Past exam of the mathematics course of the University of Cambridge 2006 iii Paper 67 6 c Solution 2026-10-14
We first derive the recurrence needed for the partition of unity. Denote an order- partition-normalized spline by . For , set , so . The elementary product identities for divided differences areThey follow from the Leibniz rule for divided differences because the divided differences of the linear factor vanish above first order. Use the first identity on and the second on . Subtracting cancels the common middle-block term. The defining divided-difference recurrence for then givesThis is the Cox-de Boor recurrence, derived here from the specified normalization rather than assumed as a partition theorem.
Extend the distinct knot sequence in both directions to a strictly increasing sequence tending to . For order one, , so the full sum equals one. All higher-order splines are supported on , making the following sums locally finite. Reindexing the second term of the recurrence yieldsInduction proves that the full sum is one for every order. The same recurrence gives nonnegativity: each coefficient is nonnegative wherever its associated lower-order spline is nonzero. Removing splines therefore gives the subpartition of unity for B-splines on the whole line.
For , a spline with index ends no later than , and one with starts no earlier than . They all vanish, leaving exactly the indices . For the splines are continuous and vanish at their support endpoints, so the equality extends to the closed basic knot interval:If order one is allowed, the standard half-open indicator convention gives equality on ; to include its final endpoint literally, assign the left-limit value there. This harmless endpoint convention does not change integrals or norms. For the continuous splines of order at least two no such convention is necessary.
Past exam of the mathematics course of the University of Cambridge 2010 iii Paper 62 4 b Solution Created 2026-10-03 Updated 2026-10-14
Use again and take . Set , so . LetThe Leibniz rule for divided differences, applied to the linear factor , readsMultiply by and substitute to get . Converting and to the lower-order normalized B-splines proves the Cox-de Boor recurrence:It starts with . The coefficients are nonnegative wherever the corresponding lower-order B-spline is nonzero, so induction proves nonnegativity.
For later use, extend the finite knot sequence to a strictly increasing sequence in both directions. The full order-one sum is one. When the Cox-de Boor recurrence is summed, the two coefficients of each fixed lower-order B-spline add to one. Induction therefore gives a full partition of unity. Removing terms proves the subpartition of unity for B-splines, . For repeated knots, the same statement follows by a limit of strictly increasing knot sequences, with the usual convention that a recurrence term with zero denominator is zero; the bound holds on each open knot interval and for the standard one-sided endpoint values. The terminology “-normalized” refers to this standard basis normalization; it does not claim that each individual B-spline has supremum norm exactly one.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 61 3 b Solution Created 2026-10-03 Updated 2026-10-07
We derive the Cox-de Boor recurrence from the Leibniz rule for divided differences. Fix , let , and setAway from the spline knots, . IntroduceThe defining divided-difference recurrence says . Since and the first factor is linear, the Leibniz rule for divided differences givesMultiplication by consequently yieldsReplace and by the corresponding lower-order B-splines divided by their support widths. This provesThe denominators are positive for the distinct spline knots of this question. The recursion starts with , the interval indicator on . For , values at the spline knots follow by the continuous extension of the left side; the usual half-open convention for the order-one factors gives the same result. This proves the recurrence rather than assuming it as a definition.
Past exam of the mathematics course of the University of Cambridge 2025 iii Paper 318 3 b Solution Created 2026-09-24 Updated 2026-09-25
Apply the Leibniz rule for divided differences to . Only the zeroth and first divided differences of the linear factor survive. After applying the normalization, this gives the Cox-de Boor recursion formula