Proof of the Sendov conjecture for polynomials of degree nine

TL;DR

This paper proves the Sendov conjecture specifically for degree nine polynomials, introducing a novel approach to establish upper bounds for the sum of distances to zeros.

Contribution

It provides the first proof of the Sendov conjecture for degree nine polynomials using a new bounding technique.

Findings

Confirmed the Sendov conjecture for degree nine polynomials.

Developed a new method to bound the sum of distances to zeros.

Enhanced understanding of polynomial zero distributions.

Abstract

In this paper, we prove the Sendov conjecture for polynomials of degree nine. We use a new idea to obtain new upper bound for the $\sigma-$sum to zeros of the polynomial.

Equations

$$p(z)=\prod _{k=1}^n(z-z_k),|z_k|\le 1(k=1,\cdots ,n)$$

$$p(z)=(z-a)\prod _{k=1}^{n-1}(z-z_k),|z_k|\le 1(k=1,\cdots ,n-1)$$

$${\rho }_1\le {\rho }_2\le \cdots \le {\rho }_{n-1},r_1\le r_2\le \cdots \le r_{n-1}.$$

$$2{\rho }_1\sin (\frac{\pi }{n})\le r_k\le 1+a,k=1,2,\cdots ,n-1.$$

$$\sum _{k=1}^N\frac{1}{c_k^2}\le \frac{N-v}{m^2}+\frac{v-1}{M^2}+\{\frac{m^{N-v}{M}^{v-1}}{C}{\}}^2,$$

$$\prod _{k=1}^{n-1}{r}_k=n\prod _{j=1}^{n-1}{\rho }_j.$$

$${\gamma }_j=\frac{{\zeta }_j-a}{a{\zeta }_j-1}and{w}_k=\frac{z_k-a}{a{z}_k-1}.$$

$$\prod _{j=1}^8|{\gamma }_j|\le \frac{\prod _{k=1}^8|w_k|}{9-4a^2/(1+a^2)-6a}.$$

$$|{\gamma }_0|>\frac{1}{\sqrt{1+\lambda (\lambda +2)-a^2\lambda (\lambda +2)}}.$$

$$(9-4a^2/(1+a^2)-6a)\frac{aR+1}{R+a}-\sqrt{1+(1-a^2)\lambda (\lambda +2)}(1+a-a^2{)}^7\ge 0.$$

$$\lambda =1-(1-aR{)}^{\frac{1}{9}},$$

$$\Delta \le -\frac{8}{a}+8a+\frac{9}{8a}(1-a^2)\sigma .$$

$$\frac{x^2-1}{(1-x^m)(a+x{)}^2}+(1-a^2)(\sigma -4)\le 0,x\in [0.46,1],a\in [0.845,1].$$

$$4^4(8-\frac{9}{8}\sigma )(1-a^2{)}^{-3}\le (\sigma -4{)}^{4}9\prod _{j=1}^8{\rho }_j.$$

$$U^{\ast }(a)=(8-v^{\ast })(\frac{2\sin \frac{\pi }{9}}{1+a}{)}^{-\frac{7}{4}}+(v^{\ast }-1)(\frac{1+a}{9^{\frac{1}{8}}}{)}^{-2}+\{(\frac{2\sin \frac{\pi }{9}}{1+a}{)}^{\frac{7}{8}(8-v^{\ast })}(\frac{1+a}{9^{\frac{1}{8}}}{)}^{v^{\ast }-1}{\}}^2,$$

$$v^{\ast }=\min \{j\in \mathbb{Z}:j\ge \{7\log (\frac{1+a}{2\sin \frac{\pi }{9}})\}(\log \{\frac{(1+a{)}^{\frac{15}{8}}}{9^{\frac{1}{8}}(2\sin \frac{\pi }{9}{)}^{\frac{7}{8}}}\}{)}^{-1}\}.$$

$$\sum _{k=1}^8\frac{1}{R_k^2}\le U^{\ast }(a).$$

$$\sigma \le U(a).$$

$$\frac{4U(a)}{U(a)-4}(\frac{8-\frac{9}{8}U(a)}{(1-a^2{)}^3}{)}^{\frac{1}{4}}\le U^{\ast }(a).$$

$$q\ge 9(\prod _{2\sin \frac{\pi k}{9}\ge 1}2\sin \frac{\pi k}{9}{)}^{-1}=(2\sin \frac{\pi }{9}{)}^2.$$

$$\prod _{r_k\ge 1}{r}_k\ge 9q^{-1}.$$

$$\prod _{r_k<1}{r}_k=q<1.$$

Full text

Proof of the Sendov conjecture for polynomials of degree nine

††thanks: 2000 Mathematics Subject Classification: primary 30C15.

