= Symmetric sixth-order two-derivative method
{title2=$y_{n+1}-\frac7{15}hf_{n+1}+\frac1{15}h^2g_{n+1}=y_{n-1}+\frac h{15}(16f_n+7f_{n-1})+\frac1{15}h^2g_{n-1}$}
For $y'=f(y)$ and $g=f'f$, inserting a smooth exact solution into the displayed recurrence gives defect $h^7y^{(7)}(t_n)/4725+O(h^9)$. 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.
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