Past exam of the mathematics course of the University of Cambridge / 2005 / ia / Paper 2 / 5B / Solution

The characteristic equation of a linear recurrence isFor $0<\theta <\pi$, these characteristic roots are distinct. Two real solutions of the second-order difference equation are $\cos (n\theta )$ and $\sin (n\theta )$. Their values at $n=0,1$ form a matrix with determinant $\sin \theta \ne 0$, proving linear independence. Consequently they span the two-dimensional vector space of recurrence solutions.

At $\theta =0$ the two roots coalesce at one. The linear recurrence becomes $X_{n+2}-X_{n+1}=X_{n+1}-X_n$, so its successive differences are constant. A basis is therefore $1,n$, and every solution is $A+Bn$.

For the specified initial conditions, the zero-angle solution is $X_n(0)=n$. For nonzero angle, it is convenient to center the basis at $n=1$:This gives $X_1=1$ and $X_2=\cos \theta +2-\cos \theta =2$, and each summand obeys the linear recurrence. For every fixed integer $n\ge 1$, the elementary limit $\sin z/z\to 1$ givesThe convergence here is for fixed $n$; it does not assert uniform convergence over all indices.