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! = {(\frac{n + 0.5}{60.5})}^{60.5} \times {(\frac{n + 0.5}{e})}^{n - 60} \times 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} \times (n + 0.5)^{n + 0.5} \times 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} \times (n + 0.7914)^{n + 0.5} \times 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: $e^{- n} \times (n + 0.91893858)^{n + 0.5}$ gives a close approximation for all $n$ . The error is less than 1% when $n > 3$ , less than 0.1% when $n > 45$ , less than 0.01% 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 $n \to \infty$ of $n! / e^{- n} \times (n + 0.5)^{n + 0.5}$ ?

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} \times (n + 0.803692299)^{n + 0.5}) \times 1.12214972

is exactly equal to

n!

when

n

is 1, and less than 0.1% 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 required limit is

\sqrt{\frac{2 π}{e}}

love arun

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

Oh, thanks Arun,

\sqrt{\frac{2 π}{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

$\sqrt{\frac{2 π}{e}} e^{- n} (n + 0.5)^{n + 0.5}$ .

--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

\sqrt{2 π} e^{- n} n^{n + 0.5}

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 . ]