Re: minimum of a function into two variables

On 3 Set, 19:39, David W. Cantrell <DWCantr...@xxxxxxxxxxx> wrote:
Andrea <aciru...@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. Are we to assume that your example was not a good
one and that, in reality, your fixed P_s will always be one which can be
obtained exactly using positive integers for p and n?

David W. Cantrell

Oh yeah this is a wrong example , I want to say that i want to choose
a fixed value of P_s that was obtained by the formula and then compute
among these values of n and p from which I obtained the fixed P_s the
minimum of f(n,p)=p+p\times \log n.

.

Relevant Pages