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Consider expanding the differentiation operator in terms of the forward difference operator as $f' = \log(1 + \Delta)f = \displaystyle \sum_{n = 1}^{\infty} \frac{(-1)^{n + 1}\Delta^n f}{n}$. For some functions, this series does indeed converge to the derivative of $f$ (e.g., polynomials, and apparently even functions of the form $f(x) = x^p$ for non-natural $p$). Alas, however, this formula cannot hold in complete generality (there's no reason the local data of $f'(x)$ for a particular $x$ should be constrained at all by the global data of $f(x), f(x + 1), f(x + 2), ...$).

So one might ask (and I do!): is there a nice characterization of conditions under which this identity does hold?

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  • $\begingroup$ i think for f=0 ,"and it is nice question " $\endgroup$
    – zeraoulia rafik
    Commented Jul 27, 2015 at 21:50
  • $\begingroup$ Have you checked Jordan's classic book The Calculus of Finite Differences? $\endgroup$
    – Tom Copeland
    Commented Jul 27, 2015 at 23:11
  • $\begingroup$ Are you familiar with Newton series? $\endgroup$
    – user13113
    Commented Jul 28, 2015 at 0:20
  • $\begingroup$ f is a Lipschitz continuous function in all the arguments $\endgroup$
    – zeraoulia rafik
    Commented Jul 28, 2015 at 0:39
  • $\begingroup$ Link to Jordan's book: books.google.com/… $\endgroup$
    – Todd Trimble
    Commented Aug 3, 2015 at 23:48

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This is a special case of well-known general formula for the n-th derivative via differences and Stirling numbers. Cf. good books on finite differences, e.g. of Gelfond or Jordan.

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  • $\begingroup$ I see the identity mentioned in Jordan's "Calculus of Finite Differences", but there does not seem to me to be much discussion of the conditions under which the identity holds, which is really the question I am interested in. (For example, is there simple reason by which one could know a priori that this identity is indeed valid at all points for all functions of the form $f(x) = x^p$, even for non-natural $p$?) $\endgroup$
    – Sridhar Ramesh
    Commented Jul 29, 2015 at 0:05
  • $\begingroup$ I suppose my example question may boil down to: Is there simple reason by which one could know that the Newton series for $x \mapsto x^p$ based at any point converges to the correct value within a neighborhood of that point? $\endgroup$
    – Sridhar Ramesh
    Commented Aug 4, 2015 at 17:29
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    $\begingroup$ In the book of Gelfond there is a special chapter on convergence of Newton series. $\endgroup$
    – Sergei
    Commented Aug 4, 2015 at 18:49
  • $\begingroup$ math.stackexchange.com/questions/892961/… $\endgroup$
    – Tom Copeland
    Commented Aug 4, 2015 at 23:52

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