Taylor moment conditions characterize the order of a linear multistep method

Fix $t_n$ and expand each $y(t_n+jh)$ through degree $p$ and each $y^{\mathrm{\prime }}(t_n+jh)$ through degree $p-1$ by [F2], on the interval containing the finitely many nodes. Bounded derivatives on the compact interval make the remainders uniform over $n$ and $j$.

Substituting those expansions into [F1] and collecting the coefficient of $y^{(m)}(t_n)h^m\mathrm{/}m!$ gives ${\tau }_n(h;y)={\sum }_{m=0}^p{\mu }_m\frac{h^m}{m!}{y}^{(m)}(t_n)+O(h^{p+1})$. The derivative term contributes $m{\sum }_j{\beta }_j{j}^{m-1}$ when $m\ge 1$ and contributes nothing for $m=0$.

If ${\mu }_0,\dots ,{\mu }_p$ vanish, step 2.1 gives ${\tau }_n=O(h^{p+1})$ uniformly, so the method has order at least $p$.

Conversely, suppose the method has order at least $p$. Induct on $m=0,\dots ,p$, testing the residual on the polynomial solution $y(t)=t^m$ of $y^{\mathrm{\prime }}=m{t}^{m-1}$ (with $y=1$, $y^{\mathrm{\prime }}=0$ when $m=0$). Lower moments vanish by the induction hypothesis, so the residual is exactly ${\mu }_m{h}^m$; since $m\le p$, it can be $O(h^{p+1})$ only if ${\mu }_m=0$. Thus every required moment vanishes.

Finally, ${\mu }_0=\rho (1)$ and ${\mu }_1={\rho }^{\mathrm{\prime }}(1)-\sigma (1)$ by the polynomial definitions. Their vanishing is exactly consistency.