Posted by Anonymous on Monday, February 4, 2008 at 7:18pm.
There exists a method that allows you to compute this very accurately (and very easily). The formula in this case is:
n! = n^n exp(-n) sqrt(2 pi n)
exp[1/(12 n) - 1/(360 n^3)
+ 1/(1260 n^5) -1/(1680 n^7)+... ]
Here n! = 1 x 2 x 3 x...x n
So, let's test this formula for n = 50. The exact answer you get by multiplying all the factors is:
50! = 3.04140932017..... x 10^(64)
If you evaluate n^n exp(-n) sqrt(2 pi n)
(this is known as Stirling's approximation) for n = 50 you get:
3.03634459394...x 10^64.
The asymptotic series in the exponential is:
1/(12 n) - 1/(360 n^3)
+ 1/(1260 n^5) -1/(1680 n^7)+...
for n = 50 this is
1.66664444699...x10^(-3)
The exponential of this number is
1.00166803407...
3.03634459394...x 10^64 times 1.00166803407... equals 50! to 12 significant figures!
1x2=2x3=6x4=24x5=120
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