Zaizhao Meng

Abstract In this paper, we prove the Sendov conjecture for polynomials of degree nine. We use a new idea to obtain new upper bound for the $\sigma-$sum to zeros of the polynomial.

Keywords: critical points, extremal polynomial, derivative.

1 Introduction

We continue the work of [5]. Let $\mathcal{P}_{n}$ denote the set of all monic polynomials of degree $n(\geq 2)$ of the form

[TABLE]

with

$p^{\prime}(z)=n\prod\limits_{j=1}^{n-1}(z-\zeta_{j}),\ \ |\zeta_{j}|\leq 1(j=1,\cdots,n-1).$

Write $I(z_{k})=\min\limits_{1\leq j\leq n-1}|z_{k}-\zeta_{j}|,\ \ I(p)=\max\limits_{1\leq k\leq n}I(z_{k})$, and $I(\mathcal{P}_{n})=\sup\limits_{p\in\mathcal{P}_{n}}I(p)$.

It was showed that there exists an extremal polynomial $p^{\ast}_{n}$, i.e., $I(\mathcal{P}_{n})=I(p^{\ast}_{n})$ and that $p^{\ast}_{n}$ has at least one zero on each subarc of the unit circle of length $\pi$(see [1]).

It will suffice to prove the Sendov conjecture assuming $p$ is an extremal polynomial of the following form,

[TABLE]

with

$p^{\prime}(z)=n\prod\limits_{j=1}^{n-1}(z-\zeta_{j}),\ \ |\zeta_{j}|\leq 1(j=1,\cdots,n-1),\ \ a\in[0,1]$.

Let $r_{k}=|a-z_{k}|,\ \rho_{j}=|a-\zeta_{j}|$ for $k,j=1,2,\cdots,n-1$. By relabeling we suppose that

[TABLE]

We have(see [3])

[TABLE]

Sendov conjecture. The disk $|z-a|\leq 1$ contains a zero of $p^{\prime}(z)$.

If $a=0$ or $a=1$, Sendov conjecture is true(see [4]), we suppose $a\in(0,1)$.

In this paper, we obtain the following theorem.

Theorem. For $n=9$, the disk $|z-a|\leq 1$ contains a zero of $p^{\prime}(z)$.

Throughout the paper, we assume that $z_{k}\neq 0,\ \ z_{k}\neq a,\ k=1,\cdots,n-1$ and that $p(z)$ is extremal in the form (1.1): $I(\mathcal{P}_{n})=I(p)=I(a)=\rho_{1}$.

In [5], we have proved the theorem for $a<0.845$, it will suffice to prove the theorem in the case $a\geq 0.845$.

2 Basic Lemmas

Lemma 2.1. If $\ \ 1-(1-|p(0)|)^{\frac{1}{n}}\leq\lambda\leq\sin(\frac{\pi}{n})$ and $\lambda<a,\ \rho_{1}\geq 1$, then there exists a critical point $\zeta_{0}=a+\rho_{0}e^{i\theta_{0}}$ such that $Re\zeta_{0}\geq\frac{1}{2}(a-\frac{\lambda(\lambda+2)}{a})$.

This is the Theorem 1 of [5].

Lemma 2.2. For $\rho>0$, we have

$\prod\limits_{r_{k}\geq\rho}r_{k}\rho^{-1}\leq\prod\limits_{\rho_{j}\geq\rho}\rho_{j}\rho^{-1}\prod\limits_{2\sin\frac{\pi k}{n}\geq 1}2\sin\frac{\pi k}{n}.$

This is the Theorem 3 of [5] by taking $m=1$.

