= Solution
Let $h=\Delta x$, $k=\mu h$ and $M_r=\sum_{j\in\{-2,-1,1,2\}}a_jj^r$. Along a smooth solution, $\partial_t^r u=\partial_x^r u$. The exact difference $u(x,t+k)-u(x,t-k)$ contains only odd <derivatives>. Matching terms through degree four therefore requires
$$
M_0=M_2=M_4=0,\qquad M_1=2\mu,\qquad M_3=2\mu^3.
$$
The first two conditions give $a_{-1}+a_1=a_{-2}+a_2=0$, which also implies $M_4=0$. With $a_{-j}=-a_j$, the remaining equations are $a_1+2a_2=\mu$ and $a_1+8a_2=\mu^3$. Hence
$$
\boxed{a_1=\frac{\mu(4-\mu^2)}3,\qquad a_2=\frac{\mu(\mu^2-1)}6,\qquad a_{-1}=-a_1,\quad a_{-2}=-a_2.}
$$
The fifth spatial moment is $M_5=10\mu^3-8\mu$. After dividing the exact-minus-scheme residual by $2k$, its first possible nonzero term is
$$
\frac{h^4}{120}(\mu^2-1)(\mu^2-4)u_{xxxxx}+O(h^6).
$$
This is the <fourth-order two-step advection stencil>. Thus the normalized <local truncation error> is $O(h^4)$ at fixed positive <Courant number>, as required. At $\mu=1$ or $2$ the formula translates the physical advection branch exactly, but <consistency of a numerical method> alone does not decide <stability> of all two-level perturbations.
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