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The falling factorial is defined as:

$x^\underline n = \prod_{k=0}^{n-1}(x-k),\quad n\in\mathbb N$

It can be used to define the binomial:

$\binom n k = \frac{n^\underline k}{k!}$

And it satisfies identities for discrete differential calculus as $x^a$ does for continuous differential calculus. $\Delta$ is the difference operator $\left(\Delta a_n = \frac{a_{n+1}-a_n}1\right)\colon$

$\Delta x^\underline k = kx^\underline{k-1}.$

One can easily find a reasonable generalization for arbitrary $n\in\mathbb Z$, but I wasn't able to come up with a generalization for $n\in\mathbb R$ or even $n\in\mathbb C$. Is there one?

2 Answers 2

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The gamma function naturally generalizes the factorial to complex values. It satisfies the functional equation $x\Gamma(x)=\Gamma(x+1)$ for any $x$ (when both sides exist anyway). Hence

$\Gamma\big(x-(n-1)\big)\prod_{k=0}^{n-1}(x-k)=\Gamma(x+1)$

by induction. Divide by the $\Gamma$ on the left and we're done.

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$ x^\underline n = \prod_{k=0}^{n-1}(x-k)= \frac{\Gamma(x+1)}{\Gamma(x+1-n)} $

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    this is not totally correct. Observe that the product is well defined for any $x\in\Bbb C$, however the fraction is not defined for $x\in\Bbb Z\cap(-\infty,0)$. But probably the analytic continuation of the fraction is equal to the product.2017-05-22