Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/iii/paper-66/3/i/solution
Past exam of the mathematics course of the University of Cambridge 2005 iii Paper 66 3 i Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-14
Fix the indexing conventionwhere the subdivision arity is an integer. A change of the initial control can influence only indices after one step, where . After steps the possible descendant indices areLet be the smallest and largest active subdivision mask indices. Dividing by gives the enclosureTherefore the support of a stationary subdivision scheme, in initial control-grid units, is enclosed byFor the stated symmetric subdivision mask with active extremes , the width is . For a nonzero compactly supported basic limit function , this is an exact endpoint calculation: its refinement equation is . If is its right support endpoint, only the largest active subdivision mask index can contribute sufficiently close to the rightmost point of this sum; all smaller indices have argument beyond . Hence , giving . The left endpoint follows in the same way. This argument also excludes cancellation at an extreme for signed subdivision masks. The support for initial control is translated by . If the nominal subdivision mask was padded by zero coefficients, use its actual extremes; and a limiting subdivision curve has to exist before its support is meaningful. This calculation describes the entire possible influence region, rather than the width of a single refinement stencil on its finer grid.
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