Limited polynomials and sendov's conjecture

TL;DR

This paper investigates a specific class of polynomials, analyzing their zeros and derivatives, and proves a weak version of Sendov's conjecture for real, same-sign zeros.

Contribution

It introduces a study of a particular polynomial class and establishes a partial proof of Sendov's conjecture under specific conditions.

Findings

Zeros and derivatives distribution analyzed

Weak Sendov's conjecture proved for real, same-sign zeros

Interaction between zeros and derivatives explored

Abstract

In this paper we study a particular class of polynomials. We study the distribution of their zeros, including the zeros of their derivatives as well as the interaction between this two. We prove a weak variant of the sendov conjecture in the case the zeros are real and are of the same sign.

Equations

$$|a_i-b_k|<1.$$

$$\mathcal{M}(P_n(z))=\prod _{i=1}^n|a_i|.$$

$$\prod _{i=1}^n|a_i|<ϵ\mathrm{and}\prod _{j=1}^m|b_j|<\delta$$

$$\mathcal{M}(P(z)Q(z))$$

$$<ϵ\delta$$

$$\prod _{j=1}^n|b_j|$$

$$<\frac{ϵ}{\prod _{j=1}^n|{\lambda }_j|},$$

$$\prod _{i=1}^n(x-a_i)$$

$$=\sum _{k=1}^n{s}_k(x-a_j{)}^k,$$

$$P(x)$$

$$\prod _{i=1}^n(x-a_i)$$

$$=(x-a_j)(x-a_j-r_1)(x-a_j-r_2)\cdots (x-a_j-r_{n-2})(x-a_j-r_{n-1})$$

$$=(x-a_j)((x-a_j)-r_1)((x-a_j)-r_2)\cdots ((x-a_j)-r_{n-2})((x-a_j)-r_{n-1})$$

$$=\sum _{k=1}^n{s}_k(x-a_j{)}^k$$

$$\prod _{i=1}^n(x+a_i)$$

$$=\sum _{k=1}^n{s}_k(x+a_j{)}^k,$$

$$P(x)$$

$$\prod _{i=1}^n(x+a_i)$$

$$=(x+a_j)(x+a_j-r_1)(x+a_j-r_2)\cdots (x+a_j-r_{n-2})(x+a_j-r_{n-1})$$

$$=(x+a_j)((x+a_j)-r_1)((x+a_j)-r_2)\cdots ((x+a_j)-r_{n-2})((x+a_j)-r_{n-1})$$

$$=\sum _{k=1}^n{s}_k(x+a_j{)}^k$$

$$P(x)$$

$$|a_j-b_i|<1$$

$$P(x)$$

$$\prod _{i=1}^n(x-a_i)$$

$$\frac{dP(x)}{dx}$$

$$=\sum _{k=2}^{n}k{s}_k(x-a_j{)}^{k-1}+s_1$$

$$\displaystyle=(x-a_{j})\bigg{(}\sum\limits_{k=2}^{n}ks_{k}(x-a_{j})^{k-2}\bigg{)}+s_{1}.$$

$$\displaystyle\bigg{|}(x-a_{j})\bigg{(}\sum\limits_{k=2}^{n}ks_{k}(x-a_{j})^{k-2}\bigg{)}\bigg{|}$$

$$\displaystyle\bigg{|}(b_{i}-a_{j})\bigg{(}\sum\limits_{k=2}^{n}ks_{k}(b_{i}-a_{j})^{k-2}\bigg{)}\bigg{|}$$

$$\displaystyle|b_{i}-a_{j}|\bigg{|}\bigg{(}\sum\limits_{k=2}^{n}ks_{k}(b_{i}-a_{j})^{k-2}\bigg{)}\bigg{|}$$

$$|b_i-a_j|<1.$$

$$1$$

$$<1,$$

$$\displaystyle\sum\limits_{s=1}^{n}\bigg{|}\frac{d^{s}P(a_{j})}{dx^{s}}\bigg{|}$$

$$P(x)$$

$$\frac{dP(x)}{dx}$$

Full text

1. Introduction and motivation

Let $P_{n}$ be a complex coefficient polynomial of degree $n$ and $P_{n}(a_{i})=0$ for each $1\leq i\leq n$. Then Sendov’s conjecture is the assertion that, if $|a_{i}|<1$ with $P_{n}(a_{i})=0$, then there exists some $b_{k}$ with $P_{n}^{\prime}(b_{k})=0$ such that