Lemma 2.3. If $c_{k}(k=1,\cdots,N),\ m,\ M,\ C$ are positive constants with

$m\leq c_{k}\leq M,\ \prod\limits_{k=1}^{N}c_{k}\geq C$ and $m^{N}\leq C\leq M^{N}$, then

[TABLE]

where $v=\min\{j\in\mathbb{Z}:M^{j}m^{N-j}\geq C\}.$

Proof. See Lemma 7.3.9 of [6] and Lemma B of [3].

Lemma 2.4. We have

[TABLE]

Proof. See [3] .

Let

[TABLE]

We take $n=9$, then

[TABLE]

Lemma 2.5. If \rho_{1}>1,\ and $\ \zeta_{0}=a+\rho_{0}e^{i\theta_{0}}$ is the critical point in Lemma 2.1, $\gamma_{0}=\frac{\zeta_{0}-a}{a\zeta_{0}-1}$, then

[TABLE]

Proof. See Lemma 3.4 of [5].

Lemma 2.6. If $|\gamma_{j}|\leq\frac{1}{1+a-a^{2}}$, then $\rho_{j}\leq 1$.

Proof. See Lemma 1 of [2].

Write $B=B(R)=\frac{R+a}{aR+1}$.

Lemma 2.7. If $a\in[0.845,1]$, and there exists $|z_{k}|\leq 0.46$, then $I(a)=\rho_{1}\leq 1$.

Proof. If $I(a)>1$ and there exists $|z_{k}|\leq R$, then, by (2.1) and Lemma 2.5, there exists some $\gamma_{j_{0}}$,

$\frac{|\gamma_{j_{0}}|^{7}}{\sqrt{1+\lambda(\lambda+2)-a^{2}\lambda(\lambda+2)}}\leq\prod\limits_{j=1}^{8}|\gamma_{j}|\leq\frac{B}{9-4a^{2}/(1+a^{2})-6a},$

hence

$|\gamma_{j_{0}}|^{7}\leq\frac{\sqrt{1+\lambda(\lambda+2)-a^{2}\lambda(\lambda+2)}}{9-4a^{2}/(1+a^{2})-6a}B$.

By Lemma 2.6, it suffices to show

[TABLE]

We consider the conditions for $\lambda$,

$|p(0)|=a\prod\limits_{k=1}^{8}|z_{k}|\leq aR,$ $\ \ 1-(1-|p(0)|)^{\frac{1}{9}}\leq\lambda\leq\sin(\frac{\pi}{9}),$

we choose

[TABLE]

and $R$ satisfies $R\leq a^{-1}(1-(1-\sin(\frac{\pi}{9}))^{9}).$

Taking $R=0.46$, we obtain (2.2), the lemma follows.

We have

$\frac{p^{\prime}(0)}{p(0)}=-(\frac{1}{a}+\sum\limits_{k=1}^{8}\frac{1}{z_{k}}),$ and $9\prod\limits_{j=1}^{8}|\zeta_{j}|=(a\prod\limits_{k=1}^{8}|z_{k}|)|\frac{1}{a}+\sum\limits_{k=1}^{8}\frac{1}{z_{k}}|.$

Let $\Delta=Re(\frac{1}{a}+\sum\limits_{k=1}^{8}\frac{1}{z_{k}}),\ \sigma=\sum\limits_{k=1}^{8}\frac{1}{r_{k}^{2}}$, then

$9\prod\limits_{j=1}^{8}|\zeta_{j}|\geq-\Delta a\prod\limits_{k=1}^{8}|z_{k}|$.

Lemma 2.8. If $\rho_{1}>1,\ a\in[0.845,1]$, and $|z_{k}|\in[0.46,1],\ k=1,\cdots,8$, then

[TABLE]

Proof. See Lemma 3.9 of [5].

Lemma 2.9. Let $m=\frac{1}{4},\ a\in[0.845,1],\ f(x)=\frac{x^{2}-1}{(1-x^{m})(a+x)^{2}}$, then $f^{\prime}(x)>0$, for $x\in[0.46,1)$.

Proof. We have $f^{\prime}(x)=\frac{Y}{(1-x^{m})^{2}(a+x)^{3}}$, where

$Y=((m-2)x^{m+1}-mx^{m-1}+2x)a+mx^{m+2}-(2+m)x^{m}+2$.

If $Y=0$, we will obtain $a>1$ for $x\in[0.46,1)$, hence $Y\neq 0$ for $a\in[0.845,1],x\in[0.46,1)$. When $a=x=0.9$, $Y>0$, the lemma follows.

