Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 1 / 18D / d

With ${\rho }_0=a$, ${\rho }_1=\theta$, ${\rho }_2=1$, the constant order condition gives $a=-1-\theta$. The next three order conditions for a linear multistep method giveSolving this linear system yields the unique coefficients of order at least three:The characteristic polynomial controlling zero-stability factors asThe other root is $-1-\theta$. Requiring its modulus to be at most one gives $-2\le \theta \le 0$, but $\theta =-2$ produces a repeated root at one and violates the root condition for a multistep method. At $\theta =0$, the roots $1,-1$ are both simple, so that endpoint is allowed. By the Dahlquist equivalence theorem, the method converges precisely forThe fourth defect coefficient is $C_4=\theta$, so the method has order exactly three for nonzero $\theta$. At $\theta =0$ it is the two-step Simpson method, of order four; thus the extra-order endpoint has correctly remained in the answer.