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An amplification root is a root of a polynomial governing a geometric mode $y_n={\xi }^n$ of a numerical recurrence on the Dahlquist test equation. Its modulus measures damping or growth per step. A repeated root produces additional modes such as $n{\xi }^n$; therefore a unit-circle root can be permitted by a bounded-root condition only when it is simple. The principal root approximates $e^z$ near zero, while additional roots can describe parasitic amplification roots.

Replacing the derivative in $x^{\prime }=-Kx$ by a central finite difference gives $x_{n+1}+2Kh{x}_n-x_{n-1}=0$. Its principal root is ${\rho }_{+}=e^{-\mathrm{arsinh}(Kh)}$, while the parasitic amplification root is ${\rho }_{-}=-e^{\mathrm{arsinh}(Kh)}$. On a fixed time interval, $A_h{\rho }_{+}^n+B_h{\rho }_{-}^n$ converges uniformly to $C{e}^{-Kt}$ exactly when $A_h\to C$ and $B_h\to 0$. A fixed nonzero parasitic amplitude alternates between two incompatible limits, but a vanishing starting error can converge despite the root's modulus exceeding one. This distinguishes fixed-time numerical convergence from long-time absolute stability.

With $h=\delta t>0$, the central finite difference converts the decay equation toThe continuous ordinary differential equation has the general solution $x(t)=C{e}^{-Kt}$. The discrete linear recurrence is second order and requires two initial values, such as $x_0,x_1$. Trying $x_n={\rho }^n$ gives the characteristic equation of a linear recurrence ${\rho }^2+2Kh\rho -1=0$. Its roots aresoThere are no repeated roots, and the same formula extends to negative integer $n$ if required.

To demonstrate the continuum limit, put $a_h=\mathrm{arsinh}(Kh)$. Then ${\rho }_{+}=e^{-a_h}$ and ${\rho }_{-}=-e^{a_h}$, with $a_h/h\to K$. For $t_n=nh\in [0,T]$,uniformly, since $0\le t_n\le T$. More precisely $a_h/h=K-K^3{h}^2/6+O(h^4)$, giving an $O(h^2)$ error on a fixed interval. Thus $B=0$, $A=C$ selects the physical discrete mode and corresponds to $x_1={\rho }_{+}{x}_0$.

The other root is a parasitic amplification root:For fixed $B\ne 0$, even and odd mesh points approach values differing by $2B{e}^{Kt}$. Equivalently successive mesh values have a nonvanishing alternating jump despite their time separation tending to zero. This mode cannot converge uniformly to a continuous ODE solution; its limiting envelope also grows rather than decays.

For step-dependent initial data the precise result is the centered two-step discretization of exponential decay criterion:Sufficiency follows from the two uniformly bounded exponential envelopes on a fixed interval. Necessity follows from convergence at the adjacent points $0,h$: $x_0\to C$, $x_1-x_0\to 0$, and the coefficient formulas force $B_h\to 0$, $A_h\to C$. In particular a nonzero but vanishing starting error is not excluded. Starting with the exact value $x_1=C{e}^{-Kh}$ gives $B_h=C{K}^3{h}^3/12+O(h^4)$ and still converges on each fixed interval. The assertion about “remaining” solutions therefore concerns a fixed nonzero parasitic amplitude, not every nonzero step-dependent amplitude. The method is nevertheless not damping-stable for arbitrarily long times because $|{\rho }_{-}|>1$.

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.

Apply the method to the Dahlquist test equation $y^{\prime }=\lambda y$, so $g(y)={\lambda }^{2}y$, and put $z=h\lambda$. A trial numerical mode $y_n=w^n$ satisfiesFor every real $z<0$, the leading coefficient is positive, whileThe intermediate value theorem gives a real amplification root $w<-1$. Its modulus exceeds one although the exact scalar solution is decaying. ThereforeFor a concrete choice, $z=-1$ gives $23w^2+16w-9=0$, whose negative root is $(-8-\sqrt{271})/23<-1$. This proves negative-real instability of the symmetric two-derivative formula directly, without using either an order barrier or the supplied quadratic criterion. The unstable branch is the parasitic amplification root issuing from $w=-1$ at $z=0$; its small-step expansion is $w=-1+z/15+O(z^2)$, so a negative $z$ moves it outside the unit circle immediately.