Pointers for direct proof of extension of the Descartes Rule of Signs to complex polynomials? - MathOverflow most recent 30 from http://mathoverflow.net2013-05-24T06:17:41Zhttp://mathoverflow.net/feeds/question/22508http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/22508/pointers-for-direct-proof-of-extension-of-the-descartes-rule-of-signs-to-complexPointers for direct proof of extension of the Descartes Rule of Signs to complex polynomials?Blue2010-04-25T14:52:02Z2010-12-23T23:22:13Z
<p>The following describes an extension of the Descartes Rule of Signs to polynomials with complex coefficients.</p>
<p>First, I need to define the notion of a "sweep"... Given a complex polynomial p(z) := c<sub>0</sub> z<sup>m<sub>0</sub></sup> + c<sub>1</sub> z<sup>m<sub>1</sub></sup> + ... + c<sub>n</sub> z<sup>m<sub>n</sub></sup> with non-zero c<sub>i</sub> and strictly increasing m<sub>i</sub>, let {α<sub>i</sub>} be the coefficients' "non-decreasing argument sequence": α<sub>i</sub> = arg c<sub>i</sub> (mod 2π), and α<sub>0</sub> ≤ α<sub>1</sub> ≤ ... ≤ α<sub>n</sub>, with each α<sub>i+1</sub>-α<sub>i</sub> taken as small as possible; then α<sub>n</sub>-α<sub>0</sub> computes the "angular sweep" --which I'll denote <b>sweep(p)</b>-- of a needle with one end anchored at the origin of the Complex Plane that begins pointing at c<sub>0</sub> and then spins counter-clockwise to point at each c<sub>i</sub> in turn ("stalling in place" when consecutive coefficients have equal arguments).[*]</p>
<p>With this, we have:</p>
<p><b>The Descartes Rule of Sweeps.</b> The number of positive real roots of p is at most $\lfloor \frac{1}{\pi} sweep(p)\rfloor$.</p>
<p>(The Descartes Rule of Signs represents a special case: each sign change in a polynomial's real coefficient sequence contributes π to the sweep, so that $\frac{1}{\pi}sweep(p)$ exactly counts those sign changes.)</p>
<p>Now, I can prove the Rule of Sweeps using the Descartes Rule of Signs itself, but that approach sheds no light on <em>why</em> the Rule of Signs works (which is what I'm really after). The result seems to be one clever contour away from falling directly out of the Cauchy Argument Principle --and it even seems curiously appropriate that the formula provides a kind of half-winding number from the coefficient sequence-- but apparently I'm not sufficiently clever. :(</p>
<p>Suggestions?</p>
<p>[*] Note: One can also compute a sweep of the coefficients taken in reverse order (or, equivalently, spinning the needle clockwise). Differently-directed sweeps are usually not equal, so we can optimize the bound in the Rule of Sweeps by taking "the" sweep as the minimum of the two directed sweeps.</p>
http://mathoverflow.net/questions/22508/pointers-for-direct-proof-of-extension-of-the-descartes-rule-of-signs-to-complex/22557#22557Answer by Victor Miller for Pointers for direct proof of extension of the Descartes Rule of Signs to complex polynomials?Victor Miller2010-04-26T03:04:18Z2010-04-26T03:04:18Z<p>The following paper by Jeff Lagarias and Tom Richardson "Multivariate Descartes rule of signs and Sturmfels' challenge problem" <a href="http://www.math.lsa.umich.edu/~lagarias/doc/intelligencer.pdf" rel="nofollow">http://www.math.lsa.umich.edu/~lagarias/doc/intelligencer.pdf</a></p>
<p>looks like it has things relevant to your question.</p>
http://mathoverflow.net/questions/22508/pointers-for-direct-proof-of-extension-of-the-descartes-rule-of-signs-to-complex/40654#40654Answer by Greg for Pointers for direct proof of extension of the Descartes Rule of Signs to complex polynomials?Greg2010-09-30T18:24:06Z2010-09-30T18:24:06Z<p>This paper is related to generalizing Sturm's theorem rather than Descartes' rule of signs, but I think it is relevant:
<a href="http://arxiv.org/abs/0808.0097" rel="nofollow">http://arxiv.org/abs/0808.0097</a></p>