Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 70 / 2

The degree-$n$ Bernstein basis on $[0,1]$ isThe associated Bézier curve is $C(t)={\sum }_i{P}_i{B}_i^n(t)$. The basis is nonnegative and has partition of unity, since ${\sum }_i{B}_i^n(t)=(t+1-t{)}^n=1$. Thus the curve is a convex combination of its control points, giving the convex hull enclosure useful for geometric bounds. The same identity makes the construction invariant under application of affine functions.

It interpolates its endpoint controls, $C(0)=P_0$, $C(1)=P_n$, and its endpoint tangent vectors are $n(P_1-P_0)$ and $n(P_n-P_{n-1})$. These follow fromThe control polygon therefore gives direct geometric editing and tangent control. Repeated interpolation by the De Casteljau algorithm evaluates and subdivides the curve without converting to a monomial representation. Together, positive geometric weights, simple endpoint behavior and stable subdivision make this basis useful in computer aided geometric design.

A degree-$p$ B-spline basis keeps the positive partition of unity and convex hull properties while introducing local support. At a generic parameter value, only $p+1$ consecutive basis functions are nonzero. Moving one control point therefore changes only its support interval, unlike a generic interior Bézier curve control, which affects the whole polynomial segment.

Many low-degree pieces can represent a long curve with automatic continuity. At a knot of multiplicity $m$, the generic continuity is $C^{p-m}$; simple cubic knots give $C^2$. Knot positions and multiplicities control the placement of detail and joins. Knot insertion can refine the representation without changing the curve, and local refinement avoids raising the global polynomial degree. Thus B-splines combine geometric bounds, local editing and controlled smoothness with a fixed low degree. Clamped end knots can also give endpoint interpolation.

Scale the equal intervals to unit length. A nonzero compactly supported cubic B-spline with maximum continuity is $C^2$, so its fourth derivative is a sum of impulses at the knots. Equivalently it is a combination of truncated cubes,To vanish to the right of its support, the polynomial tail must be identically zero. Expanding that tail requires ${\sum }_j{a}_j{j}^r=0$ for $r=0,1,2,3$. Four or fewer distinct knots give a full-rank Vandermonde matrix and only the zero solution. Hence a nonzero function requires at least five knots, or four intervals. At knots $0,1,2,3,4$, the one-dimensional nullspace has coefficients proportional to $(1,-4,6,-4,1)$. The unscaled combination integrates to six on $[0,4]$, so unit-integral cardinal normalization gives:This is the Cardinal cubic B-spline, nonnegative with support $[0,4]$. To obtain its cubic B-spline Bézier extraction, use local parameter $t=x-j\in [0,1]$ on interval $[j,j+1]$. The four power-form numerators, with a common divisor six, areFor $a+bt+c{t}^2+d{t}^3$, the cubic Bernstein basis coefficients are $a$, $a+b/3$, $a+2b/3+c/3$, $a+b+c+d$. Applying this conversion givesEach scalar row describes a cubic Bézier curve graph piece with abscissa controls $j,j+1/3,j+2/3,j+1$. The truncated-cube construction proves value, first- and second-derivative agreement at all joins, including the zero exterior. Minimum support and maximum continuity determine the basis up to a scalar; the displayed normalization fixes that remaining freedom.