= Solution
Let $h=\Delta x$, $k=\Delta t$ and $\mu=k/h$. For the exact solution $u(x,t)=F(x+t)$ of the <advection equation>, the three left-hand samples are translations by $(\mu-1)h$, $\mu h$ and $(\mu+1)h$ relative to $(x_m,t_n)$; the right-hand samples are translations by zero and $h$.
Denote their coefficients by $a=\mu(1+\mu)/6$, $b=(2-\mu)(1+\mu)/3$, $c=(2-\mu)(1-\mu)/6$, $d=(2-\mu)/3$ and $e=(1+\mu)/3$. <Taylor expansion> compares moments
$$
M_j=a(\mu-1)^j+b\mu^j+c(\mu+1)^j-d\,0^j-e,
$$
where $0^0=1$ in the zeroth moment. Direct calculation gives
$$
M_0=M_1=M_2=M_3=0,\qquad
M_4=\frac{\mu(\mu-2)(\mu-1)(\mu+1)}3.
$$
Thus the unscaled exact-solution residual is
$$
\mathcal R_h=\frac{h^4}{72}\mu(\mu-2)(\mu-1)(\mu+1)u_{xxxx}+O(h^5).
$$
For fixed nonzero $\mu$, dividing by the time step yields
$$
\boxed{\frac{\mathcal R_h}{k}
=\frac{h^3}{72}(\mu-2)(\mu-1)(\mu+1)u_{xxxx}+O(h^4).}
$$
The scheme is therefore \b[third order under fixed-Courant refinement], with an unscaled one-step defect of order four. Stability then gives third-order accumulated error for smooth data with <periodic boundary conditions>. Calling the unscaled defect fourth order is a different convention, not fourth-order approximation of the differential equation.
At $\mu=-1,1,2$, the amplification factor in (b) becomes exactly $e^{i\mu\theta}$: the numerical step translates the grid data by that integer number of cells and is exact for this constant-speed <advection equation>. At $\mu=0$ it is the identity, a zero-time-step limit. These are exact special shifts, rather than merely a cancellation of one term of the <Taylor expansion>. This is the <implicit advection scheme with exact integer shifts>.
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