 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-66/5/b/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

A (z, w) = \sum a_{ij} z^{i} w^{j} = \frac{( 2 + z + z ^{- 1} ) ( 2 + w + w ^{- 1} )}{8} = \frac{( 1 + z ) ^{2} ( 1 + w ) ^{2}}{8 z w} .

p_{j}^{(1)} = \sum_{i} a_{j - D i} p_{i}

. A second step gives

p_{k}^{(2)} = \sum_{i, j} a_{k - D j} a_{j - D i} p_{i}

. Setting

r = j - D i

and

s = k - D j

, its fine-grid offset is

k - 2 i = s + Dr

, since

D^{2} = 2 I

. Hence the binary mask is the convolution of

a

with its

D

-transformed copy:

B (z, w) = A (z, w) A (z w, z / w) = \frac{( 1 + z ) ^{2} ( 1 + w ) ^{2} ( 1 + z w ) ^{2} ( 1 + z / w ) ^{2}}{64 z ^{3} w} .

Expanding gives the centered binary propagation subdivision mask, with row and column offsets running from

- 3

3

\frac{1}{64} 0012100026862016141814612818241882161418146102686200012100 .

The numerator entries sum to

256

, so

B (1, 1) = 4

, appropriate for a binary surface refinement. More specifically, each of the four parity classes sums to one after division by

64

, checking constant reproduction at every type of new lattice point.

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