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 z in an immediate basin of a root α to α, which leads to a better upper bound for the number of iterations of Newton's method.
1. Introduction
Let P be a polynomial of degree d, and let Np(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 Npk denote the k-iterate of Np, that is, Np1=Np,Np2=Np∘Np, and Npk=Npk-1∘Np. 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 Npk(z)→α, as k→∞, for all z∈U. Every immediate basin U is forward invariant, that is, Np(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<ɛ<1, there is a constant n(d,ɛ) such that for every root α of P, there is a point z with |z|=2 such that |Npn(z)-α|<ɛ for all n≥n(d,ɛ). Schleicher also showed that n(d,ɛ) can be chosen so thatn(d,ɛ)≤9πd4fd2ɛ2log2+|logɛ|+log13log2+1
withfd:=d2(d-1)2(2d-1)(2dd).
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 fd 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 |Np(z)-z|=δ, then the distance between z and α is at most δfd.
In this paper, we give an algorithm to improve the value of fd. Even though, it is not an explicit formula, it can be easily computed. The following is our main result.
Main Theorem 1.
Let P(z) be a polynomial of degree d≥3, and let y be a positive number larger than 4d-3. If z0 is in an immediate basin of a root α and |Np(z0)-z0|=ɛ, then |z0-α|≤ɛM(d,y), where M(d,y):=max{y,Ad+y(d-1)/(y-1)} and Ad can be derived from the following iterative algorithm.
Let b=y(y-d)/(y-1), andA2=y(d-1)[2d(y-2d+3)-3y-1](y-1)(y-4d+3).
For k=2,…,d-1, set ak=1+∑j=2k-1(Ak/(Ak-Aj)).
If 2Ak<b then letAk+1=Ak((ak+d-k)Ak+b(k+1-ak-d)Ak(ak+1)-bak).
Otherwise let
Ak+1=Akak+d-kak.
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 fd/2d/2 for all d≥10 when y=d1.52(4d/3)-2. We will discuss further about this matter in Section 4.
2. 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 P. If U∩B¯(β,r)≠∅, then α is in B(β,r).
Proof.
For |z-β|=r with P(z)≠0, we have
Np(z)-β=(z-β)(1-1g(z)),
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). Since U is forward invariant under Np, Np(z) still stays in U. Since U is connected, there is a curve γ0 connecting z to Np(z) and lying entirely in U. Since Npk(γ0) converges uniformly to α as k→∞, the set ⋃k=1∞Npk(γ0)∪{α} forms a continuous curve γ joining z and α. Note that γ is contained in U because Npk(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=12(2d-1)[(2m+1)β+(2d-3)b-[(2m+1)β+b(2d-3)]2-4(2d-1)(2m-1)bβ[(2m+1)β+b(2d-3)]2-4(2d-1)(2m-1)bβ].
Proof.
Without loss of generality, we assume that α1=0. From the previous remark, it suffices to show that Re{zP′(z)/P(z)}≥1/2 for all |z|≤δ. Let P(z)=zm∏k=2d-m(z-αk). We have
zP′(z)P(z)=m+∑k=2d-mzz-αk.
Hence
Re{zP′(z)P(z)}=m+∑k=2d-mRe{zz-αk}≥m+r(d-m-1)r-β+rr-b,
where r=|z|. Note that β≤b. For r<β, we have
m+r(d-m-1)r-β+rr-b≥12,
if 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 α1 of radius β(2m-1)/(2d-1) is contained in the immediate basin of α1. 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)).
Lemma 2.3.
Let P be a polynomial of degree d. For any complex number z and any positive number y>1, if |Np(z)-z|=ɛ and there is a root αd of P with |z-αd|≥yɛ, then there is a root α of P such that |z-α|≤y(d-1)ɛ/(y-1).
Proof.
Let α1,α2,…,αd be all roots of P. Suppose that |z-αd|≥yɛ. If |z-αj|>y(d-1)ɛ/y-1 for 1≤j≤d-1, then
|Np(z)-z|≥(∑j=1d1|z-αj|)-1>(y-1y(d-1)ɛ(d-1)+1yɛ)-1=ɛ,
a contradiction.
