Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 66 / 4

The tensor-product surface basis is $b_{jk}(u,v)={\varphi }_j(u){\psi }_k(v)$. If both univariate bases consist of nonnegative functions, thenIf both factors are strictly positive at the chosen parameters, their product is strictly positive there as well. Its support is the Cartesian product of the two factor supports. With partition of unity, this nonnegativity additionally places the parametric surface point in the convex hull of the active control points.

If each univariate basis sums to one, distribute the finite sums, or use local finiteness when the basis is infinite:Thus the tensor-product surface basis is a partition of unity. In particular, translating every control point by a fixed vector translates the parametric surface by precisely that vector, and constant control data produce a constant parametric surface.

Suppose ${\varphi }_j\in C^r$ in $u$ and ${\psi }_k\in C^s$ in $v$, including their respective knot joins. The mixed derivatives of the tensor-product surface basis areThese are continuous because each factor is continuous. A finite or locally finite control sum has the same property. In the usual total-order notation the parametric surface is therefore at least $C^{\min (r,s)}$, and also has the stronger stated rectangular set of continuous mixed derivatives. Across a $u$-knot its guaranteed order is $r$, and across a $v$-knot its guaranteed order is $s$. No general improvement over a factor's smoothness is possible: setting $P_{jk}=A_j$ and using ${\sum }_k{\psi }_k=1$ recovers the univariate curve ${\sum }_j{A}_j{\varphi }_j(u)$ as a parametric surface constant in $v$.

Let the univariate linear precision sites be ${\xi }_j,{\eta }_k$, meaningThen their Cartesian-product grid is a linear precision set for the tensor-product surface basis:Indeed, the first coordinate is $({\sum }_j{\xi }_j{\varphi }_j)({\sum }_k{\psi }_k)=u$, and the second is $({\sum }_j{\varphi }_j)({\sum }_k{\eta }_k{\psi }_k)=v$. More generally, take controls $P_{jk}=A+{\xi }_{j}B+{\eta }_{k}C$. The resulting parametric surface is exactlyThis proves linear precision of a geometric basis and affine map reproduction, including arbitrary changes of the embedding plane. In fact the products also reproduce the bilinear coordinate $uv$, since ${\sum }_{j,k}{\xi }_j{\eta }_k{\varphi }_j{\psi }_k=uv$; affine precision is the requested consequence.