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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.

Use the coefficient $2-\alpha$ in the numerator, as verified in the PDF; the converted TeX lost its $-\alpha$. Multiplying the method by $2+\alpha$ gives the characteristic polynomials of a linear multistep methodFor every admissible real $\alpha$, $\rho (1)=0$ and ${\rho }^{\prime }(1)=2\alpha =\sigma (1)$, giving consistency of a numerical method. Factorization givesThe second root is $r=(2-\alpha )/(2+\alpha )$. The root condition for a multistep method requires $|r|\le 1$ and forbids a repeated root at one. For real $\alpha \ne -2$,At $\alpha =0$ both roots equal one, violating zero-stability. For $\alpha >0$ the second root lies strictly inside the unit disk. The Dahlquist equivalence theorem therefore givesTo find the order rather than only consistency, expand the exact-solution residual through the generating function:Thus its unnormalized local truncation error begins with $(\alpha /3)h^3{y}^{(3)}$ when $\alpha >0$, and the global method order is exactly two for every convergent parameter. For comparison, $\alpha =0$ has formal order three but is not convergent because of its double unit root:These orders assume a smooth solution, the usual local Lipschitz condition and starting approximations of the required accuracy.

A constant-step linear multistep method has the formwith ${\alpha }_k\ne 0$. The characteristic polynomials of a linear multistep method encode both accuracy and error propagation. The method has order of a numerical method $p$ when insertion of a sufficiently smooth exact solution leaves an unnormalized local truncation error $O(h^{p+1})$. The exponential-symbol order criterion for a multistep method expresses this as $\rho (e^z)-z\sigma (e^z)=O(z^{p+1})$. In particular, first-order consistency of a numerical method requires $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)$.

Accuracy alone is insufficient. For $f=0$, perturbations obey the recurrence $\rho (E)e_n=0$. The root condition for a multistep method requires every root of a polynomial $\rho$ to lie in the closed unit disk, with each unit-circle root simple. This is zero-stability. A root outside the unit disk causes geometric amplification as $n\sim T/h$ increases; a repeated unit-circle root permits polynomial growth in $n$. Either can destroy convergence even for starting perturbations tending to zero.

The Dahlquist equivalence theorem states that, for a fixed-coefficient method and a well-posed ordinary differential equation with a suitably Lipschitz continuous right-hand side, consistency of a numerical method and zero-stability are equivalent to convergence for all convergent starting data. If the starting errors are $O(h^p)$ and the exact solution is sufficiently smooth, the global error is $O(h^p)$ on a fixed time interval. A stability estimate bounds the propagated sum of $O(h^{p+1})$ local defects over $O(1/h)$ steps; the discrete Gronwall inequality controls the dependence of $f$ on the accumulated error. Implicit methods also require the locally consistent, uniquely solvable update branch for sufficiently small $h$.

For example, the explicit Euler method has order one and $\rho (\zeta )=\zeta -1$, satisfying the root condition for a multistep method. The two-step Adams-Bashforth method has order two and $\rho (\zeta )=\zeta (\zeta -1)$, again satisfying it. The trapezoidal rule has order two with one step. The two-step Simpson multistep method, $Y_{n+2}-Y_n=h(f_{n+2}+4f_{n+1}+f_n)/3$, has order four and the simple roots $1,-1$: it is zero-stable, although its parasitic amplification root can spoil long-time decay. These examples distinguish finite-time numerical convergence from absolute stability.

The first Dahlquist barrier gives the maximum possible order of a convergent ordinary real $k$-step method:These bounds assume only first-derivative evaluations, fixed coefficients and zero-stability; the sixth-order multiderivative multistep method in solution is outside that class.

To obtain arbitrarily high convergent order within the ordinary class, increase the number of steps. The Adams-Bashforth method integrates a polynomial interpolation of $k$ past derivative values and has order $k$. The Adams–Moulton method includes the new endpoint and has order $k+1$ for $k$ steps. In both families $\rho (\zeta )={\zeta }^{k-1}(\zeta -1)$, so the root condition for a multistep method holds for every fixed $k$. Starting values can be generated by a Runge-Kutta method of sufficient accuracy. The backward differentiation formula instead differentiates an interpolating polynomial; its order-$k$ members are zero-stable only for $1\le k\le 6$.

