= Solution
For $|\alpha|>1$, an exterior <polynomial root> of $\rho$ persists by <continuity> for sufficiently small negative real $z$ in the <stability> <polynomial> $\rho(w)-z\sigma(w)$. Hence these parameters cannot be <A-stable>.
For $-1\le\alpha<1$, the principal <polynomial root> at $w=1,z=0$ is simple. Let $s(z)=\log w(z)$ denote its analytic <logarithm> near zero. Using the expansion in part (b), and $\rho'(1)=2(1-\alpha)$, solve the <polynomial root> equation to obtain
$$
s(z)=z+Cz^4+O(z^5),\qquad C=\frac{\alpha+5}{24(1-\alpha)}>0.
$$
Indeed the residual at $s=z$ is $-(\alpha+5)z^4/12$, and the correction in $s$ cancels it by multiplication with $2(1-\alpha)$. Choose $z=-C\varepsilon^4/2+i\varepsilon$. This lies strictly in the left half-plane, but
$$
\operatorname{Re}s(z)=\frac C2\varepsilon^4+O(\varepsilon^5)>0
$$
for small positive $\varepsilon$. Thus $|w(z)|=e^{\operatorname{Re}s(z)}>1$, proving <linear instability> within the <A-stability> domain. This is a direct <principal-root obstruction to A-stability>, rather than an appeal to an order-barrier theorem alone.
Finally, at $\alpha=1$, the full <stability> <polynomial> is
$$
\rho(w)-z\sigma(w)=(w-1)^2\bigl((1-z)w-1\bigr).
$$
It retains a double unit <polynomial root> for every negative $z$, allowing growing parasitic solutions. Therefore
$$
\boxed{\text{There is no real }\alpha\text{ for which the printed recurrence is A-stable}.}
$$
The canceled <Backward Euler method> recurrence at $\alpha=1$ is <A-stable>, but its admissible solutions and starting relations differ from those of the original method.
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