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Can anyone explain (12) in this article. Why can we write $f(n)$ as a sum of derivatives: $$f(n) = \frac{ds}{dn} - \frac{1}{2!}\frac{d^2s}{dn^2} + \frac{1}{3!}\frac{d^3s}{dn^3} - \cdots$$ where $s(n) = \sum_{i = 0}^n f(i)$.

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You can't; at least, you can't when all of the symbols involved have their usual meaning. The article is behind a paywall so I can't read it.... –  Hurkyl Feb 5 '13 at 22:17
    
I changed the link. –  glebovg Feb 5 '13 at 22:18

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up vote 1 down vote accepted

It looks like $s$ isn't defined by the sum, but is instead meant to be an analytic function that has those values for natural number arguments.

The right hand side is the Taylor series for $s(x)$ centered at $n$, and evaluated at $n-1$:

$$ s(x) = \sum_{i=0}^{\infty} s^{(i)}(n) \frac{(x-n)^i}{n!} $$ $$ s(n-1) = s(n) + \sum_{i=1}^{\infty} (-1)^i s^{(i)}(n) \frac{1}{n!} $$

where $f^{(k)}$ means the $k$-th derivative of $f$.

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Which nicely explains the intermediate steps in $(12)$. –  Brian M. Scott Feb 5 '13 at 22:28

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