The Newton's series, i.e. the discrete analogue of the continuum Taylor expansion, involves classical iterated difference operators $\Delta$ defined by $\Delta f(k) = f(k+1) – f(k)$. Indeed, Newton's series writes $$f(x) = \sum_{k=0}^{\infty}\frac{\Delta^{k}f(a)}{k!}(x-a)_{k},$$
where $(x)_{k} = x(x-1)(x-2)…(x-k+1)$.
There also exist a generalized difference operator defined by
$$\Delta^{\mu}f(x) = \sum_{k=0}^{\infty}\mu_{k}f(x+k),$$
where $\mu = (\mu_{1}, \mu_{2}, …)$ is a sequence of real numbers such that $\sum_{k=0}^{\infty}\mu_{k} < \infty$.
My question: is there a "discrete Taylor expansion" like the one presented above involving $\Delta^{\mu}$ instead of the classical $\Delta$ ?
Thank you for your answers !
Best Answer
Your difference operator $\Delta^\mu$ has the property that it commutes with the shift operator $Ef(x)=f(x+1)$. It therefore falls in the purview of the work of Rota, et al., on finite operator calculus. In particular, Theorem 2 (First Expansion Theorem) of G.-C. Rota, D. Kahaner, and A. Odlyzko, On the foundations of combinatorial theory. VIII. Finite operator calculus, J. Math. Anal. Appl. 42 (1973), 684-760, gives the discrete Taylor expansion you are looking for. For reasons of formal convergence, you need the additional condition that $\Delta^\mu(x)$ is a nonzero constant.