We are now ready to prove our main theorem.
3. Proof of Main Theorem
Let α1,α2,…,αd be all roots of P such that α1 is the nearest root to z0 and |α1-αk|≤|α1-αk+1| for k=2,…,d-1. Suppose that |z0-αd|≥yɛ. By Lemma 2.3, we have |z0-α1|≤y(d-1)ɛ/(y-1). Note that |α1-αd|≥bɛ. If α=α1, we are done. Otherwise, z is not in the immediate basin of α1; thus by Lemma 2.2 with m=1, we get that |z0-α1|>δ, where δ is defined in Lemma 2.2, that is,δ=3r2+bɛ(2d-3)-[3r2+bɛ(2d-3)]2-4(2d-1)bɛr22(2d-1),
where r2=|α1-α2|. Thus z0 satisfies the inequalitiesδ<|z0-α1|≤y(d-1)ɛy-1,
which holds if |α1-α2|<A2ɛ. If α=α2, we are done. Suppose next that α≠α2.
Now let |α1-αk|=ɛrk. If |z-α1|=A2ɛ and r3>A3, then Re{(z-α1)P′(z)P(z)}≥1+A2A2+r2+A2(d-3)A2-r3+A2A2-rd>1+12+A2(d-3)A2-r3+A2A2-b>12.
hence by Lemma 2.1α must be either α1 or α2 which is not the case. Therefore r3≤A3, and if α is α3 we are done. Otherwise, let |z-α1|=A3ɛ and suppose r4>A4; then Re{(z-α1)P′(z)/P(z)}>1/2, and by Lemma 2.1 we get a contradiction. Thus we obtain r4≤A4, and if α is α4 we are done. Continuing this process, finally we get rd≤Ad which gives |z0-αd|≤ɛ(Ad+y(d-1)/(y-1)).
Note that if Ad<b, it is a contradiction to the fact that ɛrd=|α1-αd|≥bɛ, which implies that assumption |z0-αd|≥yɛ is false. Hence in this case we have |z0-αd|<yɛ. The proof is now complete.
4. Discussion
For a fixed d, M(d,y) depends on only y. If we choose y too large (for instance, y≥fd), the value of M(d,y) is useless when it is compared to fd. 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=d1.524d/3-2. It seems that this method can reduce upper bounds for the distance of z0 to the root it converges to at least 2d/2 times compared to fd. If we replace fd in (1.1) by M(d,y), we derive a new upper bound for the number of iterations.
Examples of values of M(d,y) compared to fd when y=d1.524d/3-2.
d=
M(d,y) is less than
fd is greater than
fd/2d/2M(d,y) is greater than
10
1.3385×105
4.3758×106
1.0216
20
1.0131×1010
1.343×1013
1.2946
30
4.4559×1014
2.6158×1019
1.7915
40
1.5878×1019
4.2458×1025
2.5502
50
5.0059×1023
6.2420×1031
3.7162
60
1.1486×1028
8.6222×1037
5.4054
70
4.2054×1032
1.1410×1044
7.8967
80
1.1429×1037
1.4634×1050
11.6467
90
3.0424×1041
1.8327×1056
17.1212
100
7.9376×1045
2.2523×1062
25.2027
110
2.0274×1050
2.7262×1068
37.3244
120
5.1302×1054
3.2588×1074
55.0978
130
1.2839×1059
3.8546×1080
81.3792
140
3.1697×1063
4.5186×1086
120.7511
150
7.7889×1067
5.2563×1092
178.6315
160
1.8954×1072
6.0735×1098
265.0635
170
4.5932×1076
6.9764×10104
392.6175
180
1.1074×1081
7.9718×10110
581.5469
190
2.6450×1085
9.0669×10116
863.7282
200
6.3268×1089
1.0269×10123
1280.4536
Acknowledgment
This research is supported by the Centre of Excellence in Mathematics, the Commission on Higher Education, Thailand.
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