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Interpret the printed missing parenthesis as ending $f(y_{n+2})$ before the next summand. The characteristic polynomials of a linear multistep method are thenThey satisfy $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)=2(1-\alpha )$, so the consistency identities hold. The two nonprincipal roots are reciprocal. If $|\alpha |<1$, they are $\alpha \pm i\sqrt{1-{\alpha }^2}$, distinct unit roots also distinct from the principal root one. If $|\alpha |>1$, one reciprocal root lies outside the unit circle. At $\alpha =1$ the root one is triple, and at $\alpha =-1$ the root minus one is double. The root condition for a multistep method therefore holds exactly when $-1<\alpha <1$. The Dahlquist equivalence theorem, with standard smoothness and consistent starting data, givesTo determine order rather than guess it from the number of steps, the exponential-symbol order criterion for a multistep method givesThus the formal order is three for $\alpha \ne -5$, and four for $\alpha =-5$. At the exceptional value the $z^5$ coefficient is $1/10$. That fourth-order formula is not zero-stable. Every convergent member consequently has global order three. At $\alpha =1$ the third-order Taylor cancellation is only a formal defect property of a degenerate, nonconvergent recurrence.

Here is a direct A-stability test for the convergent parameters. For a nonprincipal unit root $r=\alpha +i\sqrt{1-{\alpha }^2}$, the simple root of the amplification polynomial of a multistep method satisfiesReduction using $r^2-2\alpha r+1=0$ givesFor small negative real $z$, it follows thatSo even a small negative real test parameter has an amplifying parasitic mode. None of the convergent parameters is A-stable. Parameters outside this range already fail the necessary zero-step root condition, so there is no real $\alpha$ giving an A-stable original method.

The endpoint $\alpha =1$ is worth keeping separate from a misleading cancellation: there $\rho =(\zeta -1{)}^3$ and $\sigma =\zeta (\zeta -1{)}^2$. The original amplification polynomial retains the double root one for every $z$, giving unbounded polynomial parasitic modes for general starting values. Canceling $(\zeta -1{)}^2$ would yield a different, reduced Backward Euler method; it does not stabilize the original recurrence. This is the common-factor cancellation defect in a multistep recurrence. The family is summarized by the reciprocal-root three-step multistep family.

The first characteristic polynomial factors asAlso $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)=2(1-\alpha )$, so the order conditions for a linear multistep method hold for every $\alpha$. Convergence of the full linear multistep method recurrence requires, in addition, the root condition for a multistep method: all polynomial roots of $\rho$ lie in the closed unit disk and its polynomial roots on the unit circle are simple.

For $-1<\alpha <1$, write $\alpha =\cos \theta$ with $0<\theta <\pi$. The other polynomial roots are $e^{\pm i\theta }$, distinct from each other and from 1, so the root condition for a multistep method holds. For $|\alpha |>1$, the two quadratic polynomial roots are real and reciprocal, and one has modulus greater than one. At $\alpha =1$, $\rho =(w-1{)}^3$; at $\alpha =-1$, $\rho =(w-1)(w+1{)}^2$. Both endpoints violate simplicity. Consistency of a numerical method plus zero-stability therefore givesThe conclusion applies to the recurrence as printed, including its starting data. Canceling common factors is not innocuous: at $\alpha =1$ the polynomials contain $(w-1{)}^2$, and at $\alpha =-13/5$ they contain $w+5$. The reduced recurrences exclude parasitic solutions of the original one. This is the common-factor cancellation defect in a multistep recurrence, so those parameter values are not added to the interval.

For this two-step family with a third-order member, use the exponential-symbol order criterion for a multistep method. Expanding the exact-solution residual symbol givesThe constant, linear and quadratic coefficients vanish for every parameter. ThusAt the exceptional value the fourth-degree coefficient is $-1/30$, so there is no further increase. This means an unscaled exact-solution residual of order $h^{p+1}$, or a residual divided by the step of order $h^p$.

The first characteristic polynomial factors as $\rho (w)=(w-1)(w-a)$. The root condition for a multistep method requires every root to have modulus at most one and each unit-modulus root to be simple. The root $a=-1$ is allowed because it is distinct from the simple root one; $a=1$ is not, since it makes one a double root. Hence zero-stability holds exactly forOn this interval, $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)=1-a\ne 0$ establish consistency of a numerical method. The Dahlquist equivalence theorem, for a fixed-step consistent linear multistep method applied to an ordinary differential equation with a Lipschitz continuous right-hand side with convergent starting values and well-defined small-step updates, says that convergence of a numerical method is equivalent to zero-stability. Therefore the same interval is exactly the convergence of a numerical method range. With sufficiently accurate starting values the convergent members have the orders just computed.

At $a=1$ the Taylor expansion still gives a formal order-two residual, but ${\rho }^{\prime }(1)=\sigma (1)=0$ and the double root prevents a convergent method. Canceling the common factor $w-1$ produces the common-factor cancellation defect in a multistep recurrence and changes the starting relations and cannot repair convergence of a numerical method of the original recurrence. At $a=0$, cancellation of a zero root instead reveals the ordinary trapezoidal rule; it is consistent with the order-two result.

The characteristic polynomials of a linear multistep method areExpanding the exponential-symbol order criterion for a multistep method givesThus the formal local consistency order is two for $\alpha \ne 2$, and three at $\alpha =2$, since the fourth-degree coefficient there is $-1/12$. This exceptional formal order is degenerate: ${\rho }^{\prime }(1)=\sigma (1)=0$, so it is not a convergent third-order time integrator.

For zero-stability, factor $\rho (\zeta )=(\zeta -1)(\zeta -(\alpha -1))$. The root condition for a multistep method requires every root to have modulus at most one and every unit-modulus root to be simple. For real $\alpha$ it holds exactly forAt zero the second root is $-1$, simple and admissible; at two the root one is double and inadmissible. Outside the closed interval the second root has modulus greater than one. The Dahlquist equivalence theorem says that a consistent linear multistep method is convergent exactly when it is zero-stable, for a locally Lipschitz initial-value problem with consistent starting values and solvable implicit steps. Hence the displayed interval is also exactly the convergence range, and every convergent member has order two.

The common-factor cancellation defect in a multistep recurrence explains why the $\alpha =2$ exception cannot be repaired just by canceling $\zeta -1$. For $y^{\prime }=0$, its recurrence allows linear growth from a vanishing but insufficiently small starting perturbation: $y_0=0$, $y_1=\sqrt{h}$ gives $y_n=n\sqrt{h}$, diverging at a fixed positive time. Canceling the factor silently imposes an additional starting relation and changes the method.