Chain rule for discrete/finite calculus

Question

In the context of discrete calculus, or calculus of finite differences, is there a theorem like the chain rule that can express the finite forward difference of a composition $∆(f\circ g)$ in simplified or otherwise helpful terms?

It's probably not possible for a general function, but it might be possible with some restrictions. I'm also interested in the reverse -- a substitution rule for indefinite sums, if such exists. Or, to be honest, in any strong wealth of technical information related to this topic.

Answer 1

Provided the values of $g$ lie in the domain of $f$ and $\Delta g(n)$ is an integer, you have the obvious rule $$ \Delta(f\circ g)(n)=\sum_{d=0}^{\Delta g(n)-1}\Delta f\bigl(g(n)+d\bigr), $$ where the summation must be interpreted as a sum of negated terms in case $\Delta g(n)<0$, similarly to integrals whose upper limit is lower than their lower limit. The formula is probably not very useful though.

Added: I just found out that for such summations that might go in the wrong direction, the book Concrete Mathematics uses the notation $$ \Delta(f\circ g)(n)=\sum\nolimits_0^{\Delta g(n)}\Delta f\bigl(g(n)+i\bigr)\,\delta i $$ (the factor $\delta i$ is always $1$, but serves to indicate the sum is over $i$; also the upper bound, here $\Delta g(n)$, is omitted from the range of $i$) to emphasize even more the analogy with an integral.

Answer 2

While I doubt I'll ever find any general formulas, I've noticed that you can derive such a formula from many functions. For example, for $sin(x)$ we have: $$\sin{f(x)}⇒Δ\big({\sin{f(x)}}\big)=\sin{f(x+1)}-\sin{f(x)}=\sin{\big(f(x) + Δf(x)\big)}-\sin{f(x)}=$$ Using a trig identity: $$\sin{\big(\frac{1}{2}Δf(x)\big)}·\cos{\big(f(x)+\frac{1}{2}∆f(x) \big)}$$ For $2^{x}$, the discrete analog to $e^{x}$: $$Δ2^{f(x)}=2^{f(x)+Δf(x)}-2^{f(x)}=2^{f(x)}(2^{∆f(x)}-1)$$

While there is no chain rule to work with in discrete calculus, it seems that finding the differences (and hence, the discrete integrals) is somewhat easier.