Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 1 / 5C (Numerical Analysis)

Successive substitution shows every stage is a polynomial in $z=h\lambda$ times $y_n$, of degree at most its stage number. Hence $y_{n+1}=r(z)y_n$ with $r$ a degree-$\nu$ polynomial.

Order $\nu$ requires $r(z)-e^z=O(z^{\nu +1})$ on the test equation, fixing all $\nu +1$ coefficients: $r(z)={\sum }_{l=0}^{\nu }{z}^l/l!$.

For $\nu =2$, $|r(iy){|}^2=1+y^4/4$, so only $y=0$ is stable. For $\nu =3$, $|r(iy){|}^2=1-y^4/12+y^6/36$, so the stable imaginary segment is $|y|\le \sqrt{3}$.