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Discrete Taylor duality
The forward difference \(\Delta\) and translation \(T\) E0004 are Möbius duals: each determines the other by inclusion–exclusion.
(1) Proposition. Let \(X, Y\) be abelian groups, \(g: X \to Y\), \(x \in X\), \(u_\bullet = (u_1, \dots, u_k) \in X^k\), and \(y = g(x)\).
-
Boolean Taylor duality.
\[ g(x + {\textstyle\sum_{i=1}^k} u_i) = y + \sum_{\emptyset \neq S \subseteq [k]} \Delta(g; x; u_S), \qquad \Delta(g; x; u_\bullet) = \sum_{S \subseteq [k]} (-1)^{k-|S|}\, g(x + {\textstyle\sum_{i \in S}} u_i). \] -
Binomial Taylor duality. For \(\alpha \in \IN_0^k\):
\[ g(x + {\textstyle\sum_{i=1}^k} \alpha_i u_i) = y + \sum_{0 < \beta \leq \alpha} \frac{(\alpha)_\beta}{\beta!}\, \Delta(g; x; u_\bullet^\beta), \qquad \Delta(g; x; u_\bullet^\alpha) = \sum_{\beta \leq \alpha} (-1)^{|\alpha|-|\beta|}\, \frac{(\alpha)_\beta}{\beta!}\, g(x + {\textstyle\sum_{i=1}^k} \beta_i u_i). \]
Proof. Ad 1) The second identity is the definition of \(\Delta\) (E0004). The first is its Boolean Möbius inverse E0002, applied to the cube \(S \mapsto \Delta(g; x; u_S)\) whose zeta transform is \(S \mapsto T^S(g; x; u_\bullet)\).
Ad 2) The second identity is the profile-grouped form of \(\Delta^\alpha\) (E0004). The first is its binomial Möbius inverse E0003.
(2) Validation (AI review, 2026-07-19, claude-fable-5, pass). Both parts as Möbius inverses (\(\Delta = \mu(T)\), so \(\zeta(\Delta) = T\) by E0002/E0003); separated-\(y\) terms (\(\Delta\) at \(\emptyset\), \(\beta = 0\)); edge cases \(k = 0\), \(\alpha = 0\), height-one restriction; numeric spot check \(k = 1\), \(\alpha = 2\). No issues found.
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