Re: [math-fun] Polynomial puzzle I heard today
If n = 1, then Andy's extension of P: R^n -> R to the one-point compactifications P*: (R^n)* -> R* would indeed be continuous. But for n >= 2, this is not necessarily the case, since (say n=2) it may not be true that |P(x,y)| -> oo as |(x,y)| -> oo. E.g., P(x,y) := xy. --Dan Andy wrote: << 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.
________________________________________________________________________________________ It goes without saying that .
On Tue, Jul 10, 2012 at 4:35 AM, Dan Asimov <dasimov@earthlink.net> wrote:
If n = 1, then Andy's extension of P: R^n -> R to the one-point compactifications P*: (R^n)* -> R* would indeed be continuous.
But for n >= 2, this is not necessarily the case, since (say n=2) it may not be true that |P(x,y)| -> oo as |(x,y)| -> oo. E.g., P(x,y) := xy.
Oops. Right you are. Back to the drawing board. Maybe this proof would work with some larger compactification? But you'd have to prove that if some of the added "points at infinity" map to 0 (as the P(X,y) = xy example shows they would have to), then some of the finite points would also map to 0. hmmm.
--Dan
Andy wrote:
<< 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.
________________________________________________________________________________________ It goes without saying that .
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Andy Latto -
Dan Asimov