Third-order convergence theorem by using majorizing function for a modified Newton method in Banach space

Applied Mathematics and Computation 175 (2006) 1515–1524

www.elsevier.com/locate/amc

Third-order convergence theorem by using majorizing function for a modified Newton method in Banach space q Qingbiao Wu a

a,*

, Yueqing Zhao

a,b

Department of Mathematics, Zhejiang University, Hangzhou 310028, Zhejiang, PR China b Department of Mathematics, Taizhou University, Linhai 317000, Zhejiang, PR China

Abstract Following the ideas of Frontini and Sormani, we present a modified Newton method in Banach space which is used to solve the nonlinear operator equation. We establish the Newton–Kantorovich convergence theorem for the modified Newton method with thirdorder convergence in Banach space by using majorizing function. We also get the error estimate. Finally, two examples are provided to show the application of our theorem. Ó 2005 Elsevier Inc. All rights reserved. Keywords: Newton–Kantorovich theorem; Banach space; Nonlinear operator equation; Modified Newton method; Majorizing function

1. Introduction Solving the nonlinear operator equation is a important issue in the engineering and technology field. Determining the roots of an equation has attracted q *

This work is supported by National Natural Science Foundation of China (No. 10471128). Corresponding author. E-mail addresses: [email protected] (Q. Wu), [email protected] (Y. Zhao).

0096-3003/$ - see front matter Ó 2005 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2005.08.043

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the attention of pure and applied mathematicians for centuries. Many problems may be formulated in terms of finding zeros. These roots cannot in general be expressed in closed form. Thus, in order to solve nonlinear equations, we have to use approximate methods (see [1]). In this study, we consider to establish the Newton–Kantorovich convergence theorem for a modified Newton method with three orders in Banach space by using majorizing function which is used to solve the nonlinear operator equation F ðxÞ ¼ 0;

ð1Þ

where F is defined on an open convex X of a Banach space X with values in a Banach space Y. There are kinds of method to find a solution of (1). Iterative methods are often used to solve this problem (see [1]). If we use the famous NewtonÕs method, we can do as 1

xnþ1 ¼ xn  F 0 ðxn Þ F ðxn Þ

ðn P 0Þ ðx0 2 XÞ.

ð2Þ

Under the reasonable hypothesis, NewtonÕs method is two orders convergence. Since Kantorovich present the famous convergence result (see [2]), many Newton–Kantorovich type convergence theorems were gotten (see [3–10]). To improve the convergence order, many deformed methods have been present. The famous HalleyÕs method and Chebyshev method are three orders convergence which are the modified NewtonÕs method (see [11–15]). Recently, Frontini i.e., presented a new modified Newton method with Z x x  xn 0 ½f ðxn Þ þ f 0 ðxÞ; f 0 ðtÞ dt ’ 2 xn the modified Newton method is as follow: 2f ðxn Þ  i ; xnþ1 ¼ xn  h f 0 ðxn Þ þ f 0 xn  ff0ðxðxnnÞÞ

ð3Þ

where f is real or complex function. They also established the local convergence theorem (see [16–19]). They discussed the modified method in real or complex space. In the paper, following the ideas of Frontini and Sormani we generalize the modified Newton method in Banach space. We define the modified Newton method as follow: suppose F is defined on an open convex X of a Banach space X with values in a Banach space Y, and F(x) is Fre´chet derivatives in X, F 0 (x)1 exists, the new modified Newton method is ( y n ¼ xn  F 0 ðxn Þ1 F ðxn Þ; ð4Þ 1 xnþ1 ¼ xn  2½F 0 ðxn Þ þ F 0 ðy n Þ F ðxn Þ. We establish Newton–Kantorovich type convergence theorem for the modified Newton method with third-order by using majorizing function and get the

Q. Wu, Y. Zhao / Appl. Math. Comput. 175 (2006) 1515–1524

1517

error estimate. Finally, two example is provided to show the application of our theorem. 2. Main results In this section, we establish a Newton–Kantorovich convergence theorem and present the error estimate. Denote gðtÞ ¼ 12 Kt2  bt þ bg, where K, g, b are pffiffiffiffiffiffiffiffi the positive real numbers, while h = Kgb, Then, when h 6 12, t ¼ 1 h12h g, pffiffiffiffiffiffiffiffi t ¼ 1þ h12h g are two positive roots of g(t). Let ( 1 sn ¼ tn  g0 ðtn Þ gðtn Þ; t0 ¼ 0; ð5Þ tnþ1 ¼ tn  2½g0 ðtn Þ þ g0 ðsn Þ1 gðtn Þ. Firstly, we get some lemmas. Lemma 1. Suppose the sequences {tn}, {sn} are generated by (5), if h 6 12, then the sequences {tn}, {sn} are the increase and converge to t*, and for all natural number n, t  tnþ1 ¼

