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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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2 Answers 2

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 at 0:11
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Thanks for the correction, but copying my answer and downvoting it is not a good behavior. –  Anixx Jun 25 at 15:22
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@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 at 15:26
    
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Many users would consider continuing to comment on main after opening an accusatory meta thread to be rather close to harassment... –  T. Bongers Jun 26 at 5:05
up vote 1 down vote accepted

The answer by Did is incorrect, the correct answer is

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

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Can anyone explain the downvote? –  Anixx Jun 25 at 15:18
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Sure: I downvoted (I very rarely downvote answers to some question I also answered) because your behaviour is despicable. Satisfied? –  Did Jun 25 at 15:27
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And now you silently open a meta question with the transparent goal, not to ask a question, but to vilify me? Cuter and cuter... Please remind me to never again answer any of your questions. –  Did Jun 25 at 15:37
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+1 for your answer and -1 for did's answer. –  Zlatan der Zechpreller Jul 8 at 19:18
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@ZlatanderZechpreller Interesting take. Care to explain? –  Did Jul 8 at 21:21

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