Let $A=(a_{ij})$, $b=(b_i)$, $\mathbf1=(1,1,1)^T$, and $c=A\mathbf1$. From the displayed tableau,
$$
c=\left(0,\frac35-\frac{\sqrt6}{10},
\frac35+\frac{\sqrt6}{10}\right)^T.
$$
The quadrature moments satisfy
$$
b^Tc^q=\frac1{q+1},\qquad q=0,1,2,3,4.
$$
Direct substitution into the remaining <Butcher order conditions> gives, for example,
$$
b^TAc=\frac16,\quad
b^T(c\circ Ac)=\frac18,\quad
b^TA(c\circ c)=\frac1{12},\quad
b^TA^2c=\frac1{24},
$$
and all rooted-tree conditions of orders at most five are satisfied. The order-six moment already fails:
$$
b^Tc^5=\frac{33}{200}\ne\frac16.
$$
The <Runge-Kutta method> therefore has
$$
\boxed{\text{order }5}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 341 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-341/2/a/solution