ðt  tn Þ

3

2

ðt  tn þ t  tn Þ  ðt  tn Þðt  tn Þ

ð6Þ

;

3

t  tnþ1 ¼

ðt  tn Þ 2

ðt  tn þ t  tn Þ  ðt  tn Þðt  tn Þ 0 6 tn 6 sn 6 tnþ1 < t .

;

ð7Þ ð8Þ

Proof. Because gðtÞ ¼ K2 ðt  tÞðt  tÞ, g0 ðtÞ ¼  K2 ½ðt  tÞ þ ðt  tÞ, denote an = (t*  tn), bn = (t**  tn), then we get gðtn Þ an bn ; ¼ g0 ðtn Þ an þ bn 2 K2 an bn 2gðtn Þ ¼ tnþ1  tn ¼  0 g ðtn Þ þ g0 ðsn Þ K2 ðt  tn þ t  tn þ t  sn þ t  sn Þ an bn an bn ðan þ bn Þ ¼ ; ¼ an þ bn  ðsn  tn Þ ðan þ bn Þ2  an bn sn  t n ¼ 

tnþ1  sn ¼ tnþ1  tn  ðsn  tn Þ ¼

an bn ðan þ bn Þ ðan þ bn Þ2  an bn



ð9Þ

an bn a2n b2n ; ¼ 2 an þ bn ðan þ b2n þ an bn Þðan þ bn Þ ð10Þ

an bn ðan þ bn Þ a3n anþ1 ¼ t  tnþ1 ¼ t  tn  ðtnþ1  tn Þ ¼ an  ¼ ; 2 ðan þ bn Þ  an bn ðan þ bn Þ2  an bn an bn ðan þ bn Þ b3n ¼ . bnþ1 ¼ t  tnþ1 ¼ t  tn  ðtnþ1  tn Þ ¼ bn  ðan þ bn Þ2  an bn ðan þ bn Þ2  an bn

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By (6), (7), (9) and (10), t0 = 0 < t*, and by induction, we know (8) holds. So, we get {tn}, {sn} are the increase and converge to t*. That completes the proof of Lemma 1. h Lemma 2. Suppose X and Y are the Banach spaces, X is an open convex of the Banach space X, F : X  X ! Y has two orders continuous Fre´chet derivatives. The sequences {xn}, {yn} are generated by (4). Then, for all natural number n, the following formula holds: Z 1 F ðxnþ1 Þ ¼ F 00 ðy n þ tðxnþ1  y n ÞÞð1  tÞ dtðxnþ1  y n Þ2 0

1 þ 2 Z þ

Z

1

F 00 ðxn þ tðy n  xn ÞÞ dtðy n  xn Þðxnþ1  y n Þ 0

1

F 00 ðxn þ tðy n  xn ÞÞð1  tÞ dtðy n  xn Þ

2

0



1 2

Z

1 2

F 00 ðxn þ tðy n  xn ÞÞdtðy n  xn Þ . 0

Proof. By (4), we get 1 1 F 0 ðy n Þðxnþ1  y n Þ ¼ ½F 0 ðy n Þ  F 0 ðxn Þðxnþ1  y n Þ þ ½F 0 ðy n Þ þ F 0 ðxn Þðxnþ1  y n Þ 2 2 Z 1 1 00 F ðxn þ tðy n  xn ÞÞdtðy n  xn Þðxnþ1  y n Þ ¼ 2 0 " #  0 1 F 0 ðy n Þ þ F 0 ðxn Þ 0 F ðy n Þ þ F 0 ðxn Þ 1 F ðxn Þ þ F ðxn Þ F ðxn Þ  2 2 Z 1 1 00 F ðxn þ tðy n  xn ÞÞdtðy n  xn Þðxnþ1  y n Þ ¼ 2 0  0  F ðy n Þ þ F 0 ðxn Þ þ  F 0 ðxn Þ F 0 ðxn Þ1 F ðxn Þ 2 Z 1 1 F 00 ðxn þ tðy n  xn ÞÞdtðy n  xn Þðxnþ1  y n Þ ¼ 2 0 Z 1 1 00 F ðxn þ tðy n  xn ÞÞdtðy n  xn Þ2 ;  2 0 F ðxnþ1 Þ ¼ F ðxnþ1 Þ  F ðy n Þ  F 0 ðy n Þðxnþ1  y n Þ þ F ðy n Þ þ F 0 ðy n Þðxnþ1  y n Þ Z 1 ¼ F 00 ðy n þ tðxnþ1  y n ÞÞð1  tÞdtðxnþ1  y n Þ2 0

