as a sum to form a polynomial approximation. I would expect that the patches too would approximate the surface and approach it as the number of sample points goes to infinity.

Bernstein Polynomial Approximation

Let f be continuous on [0,1].
$\sum_{k = 0}^{n} f (\frac{k}{n}) B_{k}^{n} (x) \to f (x)$
as $n \to \infty$

Bezier Patches

These are described as rectangular patches. The four corners are fixed and the surface passes through these. All other points are viewed as control points.

For example a cubic bezier surface has two control points all sides. If the tangent is known at the corner points then the inner control points can be used to interpolate the tangent and the corner points.

The bezier patch is most generally defined with Bernstein polynomials.

$p (u, v) = \sum_{i = 0}^{m} \sum_{j = 0}^{n} B_{i}^{m} (u) B_{j}^{n} (v) p_{i, j}$

$p (u, v) = \sum_{i = 0}^{m} \sum_{j = 0}^{n} (\begin{matrix} m \\ i \end{matrix}) (\begin{matrix} n \\ j \end{matrix}) u^{i} (1 - u)^{m - i} v^{j} (1 - v)^{n - j} p_{i, j}$

$\frac{\partial p}{\partial u} = m \sum_{i = 0}^{m - 1} \sum_{j = 0}^{n} B_{i}^{m - 1} (u) B_{j}^{n} (v) (p_{i+1, j} - p_{i, j})$
$\frac{\partial p}{\partial v} = n \sum_{i = 0}^{m} \sum_{j = 0}^{n - 1} B_{i}^{m} (u) B_{j}^{n - 1} (v) (p_{i, j+1} - p_{i, j})$

Bezier Triangles

$B_{ijk}^{n} (u, v) = \frac{n!}{i! j! k!} u^{i} v^{j} (1 - u - v)^{k}$
where $i + j + k = n$

$p (u, v) = \sum_{i + j + k = n}^{} B_{ijk}^{n} (u, v) p_{ijk}$

This is a fancy way of iterating through the combinations and may be a little off putting. Here is the simplest way of iterating through the combinations.

$\sum_{i + j + k = n}^{} = \sum_{i = 0}^{n} \sum_{j = 0}^{n - i} \sum_{k = n - i - j}^{n - i - j}$

$\frac{\partial p}{\partial u} = \sum_{i + j + k = n - 1}^{} B_{ijk}^{n - 1} (u, v) p_{i+1jk}$
$\frac{\partial p}{\partial v} = \sum_{i + j + k = n - 1}^{} B_{ijk}^{n - 1} (u, v) p_{ij+1k}$

Bezier Curve

$p (t) = \sum_{i = 0}^{n} B_{i}^{n} (t) p_{i}$
$\frac{d}{dt} p (t) = n \sum_{i = 0}^{n - 1} B_{i}^{n} (t) (p_{i + 1} - p_{i})$

Here is the maths to derive the above result.

$p = \sum_{i = 0}^{n} \frac{n!}{(n - i)! i!} t^{i} (1 - t)^{n - i} p_{i}$
$\frac{dp}{dt} = \sum_{i = 0}^{n} p_{i} \frac{n!}{(n - i)! i!} {t^{i} (n - i) (1 - t)^{n - 1 - i} . - 1 + (1 - t)^{n - i} . i t^{i - 1}}$

Looking at the first sum.

$\sum_{i = 0}^{n} p_{i} \frac{n!}{(n - i)! i!} t^{i} (n - i) (1 - t)^{n - 1 - i} . - 1$
$= \sum_{i = 0}^{n - 1} p_{i} \frac{n!}{(n - i)! i!} t^{i} (n - i) (1 - t)^{n - 1 - i} . - 1$
$= n \sum_{i = 0}^{n - 1} p_{i} \frac{(n - 1)!}{(n - 1 - i)! i!} t^{i} (1 - t)^{n - 1 - i} . - 1$
$= n \sum_{i = 0}^{n - 1} p_{i} B_{i}^{n - 1} . - 1$

Looking at the second sum.

$\sum_{i = 0}^{n} p_{i} \frac{n!}{(n - i)! i!} (1 - t)^{n - i} . i t^{i - 1}$
$= \sum_{i = 1}^{n} p_{i} \frac{n!}{(n - i)! i!} (1 - t)^{n - i} . i t^{i - 1}$
$= \sum_{i = 1}^{n} p_{i} \frac{n!}{(n - i)! (i - 1)!} (1 - t)^{n - i} . t^{i - 1}$

Here is a trick from "Concret Mathematics" but I can not find the reference. Subtract 1 in the sum and add 1 in the body to variable i. This is the same as the operation of shifting in an integral, except in a discrete sense. This fundamental technique was never taught in my maths degree where generally only the continuous variables are considered as part of calculus.

$= \sum_{i = 0}^{n - 1} p_{i + 1} \frac{n!}{(n - 1 - i)! i!} (1 - t)^{n - 1 - i} . t^{i}$
$= n \sum_{i = 0}^{n - 1} p_{i + 1} \frac{(n - 1)!}{(n - 1 - i)! i!} (1 - t)^{n - 1 - i} . t^{i}$
$= n \sum_{i = 0}^{n - 1} p_{i + 1} B_{i}^{n - 1}$

Adding the two sums gives the derivative of the Bezier sum curve.

References

Tomas Akenine-Moller and Eric Haines (2002). Real-Time Rendering (Second Edition). ISBN 1-56881-182-9. Pages 496-504.