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Are there any known methods to formally solve equations like:

1)$x^3!+(2x^2)!-x!+3=0$

2)$x!=e^x$ ($0$ is trivial but there must be another one)

3)$(2x!)^2+x!-1=0$

4)$x!!+x!=7$

I don't need approximation (which are not so hard to find) but I mean closed forms, I think that there could be a lot of more equations of this kind but these are the first that came into my mind.

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  • $\begingroup$ There might be, but they will most likely revolve around special circumstances in the equation set up. $\endgroup$ – Zach466920 Mar 31 '15 at 19:18
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I'm able to solve (2) and (3). (2) for its trivialness and (3) for its generalization properties. $$x!=e^x$$ Since, $$0!=1=e^0$$ $0$ is a solution to this equation. $$(2x!)^2+x!-1=0$$ Set $u=x!$ and then solve the resulting quadratic equation... $$4u^2+u-1=0$$ $$u=x!={{\pm \sqrt{17}-1} \over 8}$$ You mentioned that the ! could be interpreted as the gamma function, so lets do so... $$\Gamma (x+1)={{\pm \sqrt{17}-1} \over 8}$$ Since almost all evaluations needing the gamma function are numerical, this is a proof that you have to use numeric methods. Your solution is... $$x=\Gamma^{-1} \left({{\pm \sqrt{17}-1} \over 8}\right)-1$$ where $\Gamma^{-1} (t)$ is the inverse gamma function.

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  • $\begingroup$ The second example depends on if $(2x!)^2$ should mean $((2x)!)^2$ or $2^2\cdot (x!)^2$. For the first one, are there other solutions? $\endgroup$ – mickep Mar 31 '15 at 19:32
  • $\begingroup$ @mickep there is one other solution for (1), but it would have to be found numerically. $\endgroup$ – Zach466920 Mar 31 '15 at 19:35
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To some extent it seems like you'll need relationships between, for example, $x^2!$ and $(x!)^2$.

That might be a tough nut to crack, or not. I tried a few and any pattern I could see broke down quickly:

$$0^2! = (0!)^2 \\ 1^2! = (1!)^2 \\ 2^2! = (2!)^2 \cdot 3! \\ 3^2! = (3!)^2 \cdot \frac{8!}{4!} \\ 4^2! = (4!)^2 \cdot \frac{15!}{6!} \cdot 5 \cdot 4 \\ 5^2! = (5!)^2 \cdot \frac{24!}{8!} \cdot 7 \cdot 5 \cdot 2$$

There may be a closed form hidden in there somewhere, but I can't see it (or cannot find reference to it).

This is just one kind of relationship that should be determined in order to attack some of the problems you've proposed.

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