þ

1 2

Z 0

1

F 00 ðxn þ tðy n  xn ÞÞdtðy n  xn Þðxnþ1  y n Þ

Q. Wu, Y. Zhao / Appl. Math. Comput. 175 (2006) 1515–1524

þ

Z

1519

1 2

F 00 ðxn þ tðy n  xn ÞÞð1  tÞdtðy n  xn Þ 0

1  2

Z

1 2

F 00 ðxn þ tðy n  xn ÞÞdtðy n  xn Þ .

0

That completes the proof of Lemma 2.

h

Theorem 1. Suppose X and Y are the Banach spaces, X is an open convex of the Banach space X  X, F : X  X ! Y has two orders continuous Fre´chet derivatives nonlinear operator, x0 2 X, F 0 (x0)1 exists, and the following conditions satisfies: 1

kF ðx0 Þ k 6 b; kF 00 ðxÞk 6 M;

1

kF 0 ðx0 Þ F ðx0 Þk 6 g;

kðF 00 ðxÞ  F 00 ðyÞÞk 6 N kx  yk;   5N M 1þ 6 K; 3M 2 b h ¼ Kbg 6 1=2;

x; y 2 X;

ð11Þ

Sðx0 ; t Þ 2 X.

Then, the sequence {xn}nP0 generated by (4) is well defined, xn 2 Sðx0 ; t Þ and converges to the unique solution x* 2 S(x0, t**) of Eq. (1), kxn  x*k 6 t*  tn, where Sðx0 ; tÞ ¼ fxjkx  x0 k < t; x 2 Xg, Sðx0 ; tÞ is the close domain of S(x0, t). Proof. Firstly, by induction, we will prove the following formula holds: xn 2 Sðx0 ; tn Þ; 1

1

kF 0 ðxn Þ F 0 ðx0 Þk 6 g0 ðtn Þ ; ky n  xn k 6 sn  tn ;     F 0 ðx Þ þ F 0 ðy Þ1   n 1 n F 0 ðx0 Þ 6 2½g0 ðtn Þ þ g0 ðsn Þ ;    2

ð12Þ

kxnþ1  y n k 6 tnþ1  sn ; kxnþ1  xn k 6 tnþ1  tn . In fact, by Lemma 1, for any natural number n, we know tn < t*. It is easily proved that when n = 0 the above formula holds. Suppose when n P 0, the above formula also holds. Then, we get kxnþ1  x0 k 6 kxnþ1  xn k þ kxn  x0 k 6 tnþ1  tn þ tn ¼ tnþ1 ; k½F 0 ðxnþ1 Þ  F 0 ðx0 Þk 6 Mkxnþ1  x0 k 6 Mtnþ1 < Kt pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1  1  2h 1 1 6 6 0 . ¼ Kg h b kF ðx0 Þ1 k

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Q. Wu, Y. Zhao / Appl. Math. Comput. 175 (2006) 1515–1524 0

By Neumann Lemma, we get F (xn+1)1 exists, and 1

kF 0 ðx0 Þ k

kF 0 ðxnþ1 Þ1 k 6

0

1

0

0

6

1  kF ðx0 Þ kk½F ðxnþ1 Þ  F ðx0 Þk b 1 1 6 6 ¼ g0 ðtnþ1 Þ . 1  bKtnþ1 b1  Ktnþ1

b 1  bMkxnþ1  x0 k

By the conditions of Theorem 1, we get  Z 1 Z   1 1 00 00  F ðxn þ tðy n  xn ÞÞð1  tÞdt  F ðy n þ tðy n  xn ÞÞdt   2 0 0 Z 1    00 00  6  ½F ðxn þ tðy n  xn ÞÞ  F ðxn Þð1  tÞdt 0  Z 1  1  00 00  þ ½F ðy n þ tðy n  xn ÞÞ  F ðxn Þdt  2 0 N N 5N ky  xn k. 6 ky n  xn k þ ky n  xn k ¼ 6 4 12 n By (9) and (10), we know, 2