Finally, high order and favorable stiff stability are separate design goals. By the Second Dahlquist barrier, an ordinary A-stable linear multistep method has order at most two. One may accept a smaller linear stability domain, use suitable backward differentiation formula members for stiff problems, or switch to high-order implicit Runge-Kutta methods or multiderivative multistep methods. A convergent high-order formula still needs time steps appropriate to its stability properties and the problem's eigenvalues.

A fixed-step linear multistep method for $y^{\prime }=f(t,y)$ uses several previous solution and derivative values:The coefficients are fixed independently of the step size. It is explicit when ${\beta }_s=0$ and implicit otherwise. Its characteristic polynomials of a linear multistep method are $\rho (w)={\sum }_j{\alpha }_j{w}^j$ and $\sigma (w)={\sum }_j{\beta }_j{w}^j$. The contraction mapping theorem gives a uniquely solvable implicit step under a Lipschitz continuity bound $L$ whenever $k|{\beta }_s|L<|{\alpha }_s|$, by contraction of the new-value equation on the appropriate solution region. A suitable starting procedure must supply the first $s$ values; the method alone does not determine them.

Order is defined by inserting a smooth exact solution into the recurrence. If the residual is $O(k^{p+1})$ and its first nonzero coefficient occurs at that degree, the method has order $p$; dividing the residual by $k$ gives an $O(k^p)$ local truncation error. Taylor expansion gives the exact order conditions for a linear multistep methodwith the right side interpreted as zero at $q=0$, and failure at $q=p+1$ for exact order. Equivalently,The basic consistency of a numerical method conditions are $\rho (1)=0$, ${\rho }^{\prime }(1)=\sigma (1)$; for the ordinary nondegenerate consistent formulation this common value is nonzero. Exactness on constants and linear functions does not control propagation of numerical perturbations.

That propagation is the role of zero-stability. On $y^{\prime }=0$ the homogeneous recurrence is $\rho (E)y_n=0$. Its modes are $n^r{\xi }^n$, with powers of $n$ determined by root multiplicities. The theorem on the root condition for a multistep method states that zero-stability is equivalent to every root of $\rho$ having modulus at most one, with every unit-modulus root simple. Interior repeated roots are permitted, since geometric decay dominates their polynomial factors. An exterior root amplifies perturbations geometrically; a repeated unit root amplifies them polynomially and violates a step-independent perturbation bound.

The Dahlquist equivalence theorem states that a fixed-step consistent linear multistep method is convergent exactly when it is zero-stable. This concerns ordinary differential equations with a uniformly Lipschitz continuous right-hand side on the solution region over a fixed finite interval, convergent starting values, and the well-defined nearby branch of each small-step implicit solve. For order $p$ convergence of a numerical method, the solution must have the required smoothness and the starting errors must be $O(k^p)$.

The mechanism can be sketched explicitly. The root condition for a multistep method bounds the recurrence's discrete Green function. For an order-$p$ exact-solution defect $d_n=O(k^{p+1})$, the Lipschitz continuity error equation then gives an estimate of the formThe current-step term can be absorbed for small $k$. The discrete Gronwall inequality says that a bound $E_n\le A+Bk{\sum }_{j<n}{E}_j$ implies $E_n\le A(1+Bk{)}^n\le A{e}^{B{t}_n}$. There are $O(k^{-1})$ defects on a fixed interval, so their sum is $O(k^p)$ and this yields the claimed global order. Conversely, testing $y^{\prime }=0$ with vanishing starting perturbations in exterior-root or repeated-unit-root modes proves that convergence of a numerical method fails without the root condition for a multistep method.

For example, the two-step Adams-Bashforth method hasIt is explicit, order two and zero-stable, hence convergent with order-two starting values. By contrast, a recurrence with $\rho =(w-1{)}^2$ has solutions $y_n=y_0+n(y_1-y_0)$ on the zero differential equation. Starting differences $\sqrt{k}$ tend to zero yet become unbounded at a fixed positive time, since $n$ is of order $1/k$. A small local defect or formal high order cannot rescue this mode. Canceling a common factor may remove it only by changing the allowed starting relations, as the parameter endpoint in Question 2 illustrates.

Absolute stability addresses a different limit: apply the method to the Dahlquist test equation $y^{\prime }=\lambda y$ and put $z=k\lambda$. The amplification roots satisfyAll roots must satisfy the disk condition, with unit roots simple; one cannot inspect only the root that tends to one as $z\to 0$. The degree and solvability of the recurrence must also be preserved. The resulting set of $z$ is the domain of absolute stability. A-stability means that it includes the whole closed left half-plane, so the numerical solutions of every Dahlquist test equation with $\mathrm{Re}\lambda \le 0$ are stable for every positive step size. It is a linear test-equation property, not a proof of arbitrary nonlinear contractivity or high accuracy at large steps.

