Five-point stencil

Remarks

The one-dimensional weights are the finite difference coefficients for the stencil $-2, -1, 0, 1, 2$. For $f \in C^5$, the first formula has error $-\frac{h^4}{30} f^{(5)}(\xi)$ for some $\xi$ in $(x - 2h, x + 2h)$, so it has order $4$; the second has order $4$ for $f \in C^6$. Check of the second formula on $f(s) = s^2$ at $x = 0$: $\frac{-4h^2 + 16h^2 + 16h^2 - 4h^2}{12h^2} = 2 = f''(0)$. For $u \in C^4$, each one-dimensional second difference has error $\frac{h^2}{12} \partial^4 u$ at an intermediate point, so $\abs{\Delta_h u - \Delta u} \le \frac{h^2}{12} \big( \max \abs{u_{xxxx}} + \max \abs{u_{yyyy}} \big)$ and the five-point Laplacian approximates the Laplacian to second order. Non-example: the nine-point and the diagonal five-point stencil $(x \pm h, y \pm h)$ also approximate $\Delta u$, the latter as $\frac{1}{2h^2}\big(\sum u(x \pm h, y \pm h) - 4u(x,y)\big)$, but it uses different points and is not the five-point stencil defined here.