2

ðsn  tn Þ ðan bn Þ ða2n þ b2n þ an bn Þðan þ bn Þ 2 2 ¼  6 an þ bn 6 t þ t ¼ 6 . tnþ1  sn ðan þ bn Þ2 Kb Mb a2n b2n

By Lemma 2, we get M M 5N kxnþ1  y n k2 þ kðy n  xn Þðxnþ1  y n Þk þ ky  xn k3 2 2 12 n M M 5N 2 3 ðsn  tn Þ 6 ðtnþ1  sn Þ þ ðsn  tn Þðtnþ1  sn Þ þ 2 2 12 " # 2 M 5N 2ðsn  tn Þ 1 2  ðsn  tn Þðtnþ1  sn Þ ¼ ðtnþ1  sn Þ þ M þ 2 12 tnþ1  sn 2   M 5N 1 2 ðsn  tn Þðtnþ1  sn Þ 6 ðtnþ1  sn Þ þ M þ 2 3Mb 2 K K 6 ðtnþ1  sn Þ2 þ ðsn  tn Þðtnþ1  sn Þ ¼ gðtnþ1 Þ. 2 2

kF ðxnþ1 Þk 6

Hence, we get 1

1

ky nþ1  xnþ1 k ¼ k  F 0 ðxnþ1 Þ F ðxnþ1 Þk 6 g0 ðtnþ1 Þ gðtnþ1 Þ ¼ snþ1  tnþ1 ;  0  F ðxnþ1 Þ þ F 0 ðy nþ1 Þ  0   F ðx0 Þ   2

Q. Wu, Y. Zhao / Appl. Math. Comput. 175 (2006) 1515–1524

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 0  ½F ðy nþ1 Þ  F 0 ðxnþ1 Þ  0 0  þ ½F ðxnþ1 Þ  F ðx0 Þ ¼  2 M M 6 ky nþ1  xnþ1 k þ Mkxnþ1  x0 k 6 ðsnþ1  tnþ1 Þ þ Mtnþ1 2 2 M 1 1  . 6 ðtnþ1 þ snþ1 Þ < Kt 6 6 00 2 b kF ðx0 Þ1 k h 0 i1 F ðxnþ1 ÞþF 0 ðy nþ1 Þ So, we know exists, and 2    1  F 0 ðx Þ þ F 0 ðy Þ  b 2   nþ1 nþ1 6 6   1  bK ðtnþ1 þ snþ1 Þ b1  Ktnþ1 þ b1  Ksnþ1  2 2 ¼ 2½g0 ðtnþ1 Þ þ g0 ðsnþ1 Þ1 ; kxnþ2  y nþ1 k    0 1   F ðxnþ1 Þ þ F 0 ðy nþ1 Þ  0  1 ¼ F ðxnþ1 Þ F ðxnþ1 Þ  F ðxnþ1 Þ   2       F 0 ðx Þ þ F 0 ðy Þ 1 F 0 ðx Þ þ F 0 ðy Þ    nþ1 nþ1 nþ1 nþ1  F 0 ðxnþ1 Þ F 0 ðxnþ1 Þ1 F ðxnþ1 Þ ¼   2 2 M ðsnþ1  tnþ1 Þ2 2 6 K½g0 ðtnþ1 Þ þ g0 ðsnþ1 Þ1 ðsnþ1  tnþ1 Þ2  0  g ðtnþ1 Þ þ g0 ðsnþ1 Þ  g0 ðxnþ1 Þ g0 ðtnþ1 Þ1 gðtnþ1 Þ ¼ 2½g0 ðtnþ1 Þ þ g0 ðsnþ1 Þ1 2 6 2½g0 ðtnþ1 Þ þ g0 ðsnþ1 Þ1

¼ g0 ðtnþ1 Þ1 gðtnþ1 Þ  2½g0 ðtnþ1 Þ þ g0 ðsnþ1 Þ1 gðtnþ1 Þ ¼ tnþ2  snþ1 .

