# Is there a formula to quickly express delayed functions in terms of finite differences? [closed]

It is easy to express difference deltas in terms of delayed functions as follows:

$$\Delta^n [f](x)= \sum_{k=0}^n {n \choose k} (-1)^{n-k} f(x+k)$$

For example.

But what about the inverse process, is there a formula?

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## closed as unclear what you're asking by Jonas Meyer, graydad, avid19, Solid Snake, Claude LeiboviciJun 14 at 5:27

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There is: for every integer $n\geqslant0$, $$f(x+n)=\sum_{k=0}^n{n\choose k}\Delta^k[f](x).$$

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This is wrong, $y(x+2)=\Delta^2 y(x)+2\Delta y(x) +y(x)$ Where is the coefficient in your formula? –  Anixx Jun 25 '14 at 0:11
Thanks for the correction, but copying my answer and downvoting it is not a good behavior. –  Anixx Jun 25 '14 at 15:22
@Anixx Where did I copy your answer? I looked at your comment, pondered it, saw that indeed there was a problem with my answer, and corrected said answer. By the way, to downvote an answer to one of your questions AND unaccept it AND post an inflammatory comment on it AND post an answer of your own AND accept said answer, all in a few minutes, this is what I call the behaviour of an xxx-xxxx. –  Did Jun 25 '14 at 15:26
–  user61527 Jun 25 '14 at 15:32
Many users would consider continuing to comment on main after opening an accusatory meta thread to be rather close to harassment... –  user61527 Jun 26 '14 at 5:05

$$f(x+n)=\sum_{k=0}^n {n\choose k} \Delta^k[f](x).$$