Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 40B (Numerical Analysis)

Taylor expansion gives ${\Gamma }_{9}u=h^2\Delta u+(h^4/12){\Delta }^{2}u+O(h^6)=h^{2}f+(h^4/12)\Delta f+O(h^6)$. Thus the local residual is $O(h^4)$.

The error equation has right side $O(h^4)$. With $\Vert A_h^{-1}{\Vert }_2\le C/h^2$, it follows that $\Vert e{\Vert }_2=O(h^2)$.

If $\Delta f=0$, the $h^4$ term vanishes, so the local residual is $O(h^6)$ and the global error is $O(h^4)$.

The five-point expression satisfies ${\Gamma }_{5}f=h^2\Delta f+O(h^4)$. Therefore the modified right side is $h^{2}f+h^4\Delta f/12+O(h^6)$, cancelling the leading residual from part (a) and leaving the same $O(h^6)$ local error as in part (c).