Expansion of iterated differences in shifts

Remarks

Write $\Delta_h = E_h - I$, where $E_h$ is the shift operator and $I$ is the identity. The operators $E_h$ and $I$ commute, so the binomial theorem gives $(E_h - I)^n = \sum_k \binom{n}{k} (-1)^{n-k} E_h^k$, and $E_h^k f(x) = f(x + kh)$. A direct proof is an induction on $n$ with Pascal's rule. Example: for $n = 3$ the formula reads $\Delta_h^3 f(x) = f(x + 3h) - 3f(x + 2h) + 3f(x + h) - f(x)$. With $f(x) = x^3$, $h = 1$ and $x = 0$ this gives $27 - 24 + 3 - 0 = 6 = 3!$.

Going deeper

For a student

The formula is the inverse of Newton's forward difference formula, which writes shifted values in terms of differences. Together the two formulas say that the sequences $(f(x + kh))_k$ and $(\Delta_h^k f(x))_k$ are each other's binomial transform. The second difference $f(x + h) - 2f(x) + f(x - h)$, divided by $h^2$, is the standard finite difference approximation of $f''(x)$. Combined with the fact that differences of polynomials vanish beyond the degree, it gives identities. An example is $\sum_{k=0}^n (-1)^k \binom{n}{k} k^m = 0$ for $0 \le m < n$.