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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.

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up vote 3 down vote accepted

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.

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I think I might not understand your formula properly. How can you do this if $∆g$ is not constant? You would need to perform the summation on the arbitrary points $Δg(0),Δg(1),...$. –  Greg Ros Nov 12 '12 at 15:27
    
@anon Yes, I wrote the right parenthesis in the wrong place; corrected now. Thank you! –  Marc van Leeuwen Nov 12 '12 at 18:16
    
Ah I'm sorry, I misunderstood the intent at first. I understand. –  Greg Ros Nov 12 '12 at 18:20
    
Thanks! I'm sorry for the misunderstanding at first. You can actually make the sum from $g(n)$ to $g(n+1)$, or to $g(n)+Δg$, thus removing the inner function from the sum itself. Unfortunately, I'm not sure how to simplify it. It just evaluates back to $f(g(n)+Δg)-f \circ g(n)$, since Δf is going to just become f anyway. I've played around with a similar notion that also applies if Δg is not an integer; to a derivative and an integral. However, it all just ends up being similar to the normal formula. –  Greg Ros Nov 12 '12 at 18:42
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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.

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