Re: minimum of a function into two variables
- From: Robert Israel <israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Mon, 03 Sep 2007 13:55:13 -0500
Andrea <acirulli@xxxxxxxxx> writes:
Hi,
I need some suggestion,
Let P_s=p!/{n choose p}
I need to calculate fixing P_s (for example P_s=2^-12) the minimum of
this function:
f(x)=p+ p \times \log{n}
There is no x on the right side of your equation.
Well, I suppose what you mean is you want to minimize
f(n,p) = p + p log(n) subject to the constraint p!/(n choose p) = c
where c is a positive constant. I'll assume 0 < c < 1.
Now we must be careful here, because normally p! and (n choose p)
are defined for nonnegative integers n and p (and so for all but
countably many c there would be no n and p satisfying the constraint),
but perhaps you mean a version of this that would be defined for real
n and p, presumably P_s(n,p) = Gamma(p+1)^2 Gamma(n-p+1)/ Gamma(n+1)
with n > 0 and n+1 > p > -1.
For fixed p > 0, P_s(n,p) ~ Gamma(p+1)^2/n^p -> 0 as n -> infty, while
P_s(n,0) = 1, so there is a curve p = g(n) for n sufficiently large
where 0 < g(n) < 1 and P_s(n,g(n)) = c. As n -> infinity, g(n) -> 0,
but rather slowly: it seems to me (using Maple) that
1/log(n) < g(n) < log(log(log(n)))/log(n)
for n sufficiently large. Now this means that
1 + 1/log(n) < f(n,g(n)) < (1 + 1/log(n)) log(log(log(n))).
I'm not convinced that f(n,g(n)) increases as n -> infinity, but if it
does then that happens only for very large n. For c = 0.8, the minimum
seems to occur around n = 300. For c = 1/2, it seems to be somewhere
between 500000 and 600000. For c = 2^(-12), I suspect the minimum would
be at a really enormous value of n, large enough that it will be extremely
difficult to find.
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.
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