On Mon, Jul 9, 2012 at 9:52 PM, Mike Stay <metaweta@gmail.com> wrote:

On Mon, Jul 9, 2012 at 6:51 PM, Mike Stay <metaweta@gmail.com> wrote:

...

On Mon, Jul 9, 2012 at 6:50 PM, Mike Stay <metaweta@gmail.com> wrote:

...

Is there a multivariate polynomial whose image over R^n is strictly positive?

Sorry, "strictly the positive reals".

Or better: "exactly the strictly positive reals."

No. Add a point at infinity to R^n and to R, extend the multvariate polynomial by mapping the point at infinity to the point at infinity, and the resulting function is still compact. Since the range is compact, so is the image. So the image of the original polynomial is compact once the point at infinity is added, and so it cannot e the strictly positive reals, since infinity maps to infinity, not 0. Andy

-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com

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On Thu, Jul 12, 2012 at 7:53 PM, David Wilson <davidwwilson@comcast.net> wrote:

image means what?

The values that are actually outputs of the function. http://en.wikipedia.org/wiki/Image_(mathematics)

On 7/9/2012 9:52 PM, Mike Stay wrote:

...

On Mon, Jul 9, 2012 at 6:51 PM, Mike Stay <metaweta@gmail.com> wrote:

...

On Mon, Jul 9, 2012 at 6:50 PM, Mike Stay <metaweta@gmail.com> wrote:

...

Is there a multivariate polynomial whose image over R^n is strictly positive?

Sorry, "strictly the positive reals".

Or better: "exactly the strictly positive reals."

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-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com

I don't work with analysis a great deal, but based on the fact that single-variate polynomials are either constant or have a finite number of local maxima and minima points, I reasoned that this would likely be the case for multivariate polynomials as well. I couldn't prove this to myself, or come up with a counterexample, so I kept silent (Proverbs 17:28). Had I been correct, the puzzle would be answered in the negative, since the global minimum of P would have to be one of a finite number of local minima, so that the image of P would take on a global minimum value at that point, whereas the positive reals has no minimum. But the answer is interesting. I could follow the algebraic reasoning, but I wanted a geometric understanding. So I thought about it, and here is what I came up with: Suppose you have polynomial p(x, y). Then p(x, y)^2 >= 0, so in R3, z = p(x, y)^2 lies above the xy-plane, except where it intersects the xy-plane at p(x, y) = 0. In the case of p(x, y) = x, we see that z = p(x, y)^2 = x^2 lies above the xy-plane, touching it only in the line x = 0. Likewise, for q(x, y) = xy+1, z = q(x, y)^2 = (xy+1)^2 lies above the xy-plane, touching it only on the hyperbola xy+1 = 0. Since the line and hyperbola are disjoint, p(x,y)^2 and q(x,y)^2 are never simultaneously 0, so P(x,y) = p(x,y)^2 + q(x,y)^2 is never 0. Finally, there are points (x,y) on the hyperbola at any desired distance |x| from the line. For these points we have q(x,y) = 0, so P(x,y) = p(x,y)^2 + q(x,y)^2 = p(x,y)^2 = x^2 takes on any desired positive value. On 7/14/2012 3:07 PM, Mike Stay wrote:

P(x, y) = x^2 + (xy + 1)^2

This is clearly nonnegative, and can be made arbitrarily small by letting x -> 0, y = -1/x. Since y can't be infinite, P(x, y) can't be zero.