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Nov 30, 2022 at 20:09 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Aug 2, 2022 at 20:02 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Jul 3, 2022 at 19:07 answer added Jens Franke timeline score: 2
Mar 21, 2020 at 10:59 comment added SophiaA @JohannesHuisman Thank you for the article, but what is an example on which the Positivstellensatz does not hold? Thanks a lot.
Mar 17, 2020 at 11:13 comment added user44143 The paper above is at math.uni-konstanz.de/~scheider/preprints/GUIDE.pdf
Mar 16, 2020 at 21:25 comment added Johannes Huisman I recommend reading the article C. Scheiderer: Positivity and sums of squares: a guide to recent results. In Emerging applications of algebraic geometry, Volume 149 of IMA Vol. Math. Appl., pages 271–324. Springer, New York, 2009. Section 3 seems to contain results and counter-examples you are looking for.
Mar 16, 2020 at 16:02 comment added user44143 I gave the following answer, and Emil Jerabek pointed out my error: Consider the set $[-1,1]$ defined by $1+x\ge0$ and $1-x\ge0$. The polynomial $x^2$ is non-negative on the whole set. Any polynomial in $1+x$ and $1-x$ with positive coefficients will have a positive constant term. So there is no way to write $x^2$ in that polynomial form. But, as Emil pointed out, the coefficients only need to be sum-of-squares in the polynomial ring, and not in the base field, so the conclusion of the Positivstellensatz is satisfied here by $$x^2=\left(\frac{x}{\sqrt{2}}\right)^2((1+x)+(1-x))$$
S Mar 16, 2020 at 13:00 history edited Igor Khavkine CC BY-SA 4.0
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S Mar 16, 2020 at 13:00 history suggested Rodrigo de Azevedo CC BY-SA 4.0
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Mar 16, 2020 at 12:37 review Suggested edits
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Mar 16, 2020 at 11:52 history edited YCor CC BY-SA 4.0
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Mar 16, 2020 at 11:45 history edited SophiaA CC BY-SA 4.0
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Mar 16, 2020 at 11:19 history asked SophiaA CC BY-SA 4.0