[TABLE]

There has been and still is a flurry of research devoted to this problem and manifestly the current literature contains dozens of papers just for the problem. There has really been substantive progress ever since it was posed. For instance, it has been shown in [5] that the conjecture holds for zeros near the unit circle. In [1] the conjecture has been verified for degree at most six. This was improved further to polynomials of degree at most seven in [2] and polynomials of degree at most eight in [4]. The best result thus far concerning sendov conjecture is found in [3], where it is verified to hold for sufficiently large degree polynomials.

In the sequel, we will study a particular class of complex polynomial and it turns out that such polynomials almost satisfies the sendov conjecture.

2. Limited complex-valued polynomials

In this section we introduce the concept of limited polynomials. We study various settings under which polynomials of this forms are preserved.

Definition 2.1**.**

Let $P_{n}(z)$ be a complex-valued polynomial of degree $n$ and let $\mathcal{Z}(P_{n}(z))=\{a_{1},a_{2},\ldots,a_{n}\}$ be the set of zeros of $P_{n}(z)$. By the measure of $P_{n}(z)$ denoted $\mathcal{M}(P_{n}(z))$, we mean the value

[TABLE]

We say $P_{n}(z)$ is $\epsilon$-limited if $\mathcal{M}(P_{n}(z))<\epsilon$, for some $\epsilon>0$, where \epsilon=\mathrm{sup}\bigg{(}M(P_{n}(z))\bigg{)}.

The concept of limited polynomial hinges essentially on the distribution of the zeros of a polynomial. This in some sense gives some information about the distribution of the zeros of their derivatives. In particular, the very notion that a polynomial is $1$-limited, for instance, is quite suggestive in it’s own right. In such a case, we could suspect that all the zeros of the polynomial in question are small, or almost all the zeros are somewhat large and the few exceptional ones are incredibly small with the tendency to absorb the large zeros upon taking their product. Next we examine some elementary properties of the concept of limited polynomials. We examine various settings under which this concept is preserved in the following sequel.

2.1. Properties of limited polynomials

In this section we examine some properties of limited polynomials. We examine the various settings under which this concept is preserved.

Proposition 1**.**

Let $P(z)$ and $Q(z)$ be any $\epsilon$ and $\delta$ -limited polynomials respectively. If $\mathcal{Z}(P(z))\cap\mathcal{Z}(Q(z))=\emptyset$, where $\mathcal{Z}(P(z))$ and $\mathcal{Z}(Q(z))$ are the set of zeros of $P(z)$ and $Q(z)$, respectively, then the product $P(z)Q(z)$ is $\epsilon\delta$-limited.

Proof.

Specify two limited polynomials $P(z)$ and $Q(z)$, where $P(z)$ is $\epsilon$-limited and $Q(z)$ is $\delta$-limited with zeros $\mathcal{Z}(P(z))=\{a_{1},a_{2},\ldots a_{n}\}$ and $\mathcal{Z}(Q(z))=\{b_{1},b_{2}\ldots,b_{m}\}$, then it follows that

[TABLE]

for some $\epsilon,\delta>0$, where \epsilon=\mathrm{sup}\bigg{(}M(P(z)\bigg{)} and \delta=\mathrm{sup}\bigg{(}M(Q(z)\bigg{)}. Since the zeros of the product $P_{n}P_{m}$ is the union of the zeros of $P_{n}$ and $P_{m}$. That is $\mathcal{Z}(P(z)Q(z))=\mathcal{Z}(P(z))\cup\mathcal{Z}(Q(z)):=\{a_{1},a_{2},\ldots,a_{n},b_{1},b_{2},\ldots b_{m}\}$. It follows that

[TABLE]

and the result follows immediately. ∎

Proposition 2**.**

Let $\lambda\in\mathbb{C}$ and $P(x)$ be an $\epsilon$-limited polynomial. Then $\lambda P(x)$ is also $\epsilon$-limited.

