The <characteristic polynomials of a linear multistep method> are
$$
\rho(\zeta)=\zeta^2-(1+a)\zeta+a=(\zeta-1)(\zeta-a)
$$
and
$$
\sigma(\zeta)=\frac1{12}\bigl((5+a)\zeta^2+8(1-a)\zeta-(1+5a)\bigr).
$$
For the <order conditions for a linear multistep method>, put $\alpha=(a,-1-a,1)$ and $\beta=(-(1+5a)/12,,8(1-a)/12,,(5+a)/12)$. The defects
$$
C_q=\sum_{j=0}^2\alpha_jj^q-q\sum_{j=0}^2\beta_jj^{q-1}
$$
vanish for $q=0,1,2,3$, while
$$
C_4=-(a+1),
\qquad
C_5=-\frac{13a+17}{3}.
$$
Therefore
$$
\boxed{
\text{order}=\begin{cases}
4,&a=-1,\\
3,&a\ne-1.
\end{cases}}
$$
Indeed, at $a=-1$ one has $C_4=0$ but $C_5=-4/3\ne0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 341 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-341/2/a/solution