Since then, I have done a bit of searching on the subject, and have come up with some interesting knowledge.
First, the formula you have posted on your board:
x! = r(x+1) = root (2*pi*x) * (x/e)^x* [Asymptotic series]
is in fact not an equality. It is an approxiamation - but of a special kind.
What is strange about an asymptotic series is that as an *infinite* series (we let partial sum S(n) approach infinity), for a fixed value of x, this sum will diverge to infinity, but if we take a finite partial sum of that series, and let the value of x approach infinity, then it will become fully accurate. This is the difference between asymptotic series, and the kind of convergence which we think of for normal series. Thus the infinite Stirling asymptotic series does not converge.
This makes the formula only good for approxiamation purposes - however, the more terms we use, depending on the value of x we plug in, the more precise the approxiamation - but there will come a certain point in the series (this point depends on what value of x is chosen) in which further terms, instead of becoming smaller and smaller will just get bigger and bigger and bigger, thus the infinite partial sum will diverge in the end.
Also, I know how one can develop the terms of the series, though it is indeed complicated.
We can use:
x! = root (2*pi*x) ((x/e)^x)e^(u(x))
And u(x) is the function (an infinite series):
inf B(2n) sum ----------------------- n=1 2n(2n-1)*x^(2n-1)where B(2n) are the even-ordered Bernoulli numbers.
B(2n) = [(-1)^(n-1)]*2(2n)! ------------------------- * zeta (2n) (2*pi)^(2n)zeta is of course the Riemann zeta function.
Bernoulli numbers are always rational, and this for every even numbered input into zeta:
zeta (2n) is always some rational multiple of pi^(2n), thus the irrational pi's cancel out in the formula. So, a sample calculation, for the 2nd Bernoulli number, n=1, we get:
zeta (2) = (pi^2)/6 and thus
2*(2*1)! BV(2) = --------------- * (pi^2) / 6 (2*pi)^2 = 1/6Substitution of this Bernoulli number into u(x) yields the first term of the series of u(x):
1/6 -------------- = 1/(12x) 2*xThus we can create the infinite series u(x) by this method, and the first few terms are:
u(x) = 1/(12x) - 1/(360x^3) + 1/(1260x^5) ........
In creating the asymptotic series, I believe just these first three terms are taken, and substituted into the Maclaurin series for e^t
e^t = 1 + t + t^2/2! + t^3/3! .....
where t = 1/(12x) - 1/(360x^3) + 1/(1260x^5)
If you perform this substitution, and expand and collect powers, I think that the terms of the asymptotic series will appear.
Now, while the asymptotic series is an approxiamation (no asymptotic series serves to be exact, as they diverge for the infinite partial sum), I believe that the formula:
x! = root (2*pi*x) * (x/e)^x*e^(u(x)) is a precise formulation, as I have seen from one source that:
ln (x!) = 1/2*ln (2*pi) + (x + 1/2)*ln (x) - x + u(x)
which follows directly if the first statement is taken as correct.
- Ricky Derr