Abstract and Applied Analysis

Volume 2011 (2011), Article ID 495312, 6 pages

http://dx.doi.org/10.1155/2011/495312

## On the Distance to a Root of Polynomials

^{1}Department of Mathematics, Faculty of Science, Silpakorn University, Nakorn Pathom 73000, Thailand^{2}Centre of Excellence in Mathematics, Commission on Higher Education, Si Ayutthaya Road, Bangkok 10400, Thailand

Received 29 July 2011; Accepted 1 September 2011

Academic Editor: Natig Atakishiyev

Copyright © 2011 Somjate Chaiya. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

In 2002, Dierk Schleicher gave an explicit estimate of an upper bound for the number of iterations of Newton's method it takes to find all roots of polynomials with prescribed precision. In this paper, we provide a method to improve the upper bound given by D. Schleicher. We give here an iterative method for finding an upper bound for the distance between a fixed point in an immediate basin of a root to , which leads to a better upper bound for the number of iterations of Newton's method.

#### 1. Introduction

Let be a polynomial of degree , and let be the Newton map induced by . Let be the set of positive integers. For each , let denote the -iterate of , that is, , and . For a root of , we say that a set is the *immediate basin* of if is the largest connected open set containing and , as , for all . Every immediate basin is forward invariant, that is, , and is simply connected (see [1, 2]). In 2002, Schleicher [3] provided an upper bound for the number of iterations of Newton's method for complex polynomials of fixed degree with a prescribed precision. More precisely, Schleicher proved that if all roots of are inside the unit disc and , there is a constant such that for every root of , there is a point with such that for all . Schleicher also showed that can be chosen so that
with

To obtain this estimate, Schleicher employed several rough estimates which cause the bound far from an efficient upper bound. The main point that causes the extremely inefficiency is the way Schleicher used to obtain which arose when he estimated an upper bound for the distance of a point to a root . Schleicher showed that if is in the immediate basin of and , then the distance between and is at most .

In this paper, we give an algorithm to improve the value of . Even though, it is not an explicit formula, it can be easily computed. The following is our main result.

Main Theorem 1. *Let be a polynomial of degree , and let be a positive number larger than . If is in an immediate basin of a root and , then , where and can be derived from the following iterative algorithm.*

*Let , and*

*For , set .*

*If then let*

*Otherwise let*

Note that the value of in the main theorem depends only on the constant and the degree . Hence if we select appropriately the value will be optimized under this method. However this estimate is still far away from the best possible one. We believe that this new upper bound is less than for all when . We will discuss further about this matter in Section 4.

#### 2. Preliminary Results

We will use for the open ball and for the closed ball , where is the set of complex numbers. If is a subset of , we denote the boundary of by .

Lemma 2.1. *Let be a polynomial. Let be a complex number and . Suppose that whenever and . Let be an immediate basin of a root of . If , then is in .*

*Proof. *For with , we have
where . Hence, if and only if which holds if . It means that if is a point in and , then the distance of to is at most the distance of to . In other words, the image of under the map also lies inside .

Let be a root of and be its immediate basin. Suppose that and . Since is forward invariant under , still stays in . Since is connected, there is a curve connecting to and lying entirely in . Since converges uniformly to as , the set forms a continuous curve joining and . Note that is contained in because lies inside for all .

Let be the last intersection point of with (i.e., the part of the curve that connects to stays outside except at ). So must send to a point outside , otherwise is a fixed point of , which is impossible because all fixed points of are only the roots of , and here on . From the first paragraph, however, we also have . Hence we get a contradiction. Therefore if is not empty, then is in , as desired.

Remark that, from the proof of Lemma 2.1, if is a root of and for all , then the closed ball is contained in the immediate basin of .

Lemma 2.2. *Let be a polynomial of degree . Let be a root of and the nearest root to . Let , and let be the multiplicity of . Suppose that there is a root of such that for some positive number . Then the closed ball is contained entirely in the immediate basin of , where
*

*Proof. *Without loss of generality, we assume that . From the previous remark, it suffices to show that for all . Let . We have
Hence
where . Note that . For , we have
if . This shows that for all , as needed.

Note that if we set in Lemma 2.2, then the closed ball centered at of radius is contained in the immediate basin of . Furthermore, if , the radius of the ball is . (Schleicher [3, Lemma 4, page 938] made a small mistake about the radius of the ball. Indeed, he should get instead of .

Lemma 2.3. *Let be a polynomial of degree . For any complex number and any positive number , if and there is a root of with , then there is a root of such that .*

*Proof. *Let be all roots of . Suppose that . If for , then
a contradiction.

We are now ready to prove our main theorem.

#### 3. Proof of Main Theorem

Let be all roots of such that is the nearest root to and for . Suppose that . By Lemma 2.3, we have . Note that . If , we are done. Otherwise, is not in the immediate basin of ; thus by Lemma 2.2 with , we get that , where is defined in Lemma 2.2, that is, where . Thus satisfies the inequalities which holds if . If , we are done. Suppose next that .

Now let . If and , then hence by Lemma 2.1 must be either or which is not the case. Therefore , and if is we are done. Otherwise, let and suppose ; then , and by Lemma 2.1 we get a contradiction. Thus we obtain , and if is we are done. Continuing this process, finally we get which gives .

Note that if , it is a contradiction to the fact that , which implies that assumption is false. Hence in this case we have . The proof is now complete.

#### 4. Discussion

For a fixed , depends on only . If we choose too large (for instance, ), the value of is useless when it is compared to . So we have to choose carefully so that is minimal as possible. We do not know yet whether there is an explicit formula for the value that minimizes . Table 1 below shows the values of where we set . It seems that this method can reduce upper bounds for the distance of to the root it converges to at least times compared to . If we replace in (1.1) by , we derive a new upper bound for the number of iterations.

#### Acknowledgment

This research is supported by the Centre of Excellence in Mathematics, the Commission on Higher Education, Thailand.

#### References

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