Roots of a certain family of polynomials

Question

Let $\xi \in (0,1)$ and $p$ be a positive non-zero integer.Show that in the limit $p \rightarrow \infty$ the following algebraic equation:

\begin{equation} \frac{x^{p+2} }{\xi} + x^{p+1} + x^p = (-\xi)^p \end{equation}

has roots: \begin{equation} \left\{ x_\xi, \bar{x}_\xi, \left\{ \xi \exp(\imath \frac{2 \pi j}{p}) \right\}_{j=1}^{p}\right\} \end{equation}

where

\begin{equation} (1) x_\xi := - \frac{1}{2} \xi +\imath \left(\sqrt{\xi} - \frac{1}{2^3} \xi^{3/2} - \frac{1}{2^7} \xi^{5/2} - \frac{1}{2^{10}} \xi^{7/2} - \frac{5 }{2^{15}} \xi^{9/2} - \frac{7}{2^{18}} \xi^{11/2}- \frac{21}{2^{22}} \xi^{13/2} - \frac{33}{2^{25}} \xi^{15/2} - \frac{429}{2^{31}} \xi^{17/2} - \frac{715}{2^{34}} \xi^{19/2} - \frac{2431}{2^{38}} \xi^{21/2} - \frac{4199}{2^{41}} \xi^{23/2} - \frac{29393}{2^{46}} \xi^{25/2}- O\left(\xi^{27/2}\right)\right) \end{equation}

How do we find all terms in the expansion of the root $x_\xi$?

On the left, roots of the algebraic equation for $p=50$. On the right, the two complex conjugate ``non-trivial'' roots $x_\xi$ and $\bar{x}_\xi$ along with the series expansion (1).

Answer 1

I do not see why a polynomial of degree $p+2$ would have $p+3$ roots...

This is a perturbation problem, essentially you have

$q(x)x^p-(-ξ)^p=0$ with the quadratic polynomial $q(x)=\frac1ξx^2+x+1$.

The quadratic polynomial $0=4ξq(x)=(2x+ξ)^2+4ξ-ξ^2$ has roots

$$x=\frac12\left(-ξ\pm\sqrt{ξ^2-4ξ}\right)=-\frac12ξ\pm i\sqrtξ\,\sqrt{1-\tfrac14ξ}$$

and the binomial series for the square root gives the expression for $x_ξ$ and $\bar x_ξ$.

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Since $q(-ω^kξ)=1-ξω^k(1-ω^k)$ is bounded away from zero, $-ω^kξ$ is an approximation for the other roots, even if not the best. $ω$ being the primitive $p$-th root of unity.

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Keyword to look up: Newton polytope.