Calculating N FactorialDate: 10/26/1999 at 11:42:06 From: Charles R. Subject: Pauli exclusion principle and n! When one calculates n!, is there a method to do so quickly and exactly without crunching all intermediate factors? For example, when summing all numbers 1 + 2 + 3 + ... + n, one can either add all numbers from 1 to n or one can apply hte formula [(n+1)*(n/2)] to find the answer without too much effort. Can this be done for n!? Specific examples: 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = (100+1)(100/2) = 101*50 = 5050 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = (shortcut formula?) = 3628800 I've run across the Stirling approximation formulas, but they're just not sufficiently exact (predictably enough.) Granted, the inaccuracy drops off to an infinitesimally small percentage of the answer (say, 1% or 2% for n ~= 100), but when comparing the logarithms, there's still an inaccuracy of several percent for extremely large numbers (1% or 2% of the log for n ~= 10^100.) I've also run across Simpson's Rule, but I must have gone at this one too many times because I can't see how it applies to this at all (at least not in any way to make matters easier and more concise.) The Gamma function offers nothing more than (n-1)! and that's not really any kind of help. Charles Rockafellor Date: 10/27/1999 at 17:25:22 From: Doctor Mitteldorf Subject: Re: Pauli exclusion principle and n! Hi Charles! Stirling's approximation is what you want. The formula is sqrt(2pi*n)*(n^n)*e^(-n). It's much better than one percent accurate when calculating the log of the factorial of a large number. This formula is only the first term (or two) in a series in which higher order terms can provide whatever accuracy is desired. My guess, though, is that you did not include the Sqrt(2pi n) term in calculating the log, as this term is generally neglected in the formula for the log when n is REALLY huge. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ |
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