kxnþ2  xnþ1 k 6 kxnþ2  y nþ1 k þ ky nþ1  xnþ1 k 6 tnþ2  tnþ1 . Hence, the sequence {xn}nP0 generated by (4) is well defined, xn 2 Sðx0 ; t Þ and converges to the solution x 2 Sðx0 ; t Þ of Eq. (1). Now we prove the unique. Suppose y* also is the solution of Eq. (1) on S(x0, t**). Then, we have   Z 1  0  0    F ðx0 Þ1 F ðx þ tðy  x ÞÞdt  I    0 Z 1    1  0 0    0 6 kF ðx0 Þ k F ½x þ tðy  x Þ  F ðx0 Þdt  Z

0

1

kx þ tðy   x Þ  x0 kdt

6 Mb 0

Z 6 Mb

0

1

½ð1  tÞkx  x0 k þ tky   x0 kdt <

Mb  ðt þ t Þ 6 1. 2

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Q. Wu, Y. Zhao / Appl. Math. Comput. 175 (2006) 1515–1524

R1 By Neumann Lemma, we get the inverse of 0 F 0 ½x þ tðy   x Þdt exists. Because Z 1 F ðy  Þ  F ðx Þ ¼ F 0 ½x þ tðy   x Þdtðy   x Þ. 0

y*

x *.

So, = when m > n,

That completes the proof of the solution unique. Moreover,

kxm  xn k 6 kxm  xm1 k þ kxm1  xm2 k þ    þ kxnþ1  xn k 6 tm  tn and let m ! 1, we get kxn  x k 6 t  tn . That completes the proof of Theorem 1. h Theorem 2. suppose F satisfies the conditions of Theorem 1, then, (I) when h < 12, n

kxn  x k 6 t  tn ¼

h3 ðt  t Þ 1h

3n

ð1  h2 Þg

¼

1h

3n

n

h3

1

;

(II) when h ¼ 12, 2g ; 3n pffiffiffiffiffiffiffiffi 1p12h ffiffiffiffiffiffiffiffi. ¼ 1þ 12h

kxn  x k 6 t  tn ¼ 

where h ¼ tt

Proof. When h < 12, by Lemma 1, we get   3   32   3n n t  tn t  tn1 t  tn2 t  t0 ¼  ¼  ¼    ¼  ¼ h3 . t  tn t  tn1 t  tn2 t  t0 2

Because t  tn ¼ t  tn þ t  t ¼ t  tn þ ð1hh Þg, we easily get n

t  tn ¼

h3 ðt  t Þ n

1  h3

¼

ð1  h2 Þg n

1  h3

n

h3

1

.

When h ¼ 12, t* = t** = 2g. By Lemma, we know t  tn ¼

t  tn1 t  t0 2g ¼  ¼ ¼ n. 3 3 3n

That completes the proof of Theorem 2. h

Q. Wu, Y. Zhao / Appl. Math. Comput. 175 (2006) 1515–1524

1523

3. Numerical examples Example 1. Consider the root of the equation F(x) = x3  2x  5 = 0 on [1, 3]. If we select the initial point x0 = 2, then we easily get F 0 ðx0 Þ1 ¼ 0:1; g ¼ kF 0 ðx0 Þ1 F ðx0 Þk ¼ 0:1; M ¼ 1:8; N ¼ 0:6;   0:6 K ¼ 1:8  1 þ ¼ 1:85556; h ¼ Kg ¼ 0:185556 < 0:5; 6  1:82 t ¼ 0:1115; t ¼ 0:9663; h ¼ 0:11540 . That means the hypotheses of Theorem 1 are satisfied. Moreover, we get error estimate kxn  x k 6 t  tn 6

ð1  h2 Þg 1h

3n

n

h3

1

¼

0:09867 1  ð0:1154Þ

3n 1

3n

ð0:1154Þ

;

kx1  x k 6 t  t1 6 1:3168  103 ; kx2  x k 6 t  t2 6 3:1105  109 ; kx3  x k 6 t  t3 6 4:1190  1026 , ... Example 2. Consider the integral equation Z 1 1 s xðsÞ ¼ 1 þ xðsÞ xðtÞdt; xðsÞ  X ¼ C½0; 1. 4 s þ t 0 we define the norm kxk = max06s61jx(s)j. We define the operator F in X as following: Z 1 1 s xðtÞdt  xðsÞ þ 1. F ðxÞ ¼ xðsÞ 4 0 sþt We consider initial point x0 = x0(s) = 1, then we get 0

1

kF ðx0 Þ k ¼ 1:5304;

Z 1    1 s  dt M ¼ kF ðx0 Þ k  2  max  06s61 4 0 sþt 0

1

ln 2 ¼ 0:5304; 2 K ¼ M ¼ 0:5304;

¼ 1:5304  N ¼ 0;

1

g ¼ kF 0 ðx0 Þ F 0 ðx0 Þk ¼ 0:2652; h ¼ Kg ¼ 0:1407 < 0:5; t ¼ 0:2870;

t ¼ 3:485;

h ¼ 0:08235.

