2
$\begingroup$

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 !

$\endgroup$
6
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.