Can You Solve an Unknown Factorial Algebraically?

[SprucerMoose]

Is there a method to solve an unknown factorial algebraically?

Homework Statement

5040 = a!

I found the solution a = 7 through trial and error, was just wondering if a more elegant method exists.

 

[abhishek ghos]

what you can do for big n is to apply stirling approximation
ln(n!)=(n*ln(n)) - n
so you can make a handy table or chart to pluck up the corresponding n values, apart from that doing this is a real mess.

 

[SprucerMoose]

Thanks a lot

 

[HallsofIvy]

Or, if you want an "algorithm" rather than "trial and error",

5040/2= 2520

2520/3= 840

840/4= 210

210/5= 42

42/6= 7
and

7/7= 1.

 

[silverstar]

Hi, question to @HallsofIvy, what is the procedure if the number is not wholly divisible at some point. For example you take the number 5 and try to divide it by 2, you get 2 remainder 1. How would you proceed from here? I vaguely remember doing something with the remainder, but don't remember what it was. Searching on google has not produced the results I'm seeking. Please let me know if you know of a good resource for this or if you know how to proceed. Thank you in advance!

 

[SammyS]

Hi, question to @HallsofIvy, what is the procedure if the number is not wholly divisible at some point. For example you take the number 5 and try to divide it by 2, you get 2 remainder 1. How would you proceed from here? I vaguely remember doing something with the remainder, but don't remember what it was. Searching on google has not produced the results I'm seeking. Please let me know if you know of a good resource for this or if you know how to proceed. Thank you in advance!

Do you know the definition of " n! " ?

 

[HallsofIvy]

SammyS's point is that n! is, by definition, divisible by all those numbers.

 

[Ray Vickson]

what you can do for big n is to apply stirling approximation
ln(n!)=(n*ln(n)) - n
so you can make a handy table or chart to pluck up the corresponding n values, apart from that doing this is a real mess.

Don't forget the constant: n! ~ sqrt(2*pi)*n^(n + 1/2) * exp(-n). If you don't have the sqrt(2*pi) factor the results are inaccurate. Here are the results for log(n!). Column 1 is n, column 2 is log(n!), column 3 is the log of approximation with the sqrt(2pi) factor, and column 4 is the log without the factor:
1 0.0000 -0.0811 -1.0000
2 0.6931 0.6518 -0.2671
3 1.7918 1.7641 0.8451
4 3.1781 3.1573 2.2383
5 4.7875 4.7708 3.8519
6 6.5793 6.5654 5.6464
7 8.5252 8.5133 7.5943
8 10.6046 10.5942 9.6753
9 12.8018 12.7926 11.8736
10 15.1044 15.0961 14.1771

As to how to solve your equation x! = y if you are not sure whether or not there is an integer n giving n! = y: write an equation in the Gamma function, so that x! is defined for all x > 0 (whether integer or not). If you have good software available, your equation can be dealt with using standard techniques such as Newton's method, etc.

RGV