a question about products??

[deltasigma]

how do you approximate this:

Product(_from i=1 _to n)=i^i

I proved (I am not sure if this is correct) that is equal to {(n!)^n}/$
(n-1)

where $n is the super factorial

the numerator can be approximated by stirling formula but for the
denominator i dont know how to find an approximation....???

[k...@kymhorsell.com]

Maybe look up Vandermonde determinant.

--
>My sig lines fall into 2 broad categories: either wise or silly.
>I rely on the reader understanding which is which.
Congratulations. You must be a very special boy or girl.
-- tg <tgde...@earthlink.net>, 1 Dec 2010 12:35:45 PST

[Robert Israel]

deltasigma <deltas...@gmail.com> writes:

If f(n) = product_{i=2}^n i^i,
ln(f(n)) = sum_{i=2}^n i ln(i)
~= int_{3/2}^{n+1/2} x ln(x) dx
~= (ln(n)/2 - 1/4) n^2

so f(n) ~= n^(n^2/2) exp(-n^2/4)

For better approximations, try Euler-Maclaurin series. Maple's eulermac
command tells me

ln(f(n)) =
1/2*n^2*ln(n)-1/4*n^2-Zeta(1,-1)+1/12+1/2*n*ln(n)+1/12*ln(n)+1/(720*n^2)-1/(5040*n^4)+O(1/(n^6))
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada