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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
    Commented Feb 12, 2014 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
    Commented Feb 12, 2014 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
    Commented Feb 12, 2014 at 9:44
  • $\begingroup$ Note that if $x$ is a nonnegative integer, then $f_1(x)=f(x)$. $\endgroup$
    – Richard Stanley
    Commented Dec 18, 2021 at 15:26

1 Answer 1

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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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