On the Distance to a Root of Polynomials

and Applied Analysis 3 Proof. For |z − β| r with P z / 0, we have Np z − β ( z − β) ( 1 − 1 g z ) , 2.1 where g z z−β P ′ z /P z . Hence, |Np z −β| ≤ |z−β| if and only if | g z −1 /g z | ≤ 1 which holds if Re{g z } ≥ 1/2. It means that if z is a point in ∂B β, r and Re{g z } ≥ 1/2, then the distance of Np z to β is at most the distance of z to β. In other words, the image of z under the map Np also lies inside B β, r . Let α be a root of P and U be its immediate basin. Suppose that α / ∈ B β, r and z ∈ U∩B β, r . SinceU is forward invariant underNp,Np z still stays inU. SinceU is connected, there is a curve γ0 connecting z to Np z and lying entirely in U. Since N p γ0 converges uniformly to α as k → ∞, the set⋃k 1 N p γ0 ∪ {α} forms a continuous curve γ joining z and α. Note that γ is contained inU because N p γ0 lies inside U for all k ∈ N. Let w be the last intersection point of γ with ∂B β, r i.e., the part of the curve γ that connects w to α stays outside B β, r except at w . So Np must send w to a point outside B β, r , otherwise β is a fixed point of Np, which is impossible because all fixed points of Np are only the roots of P , and here P z / 0 on |z − β| r. From the first paragraph, however, we also have Np w ∈ B β, r . Hence we get a contradiction. Therefore if U ∩ B β, r is not empty, then α is in B β, r , as desired. Remark that, from the proof of Lemma 2.1, if β is a root of P and Re{ z − β P ′ z /P z } ≥ 1/2 for all |z − β| ≤ r, then the closed ball B β, r is contained in the immediate basin of β. Lemma 2.2. Let P be a polynomial of degree d ≥ 3. Let α1 be a root of P and α2 the nearest root to α1. Let β |α1 − α2|, and let m be the multiplicity of α1. Suppose that there is a root α of P such that |α1 − α| ≥ b for some positive number b ≥ β. Then the closed ball {z ∈ C : |z − α1| ≤ δ} is contained entirely in the immediate basin of α1, where δ 1 2 2d − 1 [ 2m 1 β 2d − 3 b − √ [ 2m 1 β b 2d − 3 ]2 − 4 2d − 1 2m − 1 bβ ]


Introduction
Let P be a polynomial of degree d, and let N p z z − P z /P z be the Newton map induced by P .Let N be the set of positive integers.For each k ∈ N, let N k p denote the k-iterate of N p , that is, N 1 • N p .For a root α of P , we say that a set U is the immediate basin of α if U is the largest connected open set containing α and N k p z → α, as k → ∞, for all z ∈ U. Every immediate basin U is forward invariant, that is, N p U U, 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 P are inside the unit disc and 0 < ε < 1, there is a constant n d, ε such that for every root α of P , there is a point z with |z| 2 such that |N n p z − α| < ε for all n ≥ n d, ε .Schleicher also showed that n d, ε can be chosen so that 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 f d which arose when he estimated an upper bound for the distance of a point z to a root α.Schleicher showed that if z is in the immediate basin of α and |N p z − z| δ, then the distance between z and α is at most δf d .
In this paper, we give an algorithm to improve the value of f d .Even though, it is not an explicit formula, it can be easily computed.The following is our main result.

Main
Otherwise let Note that the value of M d, y in the main theorem depends only on the constant y and the degree d.Hence if we select y appropriately the value M d, y 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 M d, y is less than f d /2 d/2 for all d ≥ 10 when y d 1.5 2 4d/3 −2 .We will discuss further about this matter in Section 4.

Preliminary Results
We will use B a, r for the open ball {z ∈ C : |z − a| < r} and B a, r for the closed ball {z ∈ C : |z − a| ≤ r}, where C is the set of complex numbers.If S is a subset of C, we denote the boundary of S by ∂S.Lemma 2.1.Let P be a polynomial.Let β be a complex number and r > 0. Suppose that Re{ z − β P z /P z } ≥ 1/2 whenever |z − β| r and P z / 0. Let U be an immediate basin of a root α of Proof.For |z − β| r with P z / 0, we have where g z z − β P z /P z .Hence, then the distance of N p z to β is at most the distance of z to β.In other words, the image of z under the map N p also lies inside B β, r .
Let α be a root of P and U be its immediate basin.Suppose that α / ∈ B β, r and z ∈ U∩B β, r .Since U is forward invariant under N p , N p z still stays in U. Since U is connected, there is a curve γ 0 connecting z to N p z and lying entirely in U. Since N k p γ 0 converges uniformly to α as k → ∞, the set ∞ k 1 N k p γ 0 ∪ {α} forms a continuous curve γ joining z and α.Note that γ is contained in U because N k p γ 0 lies inside U for all k ∈ N. Let w be the last intersection point of γ with ∂B β, r i.e., the part of the curve γ that connects w to α stays outside B β, r except at w .So N p must send w to a point outside B β, r , otherwise β is a fixed point of N p , which is impossible because all fixed points of N p are only the roots of P , and here P z / 0 on |z − β| r.From the first paragraph, however, we also have N p w ∈ B β, r .Hence we get a contradiction.Therefore if U ∩ B β, r is not empty, then α is in B β, r , as desired.
Remark that, from the proof of Lemma 2.1, if β is a root of P and Re{ z − β P z /P z } ≥ 1/2 for all |z − β| ≤ r, then the closed ball B β, r is contained in the immediate basin of β.