Proposition 3**.**

Let $P(x)=\prod\limits_{i=1}^{n}(x-a_{i})$ be an $\epsilon$-limited polynomial. Then the polynomial $Q(x)=\prod\limits_{i=1}^{n}(x-\overline{a}_{i})$ is also $\epsilon$-limited. In particular, $P(x)Q(x)$ is also $\epsilon^{2}$-limited polynomial.

Proof.

The result follows follows by applying Proposition 1. ∎

Proposition 4**.**

Let $P(x)=\prod\limits_{i=1}^{n}(x-a_{i})$ be an $\epsilon$-limited polynomial of degree $n\geq 2$. If $a_{i}=\lambda_{j}b_{j}$, then $Q(x)=\prod\limits_{j=1}^{n}(x-b_{j})$ is \bigg{(}\frac{\epsilon}{\prod\limits_{j=1}^{n}|\lambda_{j}|}\bigg{)}-limited.

Proof.

Suppose $P(x)=\prod\limits_{i=1}^{n}(x-a_{i})$ be an $\epsilon$-limited polynomial, and let $a_{i}=\lambda_{j}b_{j}$, then it follows that

[TABLE]

since $P_{n}(x)$ is $\epsilon$-limited. ∎

3. Critical points of limited polynomials

In this section we prove some preliminary results concerning the existence of zeros of limited polynomials with real zeros and the zeros of their derivatives, as well as the interplay between the two concerning their local and global distributions. We begin with the following elementary inequalities and identities. These two forms a rich tool-box for further studies in the sequel.

Lemma 3.1**.**

Let $P(x)=\prod\limits_{i=1}^{n}(x-a_{i})$ be a monic polynomial of degree $n$, with $a_{i}\in\mathbb{R}^{+}$, then there exist some $s_{j}$ for $j=1,2\ldots n-1$ such

[TABLE]

where $a_{j}=\mathrm{min}\{a_{i}\}_{i=1}^{n}$ and $|s_{j}|<\prod\limits_{\begin{subarray}{c}i=1\\ i\neq j\end{subarray}}^{n}|a_{i}|$.

Proof.

Specify the monic polynomial

[TABLE]

Now choose $a_{j}=\mathrm{min}\{a_{i}\}_{i=1}^{n}$. Then we can write

[TABLE]

thereby establishing the relation. ∎

Remark 3.2*.*

Next we prove an analogue of Lemma 3.1.

Lemma 3.3**.**

Let $P(x)=\prod\limits_{i=1}^{n}(x+a_{i})$ be a monic polynomial of degree $n$, with $a_{i}\in\mathbb{R}^{+}$, then there exist some $s_{j}$ for $j=1,2\ldots n-1$ such

[TABLE]

where $a_{j}=\mathrm{max}\{a_{i}\}_{i=1}^{n}$ and $|s_{j}|<|a_{j}|^{n}$.

Proof.

Specify the monic polynomial

[TABLE]

Now choose $a_{j}=\mathrm{max}\{a_{i}\}_{i=1}^{n}$. Then we can write

[TABLE]

thereby establishing the relation. ∎

Remark 3.4*.*

By leveraging Lemma 3.1, we prove a weak variant of the sendov conjecture. Before then we launch the following definition.

Definition 3.5**.**

Let $P(x):=\prod\limits_{i=1}^{n}(x-a_{i})$ be a polynomial of degree $n$ with $a_{i}>0$. Then by the local expansion of $P(x)$ we mean the finite sum of the form

[TABLE]

where $a_{j}=\mathrm{min}\{a_{i}\}_{i=1}^{n}$ and $|s_{j}|<\prod\limits_{\begin{subarray}{c}i=1\\ i\neq j\end{subarray}}^{n}|a_{i}|$. We call $s_{j}$ for $j=1,\ldots n-1$ the index of expansion.

