Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 341 / 1 / b

Factor the first characteristic polynomial asThe other two roots of a polynomial have product one. If $-1<\alpha <1$, they are $\alpha \pm i\sqrt{1-{\alpha }^2}$: distinct unit-modulus roots, both different from one. At $\alpha =-1$, the root $-1$ is double; at $\alpha =1$, the root one is triple. Outside $[-1,1]$, one of the reciprocal real roots has modulus greater than one. Thus the root condition for a multistep method holds exactly for $-1<\alpha <1$.

For the order conditions for a linear multistep method, expand the exponential defect:The constant, linear and quadratic coefficients vanish for every $\alpha$. The cubic coefficient vanishes only at $\alpha =11/5$, where the quartic coefficient is $-3/5\ne 0$. Consequently the formal order isFor all parameters satisfying the root condition for a multistep method, ${\rho }^{\prime }(1)=\sigma (1)=2(1-\alpha )\ne 0$, so this formal calculation gives ordinary second-order consistency of a numerical method. The Dahlquist equivalence theorem therefore yieldsThe exceptional third-order formula is not zero-stable. At $\alpha =1$, $\sigma \equiv 0$ and $\rho =(w-1{)}^3$: the recurrence contains no vector-field evaluations. It satisfies the displayed Taylor identities only formally and is a degenerate, nonconvergent formula, not a usable second-order ODE solver. This distinction avoids interpreting formal cancellation as convergence.