1) for n=1, we have:
A*(x+y)>=0, so the constant A is from S(A)={A: A>=0}
2) for n=2, we have:
A*(x^2+y^2)+B*x*y>=0, it's easy to show that constants A,B
have to satisfy inequality
abs(B)>2*abs(A), so S(A,B)={(A,B): abs(A)>2*abs(B)}, and this
set is a 2D cone with origin at (0,0)
3) for n=3, we have
A*(x^3+y^3+z^3)+B*(x^2y+y^*x+...)+C*xyz>=0,
I suppose that S(A,B,C) is also a cone with origin at (0,0,0), but I
have no idea how to prove it
4) for n=4, we have:
A(x^4+y^4+z^4+t^4)+B(xyz^2+...)+C(x^2y^2+...)+D(x^3y+...)+Exyzt>=0
I'm interesting if it's also a cone S(A,B,C,D,E) ...
and if it's a general rule. Are there some simply methods for finding
such sets?
Of course: for any set S, the functions from S to [0, infty) form
a cone. That is, if f_1 and f_2 are such functions and t_1, t_2
are in [0,infty), then t_1 f_1 + t_2 f_2 is such a function.
> and if it's a general rule. Are there some simply methods for finding
> such sets?
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Why not A*x + B*y?
> > 2) for n=2, we have:
> > A*(x^2+y^2)+B*x*y>=0, it's easy to show that constants A,B
> > have to satisfy inequality
> > abs(B)>2*abs(A), so S(A,B)={(A,B): abs(A)>2*abs(B)}, and this
> > set is a 2D cone with origin at (0,0)
Are you assuming your function is not just homogeneous but also symmetric in
the variables? How does the number of variables relate to n?
Yes, of course. I've forgotten to mention it :(
> How does the number of variables relate to n?
I don't think about it, however it could be computed in some
combinatorial way, and I think that it isn't difficult. However, I'm
interesting not in it, but how to compute the bounds of the set S (for
n=2 we have the intersection of two half planes: y>=+-2x.
i didn't note that you were asking about number of variables (I though
that you are asking about constants A,B,...), for given number n,
there are n variables, becouse I want to have term x_1*x_2*...*x_n.