2 Fixed a couple of subscripts, made minor change to Note text.

The following describes an extension of the Descartes Rule of Signs to polynomials with complex coefficients.

First, I need to define the notion of a "sweep"... Given a complex polynomial p(z) := c0 zm0 + c1 zm1 + ... + cn zmn with non-zero ci and strictly increasing mi, let {αi} be the coefficients' "non-decreasing argument sequence": αi = arg ci (mod 2π), and α0 ≤ α1 ≤ ... ≤ αmn, with each αi+1i taken as small as possible; then αni0 computes the "angular sweep" --which I'll denote sweep(p)-- of a needle with one end anchored at the origin of the Complex Plane that begins pointing at c0 and then spins counter-clockwise to point at each ci in turn ("stalling in place" when consecutive coefficients have equal arguments).[*]

With this, we have:

The Descartes Rule of Sweeps. The number of positive real roots of p is at most $\lfloor \frac{1}{\pi} sweep(p)\rfloor$.

(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 $\frac{1}{\pi}sweep$ \frac{1}{\pi}sweep(p)$exactly counts those sign changes.) Now, I can prove the Rule of Sweeps using the Descartes Rule of Signs itself, but that approach sheds no light on why the Rule of Signs works (which is what I'm really after). The result seems to be one clever contour away from falling directly out of the Cauchy Argument Principle --and it even seems curiously appropriate that the formula provides a kind of half-winding number from the coefficient sequence-- but apparently I'm not sufficiently clever. :( Suggestions? [*] Note: One can also compute a sweep of the coefficients taken in reverse order (or, equivalently, effectively spinning the needle clockwiseclockwise). Differently-directed sweeps are usually not equal, so we can optimize the bound in the Rule of Sweeps by taking "the" sweep as the minimum of the two directed sweeps. 1 # Pointers for direct proof of extension of the Descartes Rule of Signs to complex polynomials? The following describes an extension of the Descartes Rule of Signs to polynomials with complex coefficients. First, I need to define the notion of a "sweep"... Given a complex polynomial p(z) := c0 zm0 + c1 zm1 + ... + cn zmn with non-zero ci and strictly increasing mi, let {αi} be the coefficients' "non-decreasing argument sequence": αi = arg ci (mod 2π), and α0 ≤ α1 ≤ ... ≤ αm, with each αi+1i taken as small as possible; then αni computes the "angular sweep" --which I'll denote sweep(p)-- of a needle with one end anchored at the origin of the Complex Plane that begins pointing at c0 and then spins counter-clockwise to point at each ci in turn ("stalling in place" when consecutive coefficients have equal arguments).[*] With this, we have: The Descartes Rule of Sweeps. The number of positive real roots of p is at most$\lfloor \frac{1}{\pi} sweep(p)\rfloor$. (The Descartes Rule of Signs represents a special case: each sign change in a polynomial's real coefficient sequence contributes π to the sweep, so that$\frac{1}{\pi}sweep\$ exactly counts those sign changes.)

Now, I can prove the Rule of Sweeps using the Descartes Rule of Signs itself, but that approach sheds no light on why 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. :(

Suggestions?

[*] Note: One can also compute a sweep of the coefficients taken in reverse order, effectively 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.