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I basically want to find $x > 0$ such that $x! = e^x$ which turns into $ln(x!) = x$

But I'm not sure how to deal with the $ln(x!)$ on the left hand side. I know the factorial function is defined for real numbers by the Gamma distribution, but I admittedly know very little about this. Is this even solvable?

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    $e^n$n\ge 6$. – 2017-02-07
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    http://www.wolframalpha.com/input/?i=solve+gamma(x%2B1)%3De%5Ex+for+x%3E2 will do it numerically, and I doubt there's a closed-form solution – 2017-02-07
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    For natural numbers this is true, but my question is about real numbers. Would it be the same? I don't see why it would. – 2017-02-07
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    x! is not defined in the real numbers – 2017-02-07
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    but you can you a generalization of n! as @Andreas has given in his answere – 2017-02-07
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    One *could* try to use Stirlings approximation to estimate the point where they are equal. This leads to $$\frac{1}{2}\log(2\pi) \simeq 2x - \left(x+\frac{1}{2}\right)\log(x)$$ which has the approximately solution $x\sim 5.3$ (found numerically). However it's just as easy to do it numerically from the get go instead of taking this detour. The most natural approach imo would be something along what is suggested above: first show that $e^n > n!$ for some large enough $n$ and then check the smaller values by hand (or computer) to find the exact turning point. – 2017-02-07

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If you generalize $x!$ to $\Gamma(x+1)$ (if $x$ is a natural number, then you have equality), then obviously for the smallest $x_0$ there is the symbolic solution $\Gamma(x_0+1) = e^{x_0}$. So the answer to the question is $x > x_0 = 5.290316$ where the value for $x_0$ was obtained numerically by MATLAB.

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    Symbolically or numerically? This seems like a numerical approximation. – 2017-02-07
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    $\Gamma(x+1) = e^x$ is the symbolic equation. $x = 5.290316$ is the numerical solution to that equation. – 2017-02-07
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    Yes, but to "symbolically solve" seems to imply that the solution, not just the representation, is algebraic. Given that there is no closed form for the Gamma function in general, that would be surprising. – 2017-02-07
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    @David I agree. Which means it's unlikely that you can do better. There are bounds to the Gamma function you can exploit, but I don't think this is the question. – 2017-02-07
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    If you agree with David then why do you say "obviously you can symbolically solve..."? – 2017-02-07
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    @JonasMeyer You're right. I made that clearer in the answer. – 2017-02-07
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We can actually prove the following statement by induction on n.

$$ e^n

We start with the base case 6! is way bigger then $e^6$.

Then we assume that $e^k

We then wish to show $e^{k+1} <(k+1)!$

using properties of these functions we get

$$ ee^k<(k+1)k! $$ then $$ e^k<\frac{k+1}{e}k! $$ And we know that $e^k$ is always less then $k!$ by inductive assumption (also note $\frac{k+1}{e} >1)$ so we are done.

A similar argument could be made in a similar way for the gamma function? Perhaps???

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It's easy enough to check that

$$5!

$$6!>e^6$$

Thus, the solution must lie in $5Gamma function to end up with the following line:

$$\Gamma(x+1)=e^x$$

Take the log of both sides to get

$$\ln(\Gamma(x+1))=x$$

$$0=\ln(\Gamma(x+1))-x$$

Applying Newton's method to get

$$x_{n+1}=x_n-\frac{\ln(\Gamma(x_n+1))-x_n}{\psi(x_n+1)-1}$$

where $\psi(x)$ is the digamma function. With $x_0=5.5$, I get

$$\begin{array}{c|c}n&x_n\\\hline0&5.50000\\1&5.29498\\2&5.29032\\3&5.29032\end{array}$$

which is the solution out $5$ places.