Re: minimum of a function into two variables
- From: David W. Cantrell <DWCantrell@xxxxxxxxxxx>
- Date: 03 Sep 2007 23:28:46 GMT
Robert Israel <israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
David W. Cantrell <DWCantrell@xxxxxxxxxxx> writes:
Andrea <acirulli@xxxxxxxxx> wrote:
Hi,
I need some suggestion,
Let P_s=p!/{n choose p}
I assume that your n and p are positive integers; let us know if that
is not the case.
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}
I don't suppose you meant f(x), since x hasn't been mentioned
elsewhere; rather, you wish to minimize p(1 + log(n)). Is that correct?
You say that P_s is fixed and give P_s = 2^-12 as an example. But,
assuming that n and p are positive integers, I suspect it not possible
to have P_s = 2^-12 exactly.
You could take n = 2^12 and p = 1.
Oops! Of course. I should have realized that. It's the trivial case:
If P_s = 1/m for positive integer m, then we can always take n = m and
p = 1.
Now I wonder if there are any nontrivial solutions when P_s = 2^-12.
David
.
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