Forward Euler method has order one and factor $1+z$, so its stability region is the disk $|1+z|\le 1$. Backward Euler method has factor $(1-z{)}^{-1}$ and is A-stable. The trapezoidal rule is order two with factor $(1+z/2)/(1-z/2)$; its modulus is at most one throughout the left half-plane. These one-step methods are included as the simplest multistep examples. For the two-step Adams-Bashforth method, the negative-real stability interval is $-1\le z\le 0$: its polynomial is $w^2-(1+3z/2)w+z/2$, and the root reaches $-1$ at $z=-1$. It therefore imposes a step restriction when applied to strongly decaying modes.

The genuinely two-step BDF2 method isIt has order two, roots $1,1/3$ at zero step and is A-stable. For its normalized polynomials $\rho =3w^2/2-2w+1/2$, $\sigma =w^2$, the unit-circle boundary locus hasThe leading coefficient has no zero in the left half-plane, and the roots at a small negative $z$ are both inside the disk. Root continuity then proves open-half-plane stability; on the imaginary boundary the only possible unit root is the simple root one at $z=0$. This gives a direct proof of the stated A-stability rather than relying on a boundary plot.

Two sharp order barriers explain the tradeoff. The first Dahlquist barrier states that a consistent zero-stable real linear $s$-step method using only first derivatives has order at most $s+1$ for odd $s$ and $s+2$ for even $s$; an explicit method has order at most $s$. The Second Dahlquist barrier states that an irreducible A-stable real linear multistep method has order at most two. These statements concern fixed constant coefficients and the ordinary first-derivative multistep class, not multiderivative methods or arbitrary variable-step formulations. Thus higher order alone is not a route to unconditional stiff stability. Question 2 provides a concrete illustration: its third-order member is zero-stable but not A-stable.

No nontrivial consistent explicit multistep method is A-stable. Its amplification polynomial of a multistep method has a fixed highest coefficient, while at least one lower coefficient grows without bound as real $z\to -\infty$. Were all roots uniformly in the unit disk, their elementary symmetric polynomials would keep all the monic coefficients bounded, a contradiction. This is the coefficient argument behind explicit multistep methods cannot be A-stable.

In practice, selecting a method therefore involves accuracy, startup, the presence of a stiff differential equation, implicit-solve cost and step-size control. Zero-stability handles propagation as the step tends to zero; consistency of a numerical method supplies a vanishing defect; A-stability handles the numerical solutions of all decaying Dahlquist test equations at finite steps. None substitutes for the others, and variable steps or PDE discretizations require additional uniform stability arguments appropriate to their setting.

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.

The relevant one of the characteristic polynomials of a linear multistep method isThe coefficients on the right are chosen to give third order, so consistency of a numerical method is already available. By the Dahlquist equivalence theorem, convergence with convergent starting values is equivalent to zero-stability: every root of $\rho$ must satisfy $|w|\le 1$, and every root on the unit circle must be simple.

If $a<0$, the quadratic is negative at $w=-1$ and tends to $+\infty$ as $w\to -\infty$, so it has a root less than $-1$. If $a>1$, its root product is $a$, so at least one root has modulus greater than one. For $0\le a\le 1/4$, its real roots $(-1\pm \sqrt{1-4a})/2$ lie in $[-1,0]$. For $1/4<a\le 1$, its conjugate roots have modulus $\sqrt{a}\le 1$.

At $a=0$, the three roots are $1,0,-1$, with the unit-modulus roots simple. At $a=1$, they are $1,e^{2\pi i/3},e^{-2\pi i/3}$, again simple. The repeated quadratic root at $a=1/4$ is $-1/2$, strictly inside the unit circle, which the root condition for a multistep method permits. ThereforeThis is the root stability of a cubic multistep polynomial; oscillatory parasitic roots at the endpoints do not invalidate convergence when their starting amplitudes tend to zero.

Insert an exact smooth solution into the linear multistep method and define its unscaled local truncation error byIts Taylor expansion has coefficientsThe method has order at least $p$ exactly when $c_0=\cdots =c_p=0$, since then ${\mathcal{R}}_h[y]=O(h^{p+1})$ for every sufficiently smooth $y$. Necessity can be checked by inserting polynomials of successive degrees. This also corresponds to the usual truncation error ${\mathcal{R}}_h/h=O(h^p)$.