Theorem 3.6**.**

Let $P(x)$ be a monic polynomial with real coefficients with index $|s_{t}|=\frac{1}{t}$ for $t=1\ldots,n-1$. Let $\mathcal{Z}(P(x))=\{a_{1},a_{2},\ldots,a_{n}\}\subset\mathbb{R}^{+}$ and $\mathcal{C}(P(x))=\{b_{1},b_{2},\ldots,b_{n-1}\}$ be the set of zeros and the set of critical points respectively. If $\frac{P(x)}{x-a_{j}}$ is $1$-limited, then for each $b_{i}$

[TABLE]

where $a_{j}=\mathrm{min}\{a_{i}\}_{i=1}^{n}$.

Proof.

Specify the monic polynomial of degree $n\geq 2$ with real coefficients, given by

[TABLE]

Then by Lemma 3.1, we have the decomposition

[TABLE]

where $a_{j}=\mathrm{min}\{a_{i}\}_{i=1}^{n}$ and $|s_{j}|<\prod\limits_{\begin{subarray}{c}i=1\\ i\neq j\end{subarray}}^{n}|a_{i}|$. It follows that

[TABLE]

By setting $\frac{dP(x)}{dx}=0$, it follows that

[TABLE]

We remark that the values of $x$ that satisfies this relation are the zeros of the polynomial $P^{\prime}_{n}(x)$. That is, for each $b_{i}\in\mathcal{C}(P(x))=\{b_{1},b_{2},\ldots,b_{n-1}\}$ the relation is valid

[TABLE]

Since $\frac{P(x)}{x-a_{j}}$ is $1$-limited, it follows that $|s_{j}|<\prod\limits_{\begin{subarray}{c}i=1\\ i\neq j\end{subarray}}^{n}|a_{i}|<1$ for each $j=1,\ldots n-1$ so that

[TABLE]

Since $|s_{k}|=\frac{1}{k}$, it follows that

[TABLE]

For suppose $|b_{i}-a_{j}|\geq 1$, then it follows that

[TABLE]

which is absurd. That is, all the critical values are sufficiently close to the least zero of the polynomial, thereby ending the proof. ∎

Remark 3.7*.*

Theorem 3.6 can be considered as a weak variant of the result of the real case of the sendov conjecture. The only compromise in this case is the assumption that all the zeros are positive, in which case we have found that indeed the smallest zero of $P(x)$ must be close to all the zeros of the polynomial $P^{\prime}(x)$. A similar approach could be adapted for the case all the zeros are negative. Next we prove a result that in some sense exposes the distribution of the zeros of limited polynomials.

Theorem 3.8**.**

Let $P(x)=\prod\limits_{i=1}^{n}(x-a_{i})$ be a polynomial of degree $n$ for $n\geq 2$, with $a_{i}\in\mathbb{R}^{+}$ and $a_{i}>0$. If $\frac{P(x)}{x-a_{j}}$ is $\epsilon$-limited, where $a_{j}=\mathrm{min}\{a_{i}\}_{i=1}^{n}$, then

[TABLE]

Proof.

Let $P(x)=\prod\limits_{i=1}^{n}(x-a_{i})$ be a polynomial of degree $n$, with $a_{i}>0$, then by definition 3.5 we can write

[TABLE]

where $a_{j}=\mathrm{min}\{a_{i}\}_{i=1}^{n}$ and $|s_{j}|<\prod\limits_{\begin{subarray}{c}i=1\\ i\neq j\end{subarray}}^{n}|a_{i}|$. It follows that

[TABLE]

It follows that $\frac{dP(a_{j})}{dx}=s_{1}$. Again, by taking the second derivative $P^{\prime\prime}(x)$, we have

[TABLE]

Thus it follows that $\frac{d^{2}P(a_{j})}{dx^{2}}=2s_{2}$. By induction, it follows that

[TABLE]

since $\frac{P(x)}{x-a_{j}}$ is $\epsilon$-limited. The result follows immediately from the relation. ∎

The above inequality tells a lot about the relationship between the zeros of a given polynomial $P(x)$ and the higher order derivatives $P^{k}(x)$ for $k=1,\ldots n$ in terms of the distribution. Indeed for any fixed $n$, by shrinking $\epsilon>0$ to be very close zero, then it follows that the least zeros of $P(x)$ are sufficiently close to the zeros of each $P^{k}(x)$. Conversely, If we take $\epsilon>0$ some what large for a fixed $n$ then certainly, the least zero of $P(x)$ is far from at least one of the zeros of $P^{\prime}(x)$. This property is archetypal of very rare class of polynomials of which limited polynomials is a sub-class. The ensuing result is a strengthening of Theorem 3.6.

