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Is there a way to solve for an unknown in a factorial?

I was just wondering, what would be the opposite of factorial?

For example, If I had $n! = 120$. How can I then show algebraically that $n = 5$?

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Jeel Shah
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    You could repeatedly divide your number by increasing integers; and if at any point you divide by $k$ and are left with $k+1$ then you know that the number you started with is $(k+1)!$. – Clive Newstead Oct 02 '12 at 13:03
  • Are you looking for a solution that works when the result is a non-integer? For example, suppose someone asks you for $n$ such that $n! = 200$. Do you want to say "There is no such $n$," or do you want to say "$n$ would have to be between 5 and 6", or do you want to say "$n\approx 5.297$"? – MJD Oct 02 '12 at 13:05
  • @MJD If possible, I would like to know both methods. – Jeel Shah Oct 02 '12 at 13:07
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    Relevant – MJD Oct 02 '12 at 13:08

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[Added because of a question in a comment] The generalization of the factorial is the gamma-function: $n! = \Gamma(1+n) $ where we can also insert noninteger values for n: $y = \Gamma(z) $ such that we have a function over the complex numbers $z$ except the poles at the non-positive integers).[/added]

The gamma-function has two real fixpoints. If you write the power-series of the gamma around one of that fixpoints, then this power series has no constant term and can be reverted by series-reversion. From this you can then get the inverse of the gamma, and from this the inverse of the factorial. Unfortunately, the convergence-radii of that series are both small, so I cant say at the moment, how useful this process would actually be.

(I think I've seen a question concerning the inverse of the gamma here or on MO, and possibly even showed a couple of that coefficients: see here for a short discussion)

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Gottfried Helms
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