In the difference calculus, the (forward) difference operator (which is usually taken to be the inverse of a summation) is defined as: $$\Delta y(t)=y(t+1)-y(t)$$

Is it possible to express this operator as a summation of some function of $y(t)$? In other words, can we express $$\Delta y(t)=\sum{f(y(t))}$$

in the general case?

I don't want any difference operators involved in the Summation.

Additionally, it may be the case that two or more seperate summations are required. I will accept this as well.

I will accept an answer for the definite sum or the indefinite sum.

  • $\begingroup$ @IshanBanerjee: Yes. I've seen differentiation expressed as integration. So I'm wondering if there is a method for it's discreet counterparts. $\endgroup$ – Matt Groff Feb 21 '13 at 15:36
  • $\begingroup$ You seek a formula similar to $\Delta y(t)=\sum{f(y(t))}$. In the RHS, what is the range of the summation? Note that one cannot sum over $t$ since $t$ appears on the LHS. Thus, at present the formula you suggest makes no sense. $\endgroup$ – Did Feb 21 '13 at 15:45
  • $\begingroup$ @Did I think he means the indefinite sum. $\endgroup$ – Ishan Banerjee Feb 21 '13 at 16:13
  • $\begingroup$ I'm interested in both the definite and indefinite sum. If anyone can provide the definite sum I would accept that as an answer. $\endgroup$ – Matt Groff Feb 21 '13 at 16:30
  • $\begingroup$ @MattGroff You do not seem extremely interested in giving a precise meaning to your question... As I said, at the moment the formula you proposed makes no sense. $\endgroup$ – Did Feb 21 '13 at 18:17

I believe the solution is $f(y(t)) = \Delta \Delta y(t)$.

It's rather like trying to take the derivative $f'(x)$ of a function $f(x)$ and find the function that that function is the integral of. The only way to do that is to take the second derivative $f''(x)$.

  • $\begingroup$ Now, if I'm wrong, and you have seen differentiation expressed as integration in another way, I'd be obliged if you could show me, and I could modify my answer accordingly. $\endgroup$ – Joe Z. Feb 21 '13 at 15:37
  • $\begingroup$ In fact I have, and on this site too. Unfortunately, I don't remember the reference, but I will hunt for it. That is indeed what I am after. $\endgroup$ – Matt Groff Feb 21 '13 at 15:39
  • $\begingroup$ Yes, but $\Delta$ isn't a function(on N). $\endgroup$ – Ishan Banerjee Feb 21 '13 at 15:39
  • $\begingroup$ Whatever function can take $y(t)$ and return $\Delta \Delta y(t)$ for all $t$, then. $\endgroup$ – Joe Z. Feb 21 '13 at 15:40
  • $\begingroup$ It's a function on the set of functions from N to R not a function on N. $\endgroup$ – Ishan Banerjee Feb 21 '13 at 15:41

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