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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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up vote 3 down vote accepted

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

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