That means the hypotheses of Theorem 1 are satisfied. Moreover, we get error estimate

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Q. Wu, Y. Zhao / Appl. Math. Comput. 175 (2006) 1515–1524

kxn  x k 6 t  tn 6

ð1  h2 Þg 1h

3n

n

h3

1

¼

0:2634 1  ð0:08235Þ

3n 1

3n

ð0:08236Þ

;

kx1  x k 6 t  t1 6 1:7873  103 ; kx2  x k 6 t  t2 6 5:5709  1010 ; kx3  x k 6 t  t3 6 1:690  1029 , ... References [1] J.M. Ortega, W.C. Rheinbolt, Iterative Solution of Nonlinear Equations in Several Variables, Academic press, New York, 1970. [2] L. Kantorovich, On Newton method, Trudy Mat. Inst. Steklov 28 (1949) 104–144 (in Russian). [3] Z.D. Huang, A note on the Kantorovich theorem for Newton iteration, J. Comput. Appl. Math. 47 (1993) 211–217. [4] X.H. Wang, Convergence of NewtonÕs method and uniqueness of the solution of equations in Banach spaces II, Acta Math Sin. (English series) 19 (2) (2003) 405–412. [5] Ioannis K. Argyros, On NewtonÕs method under mild differentiability conditions and applications, Appl. Math. Comput. 102 (1999) 177–183. [6] Ioannis K. Argyros, An improved error analysis for Newton-like methods under generalized conditions, J. Comput. Appl. Math. 157 (2003) 169–185. [7] Ioannis K. Argyros, On the Newton–Kantorovich hypothesis for solving equations, J. Comput. Appl. Math. 169 (2004) 315–332. [8] Ioannis K. Argyros, On the comparison of a weak variant of the Newton–Kantorovich and Miranda theorems, J. Comput. Appl. Math. 166 (2004) 585–589. [9] Ioannis K. Argyros, On a theorem of L.V. Kantorovich concerning NewtonÕs method, J. Comput. Appl. Math. 155 (2003) 223–230. [10] M.A. Hernandez, A modification of the classical Kantorovich conditions for NewtonÕs method, J. Comput. Appl. Math. 137 (2001) 201–205. [11] J.M. Gutierrez, M.A. Hernandez, An acceleration of NewtonÕs method: Super-Halley method, Appl. Math. Comput. 117 (2001) 223–239. [12] I.K. Argyros, D. Chen, Results on the Chebyshev method in Banach spaces, Proyecciones 12 (1993) 119–128. [13] M.A. Herna´ndez, M.A. Salanova, Modification of the Kantorovich assumptions for semilocal convergence of the Chebyshev method, J. Comput. Appl. Math. 126 (2000) 131–143. [14] S. Amat, S. Busquier, et al., A fast ChebyshevÕs method for quadratic equations, Appl. Math. Comput. 148 (2004) 461–474. [15] X.H. Wang, Convergence on the iteration of Halley family in weak condition, Chin. Sci. Bull. 42 (7) (1997) 552–555. [16] S. Weerakoon, T.G.I. Fernando, A variant of NewtonÕs method with accelerated third-order convergence, Appl. Math. Lett. 13 (8) (2000) 87–93. [17] M. Frontini, E. Sormani, Some variant of NewtonÕs method with third-order convergence, Appl. Math. Comput. 140 (2-3) (2003) 419–426. [18] M. Frontini, E. Sormani, Modified NewtonÕs method with third-order convergence and multiple roots, J. Comput. Appl. Math. 156 (2) (2003) 345–354. [19] M. Frontini, E. Sormani, Third-order methods from quadrature formulae for solving systems of nonlinear equations, Appl. Math. Comput. 149 (3) (2004) 771–782.