Prove $f_1(x)=f_2(x)=f_3(x)$ when converge.

$$f_1(x)=\sum_{m=0}^{\infty} \binom {x}m \sum_{k=0}^m\binom mk(-1)^{m-k}f(k)$$

$$f_2(x)=\lim_{n\to\infty}\binom xn\sum_{k=0}^n\frac{x-n}{x-k}\binom nk(-1)^{n-k}f(k)$$

$$f_3(x)=\lim_{n\to\infty}\frac{\sum_{k=0}^{n} \frac{(-1)^k f(k)}{(x-k)k!(n-k)!}}{\sum_{k=0}^{n} \frac{(-1)^k }{(x-k) k!(n-k)!}}$$

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    $\begingroup$ @UwF the motivation is as follows: the first one is the Newton series, but it is not applicable to the cases where the function has poles. The third one can be easily generalized to the cases where the function has poles in integer points and also it can be made symmetric around zero. So the third one is more general, while the first one is easier to analyze convergence. The third one also converges much faster. My interest about the second one is purely curiosity. $\endgroup$ – Anixx Feb 12 '14 at 8:41
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    $\begingroup$ I don't see why this is being downvoted. This doesn't seem trivial and the motivation Anixx gives in the comment is absolutely fine. $\endgroup$ – Daniel Litt Feb 12 '14 at 9:15
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    $\begingroup$ Presumably each of these formulas came from somewhere - some method of representing $f(x)$? If so, why don't those sources already prove that the expressions are equal where defined? $\endgroup$ – Greg Martin Feb 12 '14 at 9:44

The sum in the denominator of $f_3(x)$ evaluates to $$\frac{(-1)^n}{(x-n)\binom{x}{n}n!}.$$ Hence, the equality $f_2(x) = f_3(x)$ is straightforward.


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