2.2
Proof.Without loss of generality, we assume that α 1 0. From the previous remark, it suffices to show that Re{zP z /P z } ≥ 1/2 for all |z| ≤ δ.Proof.Let α 1 , α 2 , . . ., α d be all roots of P .Suppose that |z − We are now ready to prove our main theorem.

Proof of Main Theorem
Let α 1 , α 2 , . . ., α d be all roots of P such that α 1 is the nearest root to z 0 and , we are done.Otherwise, z is not in the immediate basin of α 1 ; thus by Lemma 2.2 with m 1, we get that

3.3
hence by Lemma 2.1 α must be either α 1 or α 2 which is not the case.Therefore r 3 ≤ A 3 , and if α is α 3 we are done.Otherwise, let |z − α 1 | A 3 ε and suppose r 4 > A 4 ; then Re{ z − α 1 P z /P z } > 1/2, and by Lemma 2.1 we get a contradiction.Thus we obtain r 4 ≤ A 4 , and if α is α 4 we are done.Continuing this process, finally we get r d ≤ A d which gives

Discussion
For a fixed d, M d, y depends on only y.If we choose y too large for instance, y ≥ f d , the value of M d, y is useless when it is compared to f d .So we have to choose y carefully so that M d, y is minimal as possible.We do not know yet whether there is an explicit formula for the value y that minimizes M d, y .Table 1 below shows the values of M d, y where we set y d 1.5 2 4d/3−2 .It seems that this method can reduce upper bounds for the distance of z 0 to the root it converges to at least 2 d/2 times compared to f d .If we replace f d in 1.1 by M d, y , we derive a new upper bound for the number of iterations.
Theorem.Let P z be a polynomial of degree d ≥ 3, and let y be a positive number larger than 4d − 3.If z 0 is in an immediate basin of a root α and |N p z 0 − z 0 | ε, then |z 0 − α| ≤ εM d, y , where M d, y : max{y, A d y d − 1 / y − 1 } and A d can be derived from the following iterative algorithm.Let b y y − d / y − 1 , and

Lemma 2 . 2 .
Let P be a polynomial of degree d ≥ 3. Let α 1 be a root of P and α 2 the nearest root to α 1 .Let β |α 1 − α 2 |, and let m be the multiplicity of α 1 .Suppose that there is a root α of P such that |α 1 − α| ≥ b for some positive number b ≥ β.Then the closed ball {z ∈ C : |z − α 1 | ≤ δ} is contained entirely in the immediate basin of α 1 , where

Lemma 2 . 3 .
r |z|.Note that β ≤ b.For r < β, we have m r d − m − 1 r ≤ δ.This shows that Re{zP z /P z } ≥ 1/2 for all |z| ≤ δ, as needed.Note that if we set b β in Lemma 2.2, then the closed ball centered at α 1 of radius β 2m − 1 / 2d − 1 is contained in the immediate basin of α 1 .Furthermore, if m 1, the radius of the ball is β/ 2d − 1 .Schleicher 3, Lemma 4, page 938 made a small mistake about the radius of the ball.Indeed, he should get β/ 2d − 1 instead of β/2 d − 1 .Let P be a polynomial of degree d.For any complex number z and any positive number y > 1, if |N p z − z| ε and there is a root α d of P with |z − α d | ≥ yε, then there is a root α of P such that |z − α| ≤ y d − 1 ε/ y − 1 .
is a contradiction to the fact that εr d |α 1 − α d | ≥ bε, which implies that assumption |z 0 − α d | ≥ yε is false.Hence in this case we have |z 0 − α d | < yε.The proof is now complete.

Table 1 :
Examples of values of M d, y compared to f d when y d 1.5 2 4d/3−2 .