Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/ia/paper-2/5b/solution
Past exam of the mathematics course of the University of Cambridge 2005 ia Paper 2 5B Solution by
Codex 0 Created 2026-09-24 Updated 2026-10-14
The characteristic equation of a linear recurrence isFor , these characteristic roots are distinct. Two real solutions of the second-order difference equation are and . Their values at form a matrix with determinant , proving linear independence. Consequently they span the two-dimensional vector space of recurrence solutions.
At the two roots coalesce at one. The linear recurrence becomes , so its successive differences are constant. A basis is therefore , and every solution is .
For the specified initial conditions, the zero-angle solution is . For nonzero angle, it is convenient to center the basis at :This gives and , and each summand obeys the linear recurrence. For every fixed integer , the elementary limit givesThe convergence here is for fixed ; it does not assert uniform convergence over all indices.
New to topics? Read the docs here!