what is the value x if x factorial equals zero? (x!=0)

Question

to avoid confusion: when I'm talking about x factorial I'm talking about the factorial extension to the real (and complex) numbers, not just the naturals. (because obviously x isn't natural)

Answer 1

Note that $$ \begin{align} \sum_{k=1}^{n-1}\left[\log\left(1+\frac{z}{k}\right)-\frac{z}{k}\right] &=\log\left(\frac{\Gamma(n+z)}{\Gamma(n)\Gamma(1+z)}\right)-zH_{n-1}\\ &=\log\left(\frac{\Gamma(n+z)}{\Gamma(n)\Gamma(1+z)}\right)-z\log(n)-z\gamma+O\!\left(\frac{z}{n}\right) \end{align} $$ Taking the limit as $n\to\infty$, and using Gautschi's Inequality, we get $$ \sum_{k=1}^\infty\left[\log\left(1+\frac{z}{k}\right)-\frac{z}{k}\right] =-\log(\Gamma(1+z))-z\gamma $$ Therefore, $$ \frac1{\Gamma(1+z)}=e^{z\gamma}\prod_{k=1}^\infty\left(1+\frac{z}{k}\right)e^{-\frac{z}{k}} $$ The product on the right hand side converges for all $z$ and is $0$ at the negative integers.

Thus, $\frac1{\Gamma(1+z)}$ is an entire function and $\Gamma(1+z)=z!$ is never $0$.

Answer 2

The gamma function $\Gamma(z)$ is never zero for any complex $z$. (Zero is the only complex number that is not in the range, though).

Therefore, if you let $x!$ mean $\Gamma(x+1)$ for $x\notin \mathbb Z$, then the equation $x!=0$ has no solutions either.

Answer 3

$$x!=\Gamma(x+1)$$ $$\text{limit}_{n\to\infty} \Gamma(\frac{1}{2}-n)=0\qquad n=\text{positive integer}$$ Back to factorial : $$\text{limit}_{n\to\infty} (-\frac{1}{2}-n)!=0\qquad n=\text{positive integer}$$ The improperly so called "solution" of the equation $\quad x!=0\quad$ is not finite : $$x=-\frac{1}{2}-n\qquad n\to\infty$$