Corollary 1**.**

Let $P(x)=\prod\limits_{i=1}^{n}(x-a_{i})$ be a polynomial of degree $n$ for $n\geq 2$, with $a_{i}>0$ and let $\epsilon>0$ be arbitrary. If $\frac{P_{n}(x)}{x-a_{j}}$ is $\epsilon$-limited, where $a_{j}=\mathrm{min}\{a_{i}\}_{i=1}^{n}$, then $|a_{j}-b_{ik}|<\delta$, for any $\delta>0$ and $P^{k}(b_{ik})=0$ $(i=1,\ldots,n)$.

Proof.

The result follows by varying the magnitude of $\epsilon$ in Theorem 3.8, since $\epsilon>0$ is arbitrary. ∎

Corollary 1 reveals quite well that the zeros of any real coefficient $\epsilon$-limited polynomial $P_{n}(x)$ can be made close to the zeros of their derivative $P^{\prime}_{n}(x)$ by shrinking the size of $\epsilon$. In such a case we could take

[TABLE]

Similarly the zeros of $P_{n}(x)$ can be made to be far from the zeros of $P^{\prime}_{n}(x)$ by taking $\epsilon>0$ to be smaller than

[TABLE]

Carefully tweaking $\epsilon>0$ this way gives some information about the local distribution of the zeros of limited polynomials and the derivatives.

Corollary 2**.**

Let $P(x)=\prod\limits_{i=1}^{n}(x-a_{i})$ be a polynomial of degree $n$ for $n\geq 2$, with $a_{i}>0$. If $\frac{P(x)}{x-a_{j}}$ is $\epsilon$-limited, where $a_{j}=\mathrm{min}\{a_{i}\}_{i=1}^{n}$, then

[TABLE]

where $\prod\limits_{i\in[1,n]}(a_{i})_{k}$ denotes the product of $k$ terms with index belonging to the set $[1,n]$.

Proof.

Let $P(x)=\prod\limits_{i=1}^{n}(x-a_{i})$ be a polynomial of degree $n$ for $n\geq 2$, with $a_{i}>0$. Then we observe that we can write

[TABLE]

where the sum runs over all permutations. By taking $n=1$ in Theorem 3.8, we observe on the other hand that

[TABLE]

It follows that

[TABLE]

and the result follows immediately. ∎

Remark 3.9*.*

Corollary 2 is quite suggestive. Roughly speaking, it tells us that once a polynomial is $\epsilon$-limited for any $\epsilon>0$, then by necessity the zeros should be at most within $\epsilon$ distance away from the least zero.

Corollary 3**.**

Let $P(x)=\prod\limits_{i=1}^{n}(x-a_{i})$ be a polynomial of degree $n$ for $n\geq 2$, with $a_{i}>0$. If $\frac{P(x)}{x-a_{j}}$ is $\epsilon$-limited, where $a_{j}=\mathrm{min}\{a_{i}\}_{i=1}^{n}$, then

[TABLE]

where $\prod\limits_{i\in[1,n]}(a_{i})_{k}$ denotes the product of $k$ terms with index belonging to the set $[1,n]$.

Proof.

Let $P(x)=\prod\limits_{i=1}^{n}(x-a_{i})$ for $n\geq 2$ with $a_{i}>0$, then we can write

[TABLE]

The result follows by applying Theorem 3.8. ∎

111 .

4. Data availability statement

The manuscript has no associated data.

5. Conflict of interest

The authors declare no conflict of interest regarding the publication of this manuscript.