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Along the exact ordinary differential equation solution, the chain rule gives $g(y)=f^{\prime }(y)f(y)=y^{\prime \prime }$. Insert the exact solution into the symmetric two-step two-derivative formula and expand about the middle time $t=t_{n+1}$. Its unnormalized local truncation error isThe coefficients of $h{y}^{\prime }$, $h^3{y}^{\prime \prime \prime }$ and $h^5{y}^{(5)}$ are respectivelyThe centered residual is an odd function of $h$, so every even power cancels. The next coefficient isThus the method has order six. Its zero-stability roots at $h=0$ are the simple roots $1,-1$. With sufficiently accurate starts and smooth derivative evaluation maps, convergence of a zero-stable multiderivative method gives sixth-order error on fixed time intervals. The first Dahlquist barrier for ordinary linear multistep methods does not apply, because this is a multiderivative multistep method.

Along a smooth exact solution, the chain rule gives $g(y)=f^{\prime }(y)f(y)=y^{\prime \prime }$. Expand the unscaled defect of the multiderivative multistep method about the central time $t_n$. The antisymmetric difference of the solution and second derivative, and the symmetric sum of first derivatives, giveAll derivatives are evaluated at $t_n$. Reflection symmetry removes every even power. The fifth-order conditions require the coefficients of $h,h^3,h^5$ to vanish. The last two equations give $\beta =1/15$ and $\alpha =7/15$, and the first then gives $\gamma =16/15$. ThusThese determine the symmetric two-step two-derivative formula. Substitution into the next coefficient givesThe requested fifth-order conditions actually produce formal order six. In the usual maximal-order terminology there is no member of this symmetric family with exact order five: cancelling the fifth-power defect also cancels the automatically zero sixth-power defect. Interpreting the printed request as order at least five gives precisely the boxed parameters.

At zero step size the recurrence polynomial is $w^2-1$, with simple roots $1$ and $-1$, so it satisfies the root condition for a multistep method and is zero-stable. With sufficiently smooth locally Lipschitz derivative maps, the nearby implicit solution branch and starting values of matching accuracy, the convergence of a zero-stable multiderivative method gives global order six. Formal order by itself does not imply decay of its additional numerical mode.

For an autonomous ordinary differential equation, the chain rule gives $y^{\prime \prime }=f^{\prime }(y)f(y)$. The formula is therefore a multiderivative multistep method. Insert the exact solution and expand the two earlier values about $t=t_{n+2}$:The unscaled local truncation error isThe coefficients through $h^3$ vanish, but the fourth-order coefficient does not. Hence the method has order three, not four: the one-step residual is $O(h^{p+1})$ for an order-$p$ multistep formula. Its zero-stability polynomial is $(\zeta -1)(\zeta -1/7)$, with a simple unit root and the other root inside the unit disk. Thus the stated order is also the convergence order for smooth solutions, starting errors $O(h^3)$, and the nearby implicit solution branch, by convergence of a zero-stable multiderivative method.

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.

Expand about $t=t_{n+2}$ and use $y^{\prime \prime }=f^{\prime }(y)f(y)$ from the chain rule. The residual of the exact solution isThe coefficients through $h^3$ vanish, and the displayed fourth-order coefficient does not. Thus the multiderivative multistep method has order three.

At $h=0$, its first characteristic polynomial isIts roots are $1$ and $1/7$, with the unit-modulus root simple, so it satisfies the root condition for a multistep method and is zero-stable. Combining this with the defect estimate proves convergence of order three, assuming sufficiently smooth $f$, order-three starting values, and the nearby branch of each implicit update.

More explicitly, the right-hand side is $h{F}_h(y_{n+2})$, where $F_h=6f/7-2h{f}^{\prime }f/7$ has a uniform local Lipschitz continuity bound for small $h$. A bounded zero-stable impulse response and the discrete Gronwall inequality therefore control the accumulated $O(h^4)$ defects by $O(h^3)$ over fixed time intervals. This is convergence of a zero-stable multiderivative method, rather than a direct application of the first-derivative-only Dahlquist equivalence theorem.