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### Calculating N Factorial

```
Date: 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/
```
Associated Topics:
College Number Theory

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