= Solution
{c}
Work on a fixed finite interval $[a,a+T]$. Write $Y_n=y(x_n)$, $t=x_n+h/2$, and use the paper's <normalized local truncation error>
$$
\tau_n=\frac{Y_{n+1}-Y_n}{h}
-f\left(t,\frac{Y_{n+1}+Y_n}2\right).
$$
Assume $y$ has bounded continuous <derivatives> through order three on this interval, with $\|y''\|\leq M_2$ and $\|y'''\|\leq M_3$. Assume also that the <Jacobian matrix> $f_y$ has <operator norm> at most $L$ on the segments joining $y(t)$ to the endpoint averages. For a scalar equation this last hypothesis is simply $|\partial f/\partial y|\leq L$ there.
Two symmetric estimates separate the errors. By <Taylor's theorem> applied to $y'$ about $t$, the linear term integrates to zero on $[-h/2,h/2]$, giving
$$
\begin{aligned}
\left\|\frac{Y_{n+1}-Y_n}{h}-y'(t)\right\|
&=\left\|\frac1h\int_{-h/2}^{h/2}\bigl(y'(t+s)-y'(t)\bigr)\,ds\right\|\\
&\leq\frac{M_3}{2h}\int_{-h/2}^{h/2}s^2\,ds
=\frac{M_3h^2}{24}.
\end{aligned}
$$
Likewise the two linear terms in the endpoint expansions cancel, and their second-order remainders give
$$
\left\|\frac{Y_{n+1}+Y_n}2-y(t)\right\|\leq\frac{M_2h^2}{8}.
$$
The derivative bound on $f$ gives a <Lipschitz condition> in its second variable. Since $y'(t)=f(t,y(t))$, the <triangle inequality> now yields
$$
\boxed{\|\tau_n\|\leq
\left(\frac{M_3}{24}+\frac{LM_2}{8}\right)h^2
=Kh^2.}
$$
The constant is independent of $n$ and $h$ as long as the steps remain in the fixed interval. This <implicit midpoint consistency bound> proves <numerical consistency> of order two. The corresponding unnormalized residual is $h\tau_n=O(h^3)$; the two powers conventionally called <local truncation error> differ because the paper divides by the step size. If local error instead means the error of an exactly solved numerical step started at $Y_n$, the small-step <Lipschitz condition> gives the bound $h\|\tau_n\|/(1-hL/2)=O(h^3)$ as well.
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