Newton's forward difference formula

Remarks

The shift operator satisfies $E_h = I + \Delta_h$, and $I$ commutes with $\Delta_h$. So the binomial theorem gives $E_h^n = \sum_k \binom{n}{k} \Delta_h^k$, which is the formula applied at $x$. Example: for $f(x) = x^2$, $h = 1$ and $x = 0$, the differences at $0$ are $f(0) = 0$, $\Delta f(0) = 1$, $\Delta^2 f(0) = 2$, so $f(n) = n + 2\binom{n}{2} = n + n(n-1) = n^2$. Let $f \colon \R \to \R$ and $h \ne 0$, and replace the top entry $n$ of $\binom{n}{k}$ by a real number $s$. This gives the Newton forward interpolation polynomial $$p(x_0 + sh) = \sum_{k=0}^{n} \binom{s}{k} \Delta_h^k f(x_0), \qquad \binom{s}{k} = \frac{s(s-1)\cdots(s-k+1)}{k!}.$$ It is the unique polynomial of degree at most $n$ that agrees with $f$ at $x_0, x_0 + h, \dots, x_0 + nh$.

Going deeper

For a student

The formula and the expansion of iterated differences in shifts are inverse to each other: they express the binomial transform and its inverse. In numerical analysis it gives the interpolation polynomial at equally spaced nodes, a special case of polynomial interpolation. Letting $n \to \infty$ with real $s$ leads to the Newton series of $f$. Over the $p$-adic integers, the same expansion with infinitely many terms is Mahler's theorem.