Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 67 / 6

For this part relabel one consecutive block of $k+1$ distinct knots as $t_0,\dots ,t_k$. This is a local reindexing; no extra global knot is being introduced. Let $g_t(u)=(u-t{)}_{+}^{k-1}$. Its Lagrange interpolation polynomial through the knot values isEach quotient $\omega (u)/(u-t_i)$ is monic of degree $k$. Comparing leading coefficients givesThe same leading coefficient is the order-$k$ divided difference $[t_0,\dots ,t_k]g_t$, by the Newton interpolation polynomial representation. Multiplying by $k$ therefore gives the explicit B-spline formulaDistinct knots make every denominator nonzero. For $k=1$ the degree-zero truncated power is interpreted as an interval-indicator convention; values at individual endpoints do not affect its integral.

The order-$k$ divided difference defining $M_i$ is a finite linear combination of values at its knots. Hence integration can be interchanged with it directly. For a knot value $u\in [a,b]$,It follows thatThe final equality holds because the order-$k$ divided difference of a degree-$k$ monic polynomial is its leading coefficient, one. ThusThis proves the unit-integral normalization of a B-spline. Its support is contained in $[t_i,t_{i+k}]$: below all its knots, the sampled truncated power agrees with a degree-$k-1$ polynomial, annihilated by the order-$k$ divided difference; above all its knots, every sample is zero. No mass is omitted by integration over $[a,b]$.

Since $N_i=(t_{i+k}-t_i)M_i/k$, the corresponding partition-normalized B-spline has integral $(t_{i+k}-t_i)/k$. Its normalization is different from that of $M_i$.

We first derive the recurrence needed for the partition of unity. Denote an order-$r$ partition-normalized spline by $N_{i,r}$. For $r\ge 2$, set $h_t(u)=(u-t{)}_{+}^{r-2}$, so $(u-t)h_t(u)=(u-t{)}_{+}^{r-1}$. 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 $t_i,\dots ,t_{i+r-1}$ and the second on $t_{i+1},\dots ,t_{i+r}$. Subtracting cancels the common middle-block term. The defining divided-difference recurrence for $N_{i,r}$ 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 $(t_i{)}_{i\in \mathbb{Z}}$ tending to $\pm \infty$. For order one, $N_{i,1}=1_{[t_i,t_{i+1})}$, so the full sum equals one. All higher-order splines are supported on $[t_i,t_{i+r}]$, 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 $t_k<t<t_{n+1}$, a spline with index $i\le 0$ ends no later than $t_k$, and one with $i\ge n+1$ starts no earlier than $t_{n+1}$. They all vanish, leaving exactly the indices $1,\dots ,n$. For $k\ge 2$ 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 $[t_1,t_{n+1})$; to include its final endpoint literally, assign the left-limit value there. This harmless endpoint convention does not change integrals or $L^{\infty }$ norms. For the continuous splines of order at least two no such convention is necessary.