Lemma 2.10. If $\rho_{1}\geq 1$, we have

$\prod\limits_{j=1}^{8}|\zeta_{j}|\leq(\prod\limits_{j=1}^{8}\rho_{j})(a^{2}-1+\frac{1}{4}\sum\limits_{k=1}^{8}\frac{|z_{k}|^{2}-a^{2}}{r_{k}^{2}})^{4}.$

Proof. See Lemma 3.11 of [5].

We will use the following conditions

[TABLE]

By Lemma 2.8, Lemma 2.9, and Lemma 2.10, we have the following lemma.

Lemma 2.11. If $\sigma>4,\ \rho_{1}>1$, condition (2.3) holds with $m=\frac{1}{4}$,

and $|z_{k}|\in[0.46,1],\ k=1,\cdots,8$, then

[TABLE]

Proof. See Lemma 3.12 of [5], we use (2.3) to take the place of (3.6) there.

Write R_{k}=r_{k}\prod\limits_{j=1}^{8}r_{j}^{-\frac{1}{8}},\

[TABLE]

where

[TABLE]

Lemma 2.12. If $\rho_{1}>1$, then

[TABLE]

Proof. See Lemma 3.13 of [5].

Write

[TABLE]

By Lemma 2.12, we obtain the following lemma.

Lemma 2.13. If $\rho_{1}>1,\ 4<\sigma\leq U(a)<\frac{64}{9},\ a\in[0.845,1]$, and $|z_{k}|\in[0.46,1],\ k=1,\cdots,8$, then

[TABLE]

Proof. See Lemma 3.14 of [5], we use Lemma 2.11 to take the place of Lemma 3.12 there.

3 Proof of the Theorem

Suppose $n=9$, we want to give new upper bound to $U(a)$, this is the central part of the paper.

Write

$n_{1}=\#\{z_{k}:r_{k}<1\},\ n_{2}=\#\{z_{k}:r_{k}\geq 1\},n_{1}+n_{2}=8$, and

$q=\prod\limits_{r_{k}<1}r_{k}$, $\sigma=\sum\limits_{k=1}^{8}\frac{1}{r_{k}^{2}}=\sum\limits_{r_{k}<1}\frac{1}{r_{k}^{2}}+\sum\limits_{r_{k}\geq 1}\frac{1}{r_{k}^{2}}=\sum_{A}+\sum_{B}$, where

$\sum_{A}=\sum\limits_{r_{k}<1}\frac{1}{r_{k}^{2}},\sum_{B}=\sum\limits_{r_{k}\geq 1}\frac{1}{r_{k}^{2}}$.

As we have pointed out in [5], by combining Lemma 2.2 and Lemma 2.3, new results may be obtained.

Suppose that $\rho_{1}>1$, by making use of Lemma 2.13 and the new upper bound of $U(a)$, we will get a contradiction.

If $\rho_{1}>1$, taking $\rho=1$ in Lemma 2.2, we have

$\prod\limits_{r_{k}\geq 1}r_{k}\leq\prod\limits_{j=1}^{8}\rho_{j}\prod\limits_{2\sin\frac{\pi k}{9}\geq 1}2\sin\frac{\pi k}{9},$

by Lemma 2.4, $9\leq\prod\limits_{r_{k}<1}r_{k}\prod\limits_{2\sin\frac{\pi k}{9}\geq 1}2\sin\frac{\pi k}{9}$, hence,

[TABLE]

By Lemma 2.4, in the sum $\sum_{B}$, we have

[TABLE]

In the sum $\sum_{A}$, we have

[TABLE]

By (1.2), (1.3), and Lemma 2.4, we obtain

$\prod\limits_{k=1}^{8}r_{k}=9\prod\limits_{j=1}^{8}\rho_{j}>9,$

then

$2^{n_{2}}\geq(1+a)^{n_{2}}\geq\prod\limits_{r_{k}\geq 1}r_{k}\geq 9q^{-1}>9,$

hence $n_{2}\geq 4$, and $n_{1}\leq 4$.

We will consider four subcases for $n_{1},\ a\in[0.845,1)$.

