Given z on the unit circle, let $P(z)= \sum\limits_{k=-D}^D p_k z^k $. Can one estimate the number of self crossings of the following curve with an analytic expression in terms of the coefficients $\{p_k\}$? $$ P(e^{i * \theta}), \theta \in [0, 2pi) $$
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1$\begingroup$ Related (but also unanswered) mathoverflow.net/questions/90856 $\endgroup$– David E SpeyerCommented Mar 13, 2015 at 14:20
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Let $\gamma(x) = (\alpha(x), \beta(x)),$ where $\alpha, \beta$ are the real and imaginary parts. A self-intersection corresponds to a simultaneous zero of $(\alpha(x)-\alpha(y))/(x-y)$ and $(\beta(x)-\beta(y)/(x-y),$ If you use the rational parametrization for the circle (the $\tan t/2$ trick), both expressions become polynomials, and the number of self-intersections is basically the number of real roots of the resultant. So, when the smoke clears, you are asking for an analytic formula for the number of real roots of a polynomial. This can be expressed as a mapping degree (see, for example, Edelman and Kostlan's BAMS article on zeros of random polynomial), which is close, but not exactly, what you are asking for.
answered Mar 13, 2015 at 14:32
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$\begingroup$ Thanks for the rational paramterization idea, as this never occurred to me. In your response, you say "the number of self-intersections is basically the number of real roots of the resultant". I can interpret this in one of two ways. On e way is to fix x and consider the resulting polynomial a polynomial in y. The number of real roots in y will be the number of curve crossings through x's image-point (not exactly what I was asking). Another interpretation is to consider the resulting expression a bi-variate polynomial in (x,y) and analyze its real roots. Which did you mean? $\endgroup$– JimCommented Mar 14, 2015 at 2:10
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$\begingroup$ @Jim I mean neither. but rather en.wikipedia.org/wiki/Resultant $\endgroup$ Commented Mar 14, 2015 at 16:57