Symmetric sixth-order two-derivative method ($y_{n+1}-\frac{7}{15}h{f}_{n+1}+\frac{1}{15}{h}^2{g}_{n+1}=y_{n-1}+\frac{h}{15}(16f_n+7f_{n-1})+\frac{1}{15}{h}^2{g}_{n-1}$)

For $y^{\prime }=f(y)$ and $g=f^{\prime }f$, inserting a smooth exact solution into the displayed recurrence gives defect $h^7{y}^{(7)}(t_n)/4725+O(h^9)$. This follows by expanding $2\sinh z-(14/15)z\cosh z-(16/15)z+(2/15)z^2\sinh z$. The terms through $z^6$ vanish and the $z^7$ coefficient is nonzero. Its zero-step polynomial is ${\zeta }^2-1$, whose roots are simple and of unit modulus. Thus it is zero-stable and has order six, provided starting values and derivative evaluations have corresponding accuracy.