Solutions for $\Gamma(n+1)$ = $\alpha$

Question

There's a general method of solving the equation $n!$ = $\alpha$, where $n$ and $\alpha$ can be any real numbers?

Here's why my question may be important:

There's no solution for the equation $n! = 0$. I'm assuming that this doesn't need to be proved.

Let $\epsilon$ be a number greater than $0$, then no matter how small is $\epsilon$, the solution for $n! = \epsilon$ will always exists? According to Wolfram Alpha, yes, and the number will be a complex number and your conjugate. If does exists a "human approach" for the equation $n! = \alpha$, then maybe I can solve this puzzle.

One thing that I may state is that, if $\epsilon \in (0, {} \approx 0.8815829811...)$, then the solution will be a complex value, but if $\epsilon$ is greater than this strange constant (let's call that $\gamma$), then there is a real number that satisfies that equation.

So, here's the puzzle: Why the solutions for $n! = \epsilon$ implies that $n \in \mathbb{C}$ if $\epsilon < \gamma$?

Here's the link if someone wants to find a counter-example: http://www.wolframalpha.com

Answer 1

As Jack D'Aurizo noted, for each $\epsilon > 0$, $\Gamma(m+1) = \epsilon$ has a real solution, but $m$ is negative if $\epsilon$ is too small. In fact, for any $\alpha \ne 0$ the equation $\Gamma(m+1) = \alpha$ has infinitely many real solutions. Notice that for any $\lambda \in (0,1)$, we will have $\lim\limits_{n\rightarrow-\infty} \Gamma(n + \lambda) = 0$ (this can be seen clearly from the recursion formula), and each negative integer is a simple pole of $\Gamma$.

To show that $\Gamma(x) = \alpha$ has infinitely many solutions, let's examine the reciprocal gamma function $\gamma(x) = \frac{1}{\Gamma(x)}$. This function is continuous for all $x$, and for $\lambda \in (0,1)$, we have $\lim\limits_{n\rightarrow-\infty} |\gamma(n + \lambda)| = \infty$. Looking at the graph (and using the recursion formula of $\gamma(x+1) = \frac{\gamma(x)}{x}$, we can see that for $x \in (-n-1,-n)$, $\gamma(x)$ is positive if $n$ is odd and negative if $n$ is even. If we choose $\beta \in \mathbb{R}$, we can find infinitely many solutions for $\gamma(x) = \beta$ as follows: If $\beta > 0$, we look at odd intervals $(-2n-2,-2n-1)$, where $\gamma(x)$ is positive. Since it is zero on the endpoints of each such interval, and the maximum on the interval forms a sequence diverging to infinity as $n$ goes to infinity, it follows that for sufficiently large $n$, $\gamma(x) = \beta$ always has a solution in $(-2n-2,-2n-1)$. If $\beta < 0$, we apply the same argument but to even intevals $(-2n-1,-2n)$.

If you want to actually find these solutions, you would have to use numerical methods like Newton's method.

Answer 2

By Euler's product or Weierstrass' product, the $\Gamma$ function has no zero in $\mathbb{C}$.
Additionally $\Gamma(z)$ is (log-)convex on $\mathbb{R}^+$, and since $\Gamma(1)=\Gamma(2)$ we have $$ \Gamma(z)\geq \min_{z\in[1,2]}\Gamma(z) \approx 0.8856$$ for any $z\in\mathbb{R}^+$. The situation is pretty different on $\mathbb{R}^-$: for any $n\in\mathbb{N}$, the sign of the $\Gamma$ function on the interval $(-n-1,-n)$ (whose endpoints are simple poles) only depends on the parity of $n$. On the other hand, by the reflection formula

$$ \left|\Gamma\left(-n-\tfrac{1}{2}\right)\right| =\frac{\pi}{\Gamma\left(n+\tfrac{3}{2}\right)}$$ which converges extremely fast to zero as $n\in\mathbb{N}$ tends to $+\infty$. It follows that for any $\varepsilon>0$ the equation $$ m! = \Gamma(m+1) = \varepsilon $$ has real solutions, but they are negative solutions if $\varepsilon<0.8856$.