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$$\sum _{x\ge0}^\Re f(x)= -\sum_{k=1}^\infty \frac{c_k \, \Delta^{k-1}f(0)}{k!}$$

where $c_k=\int _0^1 \frac {\Gamma (x+1)}{\Gamma (x-k+1)} \, dx$

$$\sum _{x\ge0}^\Re f(x)=\sum _{k=1}^\infty \frac {\Delta^{k-1}f(0)}{k!} (0)_k$$

where $(x)_k=\lim_{s\to k} \frac {\Gamma (x+1)}{\Gamma (x-s+1)}$

$$\sum _{x\ge0}^\Re f(x)=-\sum _{n=1}^\infty \frac {f^{(n-1)}(0)}{n!} B_n(0)$$

$$\sum _{x\ge0}^\Re f(x)= \frac {1}{2} f(0) - \sum_{k=1}^\infty \frac {B_{2k}}{(2k)!} f^{(2k-1)}(0)$$

Are all these definitions equal? If not, in what cases they are?

Are they equal if $f(x)$ is equal to its Newton series expansion?

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  • 1
    $\begingroup$ Where is the dependence of the right-side on $R$? $\endgroup$ Commented Nov 26, 2016 at 5:55
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    $\begingroup$ @T. Amdeberhan R symbolizes Ramanujan summation. $\endgroup$
    – Anixx
    Commented Nov 26, 2016 at 15:08
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    $\begingroup$ @Anixx : Is that a "summation method" for "divergent series"? Maybe you should "remind" us of its definition. $\qquad$ $\endgroup$ Commented Nov 26, 2016 at 17:28
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    $\begingroup$ @Michael Hardy : yes it is, but the wikipedia article is a bit unclear as it mentions notations it doesn't explain. $\endgroup$ Commented Nov 26, 2016 at 17:29
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    $\begingroup$ Everything is explained there :) The definition is $g(z)\in O^{2\pi}$ if $g(z)$ is analytic and $|g(z)| < C e^{(2\pi-\epsilon) |z|}$ for $Re(z) > 1-\epsilon$.The idea is that $O^{2\pi}$ makes the contour integral $\int_\gamma \frac{g(z)}{e^{2\pi z}-1}dz$ well-defined. Then for $f(z) \in O^{2\pi}$ : $\displaystyle\overset{\mathfrak{R}}{\sum_{n\ge 1}} f(n) = R_f(1) $ where $R_f(z) \in O^{2\pi}$ and is the unique solution of $R_f(z)-R_f(x+1) = f(z)$ such that $\int_1^2 R_f(z)dz = 0$... $\endgroup$
    – reuns
    Commented Nov 27, 2016 at 7:17

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