Test $\mathbf{C}$ Let $u, f$ and $g$ be functions defined for all real numbers $x$, and satisfy the conditions

and
$$
\begin{array}{l}
\frac{u(x+1)+u(x-1)}{2}=f(x), \\
\frac{u(x+4)+u(x-4)}{2}=g(x),
\end{array}
$$

Express $u(x)$ in terms of $f$ and $g$.
(38th Putnam Competition Problem A-3)
We can prove: there are infinitely many expressions for $u(x)$ in terms of $f$ and $g$, with some of the simpler ones being
$$
\begin{aligned}
u(x)= & g(x)-f(x+3)+f(x+1) \\
& +f(x-1)-f(x-3) \\
= & -g(x+2)+f(x+5)-f(x+3) \\
& +f(x+1)+f(x-1) \\
= & g(x+4)-f(x+7)+f(x+5) \\
& -f(x+3)+f(x+1) .
\end{aligned}
$$

Let $E$ be the shift operator for the function $A$, defined by $E A(x)=A(x+1)$. Then the given conditions are $\left(E+E^{-1}\right) u(x)=2 f(x)$ and $\left(E^{4}+E^{-4}\right) u(x)=2 g(x)$, so we have $\left(E^{2}+1\right) u(x)=2 E f(x)$ and $\left(E^{8}+1\right) u(x)=2 E^{4} g(x)$. Since $E^{2} +1$ and $E^{8}+1$ are coprime polynomials in $E$, we can find
$$
\begin{array}{c}
1=\frac{1}{2}\left(E^{8}+1\right)-\frac{1}{2}\left(E^{6}-E^{4}+E^{2}-1\right)\left(E^{2}+1\right), \\
u(x)=\frac{1}{2}\left(E^{8}+1\right) u(x)-\frac{1}{2}\left(E^{6}-E^{4}+E^{2}-1\right)\left(E^{2}+1\right) u(x), \\
u(x)=E^{4} g(x)-\left(E^{6}-E^{4}+E^{2}-1\right) E f(x), \\
u(x)=E^{4} g(x)+\left(-E^{7}+E^{5}-E^{3}+E\right) f(x), \\
u(x)=g(x+4)-f(x+7)+f(x+5)-f(x+3)+f(x+1) .
\end{array}
$$

Other expressions can be obtained from
$$
\begin{aligned}
g(y) & =-g(y-2)+f(y+3)+f(y-5) \\
& =-g(y+2)+f(y+5)+f(y-3)
\end{aligned}
$$