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Work in the standard class of real constant-coefficient linear multistep methods using only first-derivative evaluations. Writewith ${\alpha }_s\ne 0$. The Dahlquist equivalence theorem states that such a method converges on each fixed finite time interval, for sufficiently regular Lipschitz initial value problems and every set of consistent starting values, if and only if it is consistent and zero-stable. Consistency means $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)$. Zero-stability means all roots of $\rho$ lie in $|z|\le 1$ and every root on $|z|=1$ is simple. In particular the consistency root at $1$ is simple.

Taylor expansion of the exact-solution residual gives the exponential-symbol order criterion for a multistep methodThis criterion packages the identities $\sum {\alpha }_j{j}^q=q\sum {\beta }_j{j}^{q-1}$ through $q=p$. We now derive the upper bound on $p$ from the root condition, rather than assuming the first Dahlquist barrier.

Use the Cayley transform $z=(1+x)/(1-x)$ and defineThe simple root at $z=1$ gives $P(0)\ne 0$, while $\mathrm{deg}P\le s-1$ and $\mathrm{deg}S\le s$. Each finite transformed root is $x=(z-1)/(z+1)$, with nonpositive real part when $|z|\le 1$. A possible root at $z=-1$ simply lowers the degree of $R$. Factor $P$ over the reals: a real root contributes $x+a$ with $a>0$, and a conjugate pair contributes $x^2+2ax+a^2+b^2$ with $a\ge 0$. Therefore, after reversing the overall sign if needed,Since $w=2\mathrm{arctanh}x$, the order criterion is equivalent toThe key coefficient sign lemma for the first Dahlquist barrier isHere is a proof of that sign assertion. Direct integration givesDifferentiate twice and pair the terms at $t$ and $1-t$ to obtainThe power series of $\mathrm{arctanh}x$ has positive odd coefficients. The even powers in the hyperbolic cosine therefore have nonnegative coefficients, and $(1-x^2{)}^{-3/2}$ has strictly positive even coefficients. Every even coefficient of $F^{\prime \prime }$ is consequently negative. Also $F$ is even and $F(0)=1$, proving the assertion.

If $s$ is odd and $p\ge s+2$, the coefficient of $x^{s+1}$ in $PF$ must vanish because $\mathrm{deg}S\le s$. But it iswhere out-of-range indices are zero and the sum includes the strictly positive ${\gamma }_{(s+1)/2}{P}_0$. This is impossible, so $p\le s+1$. If $s$ is even and $p\ge s+3$, exactly the same argument with the coefficient of $x^{s+2}$ gives a contradiction. Thus $p\le s+2$ in that case. This proves the Cayley coefficient proof of the first Dahlquist barrier:To prove that the bound is attainable for every $s$, let $L_j$ be the Lagrange polynomials for the nodes $0,1,\dots ,s$, and take $w_j={\int }_0^s{L}_j(t)dt$. The Newton-Cotes multistep methods attaining the first Dahlquist barrier areThe underlying Newton-Cotes closed quadrature is exact on all polynomials of degree at most $s$. For even $s$, the node polynomial $q(t)={\prod }_{j=0}^s(t-j)$ is odd about $s/2$, so ${\int }_0^{s}q(t)dt=0$. Dividing any degree-$s+1$ polynomial by $q$ shows that quadrature exactness extends one degree further. Integrating the exact derivative therefore gives method order at least $s+1$ for odd $s$ and at least $s+2$ for even $s$.

Finally $\rho (z)=z^s-1$ has only simple unit-circle roots, so these methods satisfy the root condition. The weights sum to $s={\rho }^{\prime }(1)$, proving consistency. The Dahlquist equivalence theorem gives convergence with consistent starts; the proved upper bound makes the attained formal orders exact; starting values accurate to those orders also give the corresponding global accuracy. Hence the highest convergent order is $2\lfloor (s+2)/2\rfloor$. This result concerns the stated linear first-derivative class, not unrestricted multiderivative or nonlinear formulas.