Newton–Gregory forward formula

Remarks

Proof idea: the Newton polynomial gives $p(x) = \sum_{k=0}^{n} f[x_0, \dots, x_k] \prod_{j<k} (x - x_j)$. For equally spaced nodes, induction on $k$ shows that the divided difference is $f[x_0, \dots, x_k] = \Delta^k f(x_0) / (k!\, h^k)$. With $x = x_0 + s h$ one has $\prod_{j<k} (x - x_j) = h^k s (s - 1) \cdots (s - k + 1)$, and the two formulas combine.

Example: $f(x) = x^2$, $x_0 = 0$, $h = 1$, $n = 2$. The values are $0, 1, 4$, so $\Delta f(0) = 1$ and $\Delta^2 f(0) = 2$. The formula gives $p(s) = 0 + s + \binom{s}{2} \cdot 2 = s + s(s - 1) = s^2$, which is $f$ itself.

The formula needs equal spacing; for unequal nodes the general Newton form with divided differences must be used.

Going deeper

For a student

The formula is the basis of classical table interpolation and of many finite difference formulas. Differentiating or integrating it term by term gives numerical differentiation rules and the Newton–Cotes formulas. The error is that of polynomial interpolation: if $f$ is $n + 1$ times differentiable, then $f(x_0 + s h) - p(x_0 + s h) = h^{n+1} \binom{s}{n+1} f^{(n+1)}(\xi)$ for some $\xi$ in the smallest interval containing $x_0, x_n$ and $x_0 + s h$. There is a matching backward formula, based at $x_n$ with backward differences.