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April 1st, 2018, 07:01 AM  #1 
Senior Member Joined: Dec 2015 From: iPhone Posts: 474 Thanks: 73  Factorial equations
How to solve without inspection , for $\displaystyle n $ natural 1) $\displaystyle n!=1$ 2) $\displaystyle n!=2$ 3) $\displaystyle n!=6$ 
April 1st, 2018, 08:16 AM  #2 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 
I have no idea what this question even means. Are you asking whether there is a mystical process whereby you can evaluate 23! without computation? One way to define the factorial operation on nonnegative integers is $n \in \mathbb Z_{\ge 0} \implies$ $n! = 1 \text { if } n = 0 \text { and }$ $n! = n * (n  1)! \text { if } n > 0.$ But to determine what number is equal to 23!, what kind of process are you looking for that does not involve any arithmetic? 
April 1st, 2018, 11:06 AM  #3  
Senior Member Joined: Sep 2016 From: USA Posts: 598 Thanks: 366 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
\[n! = \int_0^\infty x^ne^{x} \ dx \] but I am not sure how that would be easier nor whether it would avoid inspection whatever that means. I can't imagine an easier computation than multiplying $n$ numbers together.  
April 1st, 2018, 11:35 AM  #4  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,138 Thanks: 872 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  
April 1st, 2018, 12:29 PM  #5 
Global Moderator Joined: May 2007 Posts: 6,727 Thanks: 687 
For small n, brute force is simplest. For large n, use Stirling's formula and invert (may not be easy).

April 1st, 2018, 02:03 PM  #6  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551  Quote:
If we are discussing positive integers, a fairly efficient algorithm is available. To determine whether any positive integer less than 200 trillion is a factorial of an integer, a binary search against a table of the factorials of 1 through 16 will require at most 4 comparisons. The brute force involved is that of a mouse.  
April 1st, 2018, 03:35 PM  #7  
Senior Member Joined: Sep 2016 From: USA Posts: 598 Thanks: 366 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
Notice that $\log n! = \sum_{k = 1}^n \log k$ leading to a simple integral bound \[\int_2^n \log x \ dx \leq \log n! \leq \int_1^n \log x \ dx. \] Solving these integrals we get the bound \[(n1) \log (n1) + 1 \leq \log n! \leq n \log n.\] Now, given an integer $m$, solve $(x1) \log (x1) = \log m  1$ for $x_L$, and solve $x \log x = \log m$ for $x_R$, then if $m = n!$ for some $n$, it must be that $n \in [\exp(x_L), \exp(x_R)]$ which is easy to check. In fact, one can get an even faster computation by checking that $m$ is divisible by $\lceil x_L \rceil$ and if so, dividing it out and iterating.  

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