Two-dimensional relaxation calculations commonly use sites on a square Iattice with spacing $\Delta x=\Delta y=h$, and label the sites by $(i, j)$, where $i, j$ are integers and $x_{i}=$ $i h+x_{0}, y_{j}=j h+y_{0} .$ The value of the potential at $(i, j)$ can be approximated by the average of the values at neighboring sites. [Recall the relevant theorem about harmonic functions.] But what average?
(a) If $F(x, y)$ is a well-behaved function in the neighborhood of the origin, but not necessarily harmonic, by explicit Taylor series expansions, show that the "cross"' sum
$$
S_{e}=F(h, 0)+F(0, h)+F(-h, 0)+F(0,-h)
$$
can be expressed as
$$
S_{c}=4 F(0,0)+h^{2} \nabla^{2} F+\frac{h^{4}}{12}\left(F_{x x x}+F_{m q y}\right)+O\left(h^{6}\right)
$$
(b) Similarly, show that the "square" sum,
$$
S_{\mathrm{S}}=F(h, h)+F(-h, h)+F(-h,-h)+F(h,-h)
$$
can be expressed as
$$
S_{\mathrm{S}}=4 F(0,0)+2 h^{2} \nabla^{2} F-\frac{h^{4}}{3}\left(F_{x x x}+F_{y y y y}\right)+\frac{h^{4}}{2} \nabla^{2}\left(\nabla^{2} F\right)+O\left(h^{6}\right)
$$
Here $F_{\text {mexr }}$ is the fourth partial derivative of $F$ with respect to $x$, evaluated at $x=0, y=0$, etc. If $\nabla^{2} F=0$, the averages $S_{d} / 4$ and $S_{3} / 4$ each give the value of $F(0,0)$, correct to order $h^{2}$ inclusive. Note that an improvement can be obtained by forming the "improved" average,
$$
\langle\langle F(0,0)\rangle\rangle=\frac{1}{5}\left[S_{e}+\frac{1}{4} S_{4}\right]
$$
where
$$
\langle(F(0,0))\rangle=F(0,0)+\frac{3}{10} h^{2} \nabla^{2} F+\frac{h^{4}}{40} \nabla^{2}\left(\nabla^{2} F\right)+O\left(h^{6}\right)
$$
If $\nabla^{2} F=0$, then $S$ gives $F(0,0)$, correct to order $h^{5}$ inclusive. For Poisson's equation, the charge density and its lowest order Laplacian can be inserted for the same accuracy.