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I've been told that factorials of negative numbers doesn't exists that's what I also found while trying to calculate factorial of negative $1$. But, I can see that graph of factorial $x$ is even extended to negative side of $x$ axis. What's the reason for this?

But, Desmos is able to calculate factorial of floating point values such as $0.1 , 0.2 , 0.3, 0.4$ etc.

I'm aware of gamma function but not well verse will integrals.

Can anyone explain this?

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  • $\begingroup$ The wiki page on Gamma function contains the information you need. $\endgroup$
    – WhatsUp
    Commented Oct 6, 2021 at 2:48
  • $\begingroup$ @WhatsUp I'm unable to find out the information which I need - Why is factorial function only works with positive integers? Isn't it possible to calculate $\int t^x \cdot e^-t \,dt$ for $x < 0$ and $x \in \mathbb {N}$ $\endgroup$
    – user947346
    Commented Oct 6, 2021 at 2:52
  • $\begingroup$ The key point being that $\Gamma$ has the iterative property that $\Gamma(x+1)=\Gamma(x)\cdot x$ for all $x$ over which the function is defined. However, $\Gamma(1)=1$, so $\Gamma(0), \Gamma(-1), ...$ are indefinite. $\endgroup$
    – Graham Kemp
    Commented Oct 6, 2021 at 2:59
  • $\begingroup$ @GrahamKemp, what's the reason for factorial of negative decimal values to be exist? $\endgroup$
    – user947346
    Commented Oct 6, 2021 at 3:01
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    $\begingroup$ @ProThala The integral $$ \int_0^{ + \infty } {e^{ - t} t^x dt} $$ converges for $x>-1$ and diverges for $x \leq -1$. The problem is that the integrand behaves like $t^{x}$ near $t=0$, which is not integrable when $x\leq-1$. $\endgroup$
    – Gary
    Commented Oct 6, 2021 at 3:52

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The "usual" definition of factorial which one is first exposed to only works for positive integers. That is, if we construct the factorial as $n!=1\cdot2\cdot3\dotsm (n-1)\cdot n$ then obviously you cannot extend this to all real numbers. However, we observe that $(n+1)! = (n+1) n!$, so we construct a function $f$ defined for all real numbers* such that $f(x+1)=(x+1) f(x)$ and $f(1)=1$, by analogy with the recurrence for the factorial. This is essentially the Gamma function, and in fact works for all* complex number arguments, not just real numbers. It is useful in a variety of applications.

So, this definition of the factorial of any complex number is not the same as the usual notion of factorial: in particular it does not have a combinatorial interpretation of "all ways to permute $n$ distinct objects". But it is a generalisation which people have found to be useful.

*Technically, there is a small set of values for which this function cannot be defined (or has value infinity). These are the vertical asymptotes you see in the graph, and are called "poles".

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  • $\begingroup$ How is f(x+1) = x * f(x) ? $\endgroup$
    – user947346
    Commented Oct 6, 2021 at 4:48
  • $\begingroup$ @ProThala sorry about that, it should be $x+1$ on the RHS. However, the Gamma function for historical reasons is defined with $\Gamma(x) = f(x-1)$ which is why it satisfies $\Gamma(x+1) = f(x) = x f(x-1) = x\Gamma(x)$. $\endgroup$
    – YiFan Tey
    Commented Oct 6, 2021 at 6:44

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