 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/iii/paper-68/4/b/solution

Let

d = k - 1

and let

e_{m}

m

in the

d

interior knots

t_{i + 1}, \dots, t_{i + d}

, with

e_{0} = 1

. Expand the two sides of the Marsden identity as polynomials in

x

(x - t)^{d} = \sum_{m = 0}^{d} (- 1)^{m} (m d) t^{m} x^{d - m},

and

ω_{i} (x) = \prod_{j = 1}^{d} (x - t_{i + j}) = \sum_{m = 0}^{d} (- 1)^{m} e_{m} (t_{i + 1}, \dots, t_{i + d}) x^{d - m} .

Comparing the coefficient of

x^{d - m}

and cancelling the common sign gives the monomial B-spline coefficients

a_{i}^{(m)} = \frac{e _{m} ( t _{i + 1} , \dots , t _{i + k - 1} )}{( m k - 1 )}, 0 \leq m \leq k - 1.

Equivalently,

a_{i}^{(m)} = (m k - 1)^{- 1} \sum_{1 \leq j_{1} < \dots < j_{m} \leq k - 1} t_{i + j_{1}} \dots t_{i + j_{m}} .

In particular

a_{i}^{(0)} = 1

, giving partition of unity on the interior interval. For

k \geq 2

a_{i}^{(1)} = (t_{i + 1} + \dots + t_{i + k - 1}) / (k - 1)

, and the highest-degree coefficient is the product of those interior knots. The formulas also cover

k = 1, m = 0

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