= Solution
Keep the <Courant number> $\mu=\Delta t/\Delta x$ fixed and write $h=\Delta x$. An exact solution of $u_t=u_x$ is $u(x,t)=g(x+t)$. Substituting it into the update gives the defect
$$
g(s+\mu h)-(1-2\mu)\{g(s)-g(s+h)\}-g(s+(1-\mu)h)
=\frac{\mu(1-\mu)(1-2\mu)}6h^3g^{(3)} + O(h^4),
$$
where $s=x+t$ and $g^{(3)}$ is evaluated at $s$. The constant, first-derivative and second-derivative terms cancel. Dividing by $\Delta t=\mu h$ gives a second-order consistency error. Hence for fixed $0<\mu<1$, $\mu\ne1/2$, the scheme has \b[order two].
At the special value $\mu=1/2$, the update reduces to $u_m^{n+1}=u_{m+1}^{n-1}$. This is exactly the transport over two time steps, since $2\Delta t=\Delta x$, so the defect vanishes identically. With exact data at both starting levels it transports their values exactly; errors in those starting levels are still present, and the even and odd time subsequences evolve separately.
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