On the other hand, expansion of the characteristic polynomials of a linear multistep method givesThus the exponential-symbol order criterion for a multistep method isIf “order $p$” is meant to be the exact rather than guaranteed order, one additionally requires $c_{p+1}\ne 0$. The first two vanishing conditions are the usual consistency of a numerical method relations $\rho (1)=0$, ${\rho }^{\prime }(1)=\sigma (1)$.

The characteristic polynomials of a linear multistep method areTo determine formal order, substitute a smooth exact solution and expand about the first time level. The exponential-symbol order criterion for a multistep method collects precisely the same coefficients:The constant, linear and quadratic coefficients vanish for every $a$. The cubic coefficient vanishes only at $a=-1/5$, where the quartic coefficient is $-1/30\ne 0$. HenceHere order means the exact-solution step residual is $O(h^{p+1})$. It is a formal consistency result, not a convergence assertion. In particular at $a=1$ both ${\rho }^{\prime }(1)$ and $\sigma (1)$ vanish, and the double root at one destroys zero-stability; cancelling its common factor gives a different, first-order recurrence with an additional integration constant left unspecified by the original formula.

The characteristic polynomials of a linear multistep method areThe consistency of a numerical method conditions hold for every real $a$:Both roots of a polynomial of $\rho$, namely $1$ and $-1$, are simple and have modulus one. The root condition for a multistep method therefore gives zero-stability for every $a$. The Dahlquist equivalence theorem now gives convergence for every fixed $a\in \mathbb{R}$. As usual, this means convergence of a numerical method on each fixed finite interval for a sufficiently regular ordinary differential equation with a Lipschitz continuous vector field and starting values tending to the exact starting values. When $a\ne 0$, the implicit time-stepping method update is locally uniquely solvable for sufficiently small $h$, for example by a contraction mapping if $|ha|L<1$. Zero-stability does not assert that a large fixed step is suitable for a stiff differential equation.

The first characteristic polynomials of a linear multistep method isThe root condition for a multistep method requires $|\alpha |\le 1$ and excludes $\alpha =1$, where the root of a polynomial at one is double. At $\alpha =-1$, the two unit-modulus roots of a polynomial are distinct, so zero-stability holds. For $\alpha \ne 1$, the first-order consistency of a numerical method identity is ${\rho }^{\prime }(1)=1-\alpha$, equal to the coefficient of $hf$.

Using the permitted Dahlquist equivalence theorem in this multiderivative multistep method setting givesThis means convergence on every fixed finite time interval where the smooth exact solution of a differential equation exists, with consistent starting values and the nearby branch of each implicit solve. More specifically, the convergence of a zero-stable multiderivative method gives order $p$ when the starting errors are $O(h^p)$ and both $f$ and $f^{\prime }f$ are locally uniformly Lipschitz continuous on the relevant region.

The exclusions are genuine. For $y^{\prime }=0$ and $|\alpha |>1$, an arbitrarily small parasitic starting error is amplified as ${\alpha }^n$. At $\alpha =1$, take $y_0=0$, $y_1=h$, also for $y^{\prime }=0$. The exact solution is zero, but the numerical recurrence gives $y_n=nh$, an error of order one at fixed positive time. This proves failure even though the starting values converge.

For a fixed-step linear multistep method with a nonzero leading coefficient, applied to an ordinary differential equation with a locally Lipschitz continuous vector field, the Dahlquist equivalence theorem states thatHere convergence is uniform on each fixed finite time interval as the step size tends to zero, for every set of starting values tending to the corresponding exact values. Consistency of a numerical method means that the normalized local defect tends to zero; for the characteristic polynomials of a linear multistep method it gives $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)$. Zero-stability is equivalent to the root condition for a multistep method: every root of a polynomial $\xi$ of $\rho$ satisfies $|\xi |\le 1$, and roots with $|\xi |=1$ are simple. Order $p\ge 1$, together with zero-stability and starting errors $O(h^p)$, gives global error $O(h^p)$ under the usual smoothness assumptions.

