Chain rule for discrete/finite calculus

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.

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.

I had engineered a solution a little while back!

let

$$D_{w,x}[f(x)]= \frac{f(x+w)-f(x)}{w} $$

Then the question amounts to finding a closed form to

$$D_{w,x}[f(g(x))] = \frac{f(g(x+w)) - f(g(x))}{w}$$

We note this can be rewritten as:

$$\frac{f(g(x+w)) - f(g(x))}{w} = \frac{f \left( g(x) + w* \frac{g(x+w) - g(x)}{w} \right) - f(g(x))}{w}$$

$$ = \frac{f \left( g(x) + w* D_{w,x}[g(x)] \right) - f(g(x))}{w} $$

From here it becomes clear that

$$ D_{wD_{w,x}[g(x)],g(x)}[f(g)] = \frac{f \left( g(x) + w* D_{w,x}[g(x)] \right) - f(g(x))}{wD_{w,x}[g(x)]}$$

Thus:

$$D_{w,x}[g(x)]D_{wD_{w,x}[g(x)],g(x)}[f(g)] = \frac{f \left( g(x) + w* D_{w,x}[g(x)] \right) - f(g(x))}{w} = D_{w,x}[f(g(x))] $$

This chain rule however is very complex, as it involves now variable step size being involved in the finite difference itself. But for the sake of completeness, here it is!

Furthrmore its clear that as the step size $w$ tends to 0. You determine that

$$D_{w,x}[g(x)]D_{wD_{w,x}[g(x)],g(x)}[f(g)] \rightarrow D_{0,x}[g(x)]D_{0*D_{0,x}[g(x)],g(x)}[f(g)] \rightarrow D_{0,x}[g(x)]D_{0,g(x)}[f(g)] = g'(x)\cdot f'(g(x))$$

Which is the standard chain rule from calculus