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As we all know Stirling Approximation is giving us an approximate value of factorial, aka $\Gamma(x + 1)$.

$\sqrt{2\pi n}(\frac{n}{e}) \approx n!$

But what if we have equations with factorials. In this case we are allowed to use Stirling approximation in order to simplify our problem. But how can we find the exact value of this equation?

$ \sqrt{2\pi n}(\frac{n}{e}) = g(n)\\ g(n) \in \mathbb{R} $

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  • $\begingroup$ Would you give an example of your choice for $g(n)$ ? $\endgroup$ – Claude Leibovici Mar 21 '20 at 8:28
  • $\begingroup$ for instance: $\\ \sqrt{2\pi n}(\frac{n}{e})^n = 120$ $\endgroup$ – Alex Aramian Mar 21 '20 at 16:21
  • $\begingroup$ If $g(n)$ is a constant, you can inverse the factorial function very accurately. Have a look at math.stackexchange.com/questions/430167/… $\endgroup$ – Claude Leibovici Mar 22 '20 at 2:12
  • $\begingroup$ @ClaudeLeibovici thanks a lot $\endgroup$ – Alex Aramian Mar 22 '20 at 7:21