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Factorials and Stirling's Formula
By Paul on Saturday, December 22, 2001 - 12:17 pm:

Is there any way of calculating n! without encountering the large number created by the factorial?

By Nicholas Dean on Thursday, June 06, 2002 - 03:16 pm:

On a calculator I approximate n! = ( (n+0.5)/(60.5) )^60.5 x ( (n+0.5)/e )^(n-60) ) x 60!. When n=100 the error is less than 0.03%.

By Graeme Mcrae on Thursday, June 06, 2002 - 06:06 pm:

Nicholas, it's a good approximation. In fact, for n=60, it's perfect!

Your expression can be simplified to:
=e^(-n)×(n+0.5)^(n+0.5)× 1.51930020735084

The constant (1.519...) can be adjusted to make any particular "n" come out exact, and the others will be wrong by only a few percent at the most. It's a pretty good approximation.

By Graeme Mcrae on Thursday, June 06, 2002 - 10:07 pm:

Nicholas, just fooling around with an Excel spreadsheet, I noticed that some adjustments can be made that make the formula fit even better:

e^(-n)×(n+.7914)^(n+.5)×1.13603289

gives n! to the nearest integer for n up to 7, and the nearest multiple of 10 for n up to 10.

After 10, the formula gives n! with an error less than 0.002%

Here's another thought:
e^(-n)×(n+.91893858)^(n+.5) gives a close approximation for all n. The error is less than 1% when n > 3, less than .1% when n > 45, less than .01% when n > 459, approaching zero (up to rounding error of Excel) as n increases.

By the way, the gammaln function in Excel gives the natural log of (n-1)! which is very useful for evaluating formulas like this for large n.

Does anyone know the limit as n-->infinity of

n! / (e^(-n)×(n+.5)^(n+.5))

It seems to be about 1.52034683, but is there an exact answer, or does this number have a name?

Here's another one:
(e^(-n)×(n+.803692299)^(n+.5))×1.12214972 is exactly equal to n! when n is 1, and less than .1% off for all other n, approaching a limit of 0% error as n approaches infinity.

--Graeme

By Arun Iyer on Thursday, June 06, 2002 - 10:41 pm:

the answer of the requires limit is
Ö(2p/e)

love arun

By Graeme Mcrae on Thursday, June 06, 2002 - 11:05 pm:

Oh, thanks Arun, Ö(2p/e) is 1.52034690, proving that my numerical fooling around came very close to the right answer. Thanks for the info!

This means that a fairly good approximation of n!, for large n, is
Ö(2p/e) e(-n) (n+0.5)n+0.5

--Graeme

By Arun Iyer on Friday, June 07, 2002 - 06:52 am:

Oh!!so you require an approximation of n!....

Why don't you use Stirlings formula..
n! is asymptotic to Ö(2p)e-nnn+.5 as n tends to infinity.

This is what i used to evaluate your limit actually.

love arun

[Editor: Here's an interesting thread on large factorial calculations.]