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From the discussion here, it seems that a generalization of the factorial can be written as

$$z!\equiv\Gamma\left(z+1\right)\mid z\in\mathbb{C}-\left\{-1,-2,\dots\right\}$$

We know that in the case where $z$ is real, we can simply write out

$$z! = \prod_{j=1}^z k$$

Is there a way to do the same for $z!$ when $z$ is complex? That is, can it be said, for example that

$$z!=z\left(z-1\right)\left(z-2\right)\cdots$$

My guess is no since the above product won't "terminate" as it does for real $z$. Additionally, from the extended definition using the Gamma function above, why do we exclude only the negative integers and not the entire negative real line?

gettingmathy
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  • If $z$ is not a positive integer, $\prod_{j=1}^z k$ makes no sense. – Taladris Oct 21 '22 at 02:21
  • Also, the accepted answer in the question you link seems to answer your question. Is there anything unclear in this answer? – Taladris Oct 21 '22 at 02:23
  • @Taladris Ah, I suppose I just answered my own question. I was simply confused as to why we don't discard negative non-integers as well as negative integers when defining the generalization above but the factorial is only really "undefined" for $\mathbb{Z}^-$. The Gamma function provides a way to deal with negative non-integers. – gettingmathy Oct 21 '22 at 02:32
  • @Taladris any thoughts on the other part of the question? Related to "writing out" $z!$ for complex $z$? – gettingmathy Oct 21 '22 at 02:34
  • You simply cannot write out $z!$ as $\prod_{k=1}^z k$ (notice that there is a typo in your notation) when $z$ is not a positive integer. If you need a formula, you can use $\Gamma(z)=\int_0^\infty e^{-t}t^{z-1}; dt$. This gives you already many information about the Gamma function, for example its derivative. – Taladris Oct 21 '22 at 03:15
  • You can derive the identity $\Gamma(z + 1) = z \Gamma(z)$ over the appropriate domain. You can iterate that as often as you like to get $\Gamma(z + n) = z(z+1)\ldots(z+n-1)\Gamma(z)$, which at least means you only need to know $\Gamma(z)$ for $0 \leq \Re(z) < 1$ to be able to calculate it over the whole domain. – ConMan Oct 21 '22 at 06:32
  • $z! = z(z-1)(z-2)\dots$ only terminates when $z$ is a positive integer. For other real $z$, you also keep going down forever. This only works to completely define $z!$ for non-negative integers, For negative integers, the same formula shows they must be undefined. For everything else, you have to know $z!$ on some strip of width $1$ by some other fashion. – Paul Sinclair Oct 21 '22 at 19:18

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