To prove the necessity of the root condition for multistep convergence, use the test problem $y^{\prime }=0$, $y(0)=0$. Its discrete error satisfies $\rho (E)e_n=0$, where $E{e}_n=e_{n+1}$. Set $h=T/N$. If $|\xi |>1$, the exact recurrence solutionhas starting errors tending to zero at every fixed starting index, but $|e_N|=1$. Thus convergence fails.

If a unit-modulus root has multiplicity $m\ge 2$, the recurrence instead admitsAgain every fixed starting error tends to zero while $|e_N|=1$. These polynomial-times-exponential solutions follow from the repeated factor $(E-\xi {)}^m$. Complex modes can be interpreted as a real two-component system, or their real and imaginary parts can be used. Therefore convergence requires precisely the root condition for a multistep method. Repeated roots strictly inside the unit disk are allowed, since their polynomial factors are dominated by exponential decay.

For this Adams–Moulton method, the characteristic polynomials of a linear multistep method areExpansion at zero givesPart (a) therefore shows that the method has order exactly three.

It is consistent because its order is at least one. The roots of $\rho (w)=w(w-1)$ are $0$ and $1$; both lie in the closed unit disk, and the only unit-modulus root is simple. Thus the method is zero-stable. By the Dahlquist equivalence theorem, consistency and zero stability imply convergence. Hence the method is

For $s=1$, imposing order two givesThe method is therefore the trapezoidal ruleIts first characteristic polynomial is $\rho (\zeta )=\zeta -1$, so it satisfies the root condition for a multistep method.

For $s=2$, the four conditions through order three giveand henceHereso the root condition for a multistep method again holds. Both methods are consistent and zero-stable, and the Dahlquist equivalence theorem therefore proves that both are convergent.

The characteristic polynomials of a linear multistep method areandFor the order conditions for a linear multistep method, put $\alpha =(a,-1-a,1)$ and $\beta =(-(1+5a)/12,,8(1-a)/12,,(5+a)/12)$. The defectsvanish for $q=0,1,2,3$, whileThereforeIndeed, at $a=-1$ one has $C_4=0$ but $C_5=-4/3\ne 0$.

A linear multistep method for $y^{\prime }=f(t,y)$ has the formIts characteristic polynomials of a linear multistep method are $\rho (\zeta )={\sum }_j{\alpha }_j{\zeta }^j$ and $\sigma (\zeta )={\sum }_j{\beta }_j{\zeta }^j$. Substitution of the exact solution, or equivalently expansion of $\rho (e^z)-z\sigma (e^z)$ at $z=0$, gives the order conditions for a linear multistep method. The method has order $p$ whenwith a nonzero coefficient at the next power. Its one-step defect is then $O(h^{p+1})$, while its global error is $O(h^p)$ when stability prevents accumulation from being amplified.

Order alone does not ensure convergence because the recurrence may contain growing parasitic modes. Zero-stability is the root condition for a multistep method: every root of $\rho$ lies in $|\zeta |\le 1$, and every root on the unit circle is simple. Consistency is the first-order conditionThe Dahlquist equivalence theorem states that a consistent linear multistep method is convergent if and only if it is zero-stable. For example, the leapfrog method $y_{n+2}-y_n=2h{f}_{n+1}$ has $\rho (\zeta )={\zeta }^2-1$; its roots $\pm 1$ are simple, so this second-order method is convergent, although its parasitic $-1$ mode can oscillate.

To study stiff decay, apply the method to $y^{\prime }=\lambda y$ and put $z=h\lambda$. The amplification factors are the roots ofThe linear stability domain consists of the $z$ for which these roots satisfy the strict interior condition appropriate away from the boundary, with simple unit roots where boundary stability is admitted. The method is A-stable when this domain contains the left half-plane, so every mode with $\mathrm{Re}\lambda <0$ remains bounded for every positive step size. The Second Dahlquist barrier says that an A-stable multistep method has order at most two.

The main families illustrate the tradeoffs. The explicit Adams-Bashforth methods interpolate past derivative values and are inexpensive, but their stability domains are bounded. The implicit Adams-Moulton methods include the new derivative value; the one-step second-order member is the trapezoidal rule, which is A-stable. The backward differentiation formulas interpolate past solution values and differentiate the interpolant; the first two are A-stable, while higher orders retain useful sectors of the left half-plane. Implicit methods require a nonlinear solve at each step, commonly initialized by an explicit predictor to form a predictor-corrector pair. Thus order controls consistency error, the root condition controls convergence as $h\to 0$, and the amplification polynomial controls whether a convergent method remains useful on stiff equations.