Let \(f\colon \mathbb{Z}^d\to \mathbb{C}\). The Laplacian of \(f\) is defined by
\[ \Delta f(x):= \sum_{i=1}^{d} \partial_i^2 f(x). \]The definition of a partial derivative of \(f\) implies
\[ \Delta f(x) = \frac{1}{4} \sum_{h\in \{\pm 2e_1,\ldots ,\pm 2e_d\}} f(x+h)-f(x). \]
Proof
For a fixed \(i\in \{1,\ldots,d\}\) we obtain
\begin{align*} \partial_i^2 f(x) &= \frac{1}{2} \bigl[\partial_i f(x+e_i) - \partial_i f(x-e_i)\bigr] \\ &= \frac{1}{4} \bigl[f(x+2e_i)-f(x) - (f(x)-f(x-2e_i))\bigr]\\ &= \frac{1}{4} \bigl[f(x+2e_i)-f(x) + f(x-2e_i)-f(x)\bigr]. \end{align*}Finally, summing over all \(i\in \{1,\ldots ,d\}\) gives
\[ \Delta f(x) = \frac{1}{4}\sum_{h\in \{\pm 2e_1,\ldots ,\pm 2e_d\}} f(x+h)-f(x). \]
Remarks
- There are other definitions for the Laplacian on \(\mathbb{Z}^d\).