(i) $n_{1}=0,n_{2}=8$

$\sum_{A}=0=U_{A}$. say

In the sum $\sum_{B}$, we have $1\leq r_{k}\leq 1+a$, hence by Lemma 2.3

$\sum_{B}\leq n_{2}-v_{2}+\frac{v_{2}-1}{(1+a)^{2}}+\{\frac{(1+a)^{v_{2}-1}}{9}\}^{2}=U_{B},$ say

where $v_{2}=\min\{j\in\mathbb{Z}:(1+a)^{j}\geq 9\}$,

$U(a)=U_{A}+U_{B}$,

using the bound of $U(a)$, we get the desired contradiction to Lemma 2.13.

(ii) $n_{1}=1,n_{2}=7$

$\sum_{A}=q^{-2}=U_{A}$. say

In the sum $\sum_{B}$, we have $1\leq r_{k}\leq 1+a$, hence by Lemma 2.3

$\sum_{B}\leq n_{2}-v_{2}+\frac{v_{2}-1}{(1+a)^{2}}+\{\frac{(1+a)^{v_{2}-1}}{9q^{-1}}\}^{2}=U_{B},$ say

where $v_{2}=\min\{j\in\mathbb{Z}:(1+a)^{j}\geq 9q^{-1}\}$,

$U(a)=U_{A}+U_{B}$,

using the bound of $U(a),q\geq 2\sin\frac{\pi}{9}$, we get the desired contradiction to Lemma 2.13.

(iii) $n_{1}=2,n_{2}=6$ or $n_{1}=3,n_{2}=5$

In the sum $\sum_{A}$, we have $2\sin\frac{\pi}{9}\leq r_{k}<1$, hence by Lemma 2.3

$\sum_{A}\leq\frac{n_{1}-v_{1}}{(2\sin\frac{\pi}{9})^{2}}+v_{1}-1+\{\frac{(2\sin\frac{\pi}{9})^{n_{1}-v_{1}}}{q}\}^{2}=U_{A},$ say

where $v_{1}=\min\{j\in\mathbb{Z}:j\geq n_{1}-\frac{\log q}{\log(2\sin\frac{\pi}{9})}\}$.

In the sum $\sum_{B}$, we have $1\leq r_{k}\leq 1+a$, hence by Lemma 2.3

$\sum_{B}\leq n_{2}-v_{2}+\frac{v_{2}-1}{(1+a)^{2}}+\{\frac{(1+a)^{v_{2}-1}}{9q^{-1}}\}^{2}=U_{B},$ say

where $v_{2}=\min\{j\in\mathbb{Z}:(1+a)^{j}\geq 9q^{-1}\}$,

$U(a)=U_{A}+U_{B}$,

using the bound of $U(a),q\geq(2\sin\frac{\pi}{9})^{2}$, we get the desired contradiction to Lemma 2.13.

(iv) $n_{1}=4,n_{2}=4$

In the sum $\sum_{A}$, we have $2\sin\frac{\pi}{9}\leq r_{k}<1$, hence by Lemma 2.3

$\sum_{A}\leq\frac{n_{1}-v_{1}}{(2\sin\frac{\pi}{9})^{2}}+v_{1}-1+\{\frac{(2\sin\frac{\pi}{9})^{n_{1}-v_{1}}}{q}\}^{2}=U_{A},$ say

where $v_{1}=\min\{j\in\mathbb{Z}:j\geq n_{1}-\frac{\log q}{\log(2\sin\frac{\pi}{9})}\}$.

In the sum $\sum_{B},\ q(1+a)^{n_{2}}\geq 9$, we have $1\leq r_{k}\leq 1+a$, hence by Lemma 2.3

$\sum_{B}\leq n_{2}-v_{2}+\frac{v_{2}-1}{(1+a)^{2}}+\{\frac{(1+a)^{v_{2}-1}}{9q^{-1}}\}^{2}=U_{B},$ say

where $v_{2}=\min\{j\in\mathbb{Z}:(1+a)^{j}\geq 9q^{-1}\}$.

$U(a)=U_{A}+U_{B}$,

using the bound of $U(a),q\geq(2\sin\frac{\pi}{9})^{2}$, we get the desired contradiction to Lemma 2.13.

Finally, by Lemma 2.7, we obtain the theorem.

The method of this paper can be used to obtain new results for $n>9$, we leave this case to the reader.