= Reciprocal-root three-step multistep family
{title2=$\rho(\zeta)=(\zeta-1)(\zeta^2-2\alpha\zeta+1)$}
Take $\sigma(\zeta)=\zeta[(5+\alpha)\zeta^2-(4+8\alpha)\zeta+11-5\alpha]/6$. The <root condition for a multistep method> holds exactly for $-1<\alpha<1$. Since $\rho'(1)=\sigma(1)=2(1-\alpha)$, these and only these parameters give a convergent method. Its <exponential-symbol order criterion for a multistep method> has defect $-(\alpha+5)z^4/12+O(z^5)$, giving formal order three except at $\alpha=-5$, where the first defect is $z^5/10$. None of the convergent family is <A-stable>: for a parasitic root $r=\alpha+i\sqrt{1-\alpha^2}$, perturbation of $\rho(r(z))-z\sigma(r(z))=0$ gives $\operatorname{Re}[r'(0)/r]=(\alpha-1)/12<0$. A small negative real $z$ therefore moves it outside the unit disk. At $\alpha=1$, cancellation would remove a double unit root, but the original recurrence retains that unstable parasitic behavior; it is not legitimately replaced by the reduced Backward Euler recurrence for arbitrary starting values.
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