An example application is provided below that uses numbers that your
calculator can determine so you can compare the results to the
correct answer:
Let N(0)! = 49!; N(1)! = 23!; N(2)! = 14!
First, solve the problem to get the true answer:
49!
--------- = 2.698995 * 10^+29 the accuracy limit here calculator roundoff
23! 14!
The calculator answer is
Now, use Stirling's approximation to calculate each N!, multiply them
together, and see how close the answer this generates is to the true
answer.
You need to recall the rules for
convolution.
Continuing with the problem, use Stirling's approximation to
calculate each ln (number)! for the problem:
- ln 49!
49 (3.8918203) - 49 = 141.69919
[ln 49! actually equals 144.56574]
- ln 23!
23 (3.1354942) - 23 = 49.116367
[ln 23! actually equals 51.606676 ]
- ln 14!
14 (2.6390573) - 14 = 22.946803
[ln 14! actually equals 25.191221 ]
- exp (141.69919 - 49.116367 - 22.946803) =
exp (69.63602) =
1.7479911 * 10^+30
- is the Stirlings Approximation Answer
- Compare the Sperlings Approximation Answer to
,
the calculator answer seen
above.
- Finally, you will be working with numbers much larger than
those above: I think the trend is for the approximation to become
more accurate as the numbers become larger:
- ln 9! = 12.801827 and 9 (2.1972246) - 9 = 10.775021
- ln 29! = 71.257039 and 29 (3.3672958) - 29 = 68.651579
(96.34% of correct answer)
- ln 49! = 144.56574 and 49 (3.8918203) - 49 = 141.69919
(98.02% of correct answer)
- ln 69! = 226.19055 and 69 (4.2341065) - 69 = 223.15335
(98.65% of correct answer)
Assuming that this trend continues, the approximation is pretty
accurate for high numbers like 1000!.
Last Update, June 30, 1995- wld