Convergence analysis of the secant type methods

Convergence analysis of the secant type methods

Applied Mathematics and Computation 188 (2007) 514–524 www.elsevier.com/locate/amc Convergence analysis of the secant type methods Jinhai Chen a a,*...

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Applied Mathematics and Computation 188 (2007) 514–524 www.elsevier.com/locate/amc

Convergence analysis of the secant type methods Jinhai Chen a

a,*

, Zuhe Shen

b

Department of Mathematics, City University of Hong Kong, 83 Tat Chee Avenue, Kowloon, Hong Kong b Department of Mathematics, Nanjing University, Nanjing 210093, People’s Republic of China

Abstract In this paper, under the hypothesis that derivative satisfies some kinds of weak Lipschitz condition, the radius estimates of the convergence balls of secant type methods for operator equations are established in Banach space. Some applications and numerical experiments are also given. Ó 2006 Elsevier Inc. All rights reserved. Keywords: Secant type methods; Convergence ball; Uniqueness ball

1. Introduction Consider a Fre´chet differential nonlinear operator equation F ðxÞ ¼ 0;

ð1:1Þ

defined on a convex open subset D of a Banach space X with values in a Banach space Y. Solving the problem (1.1) is a basic and important problem in applied and computational mathematics. The most well-known method to solve (1.1) is the Newton’s method, which is quadratically convergent under the proper conditions showed by the well-known Kantorovich theorem [5,6]. Classical secant method can be used to solve the problem (1.1), instead of Newton’s method. Starting with two points, x1, x0 2 D, the secant method is described by the following procedure: 1

xnþ1 ¼ xn  ½xn1 ; xn ; F  F ðxn Þ;

n P 0;

ð1:2Þ

where the bound linear operator [x, y; F] is called a divided difference of first order for operator F on the points x and y and it satisfies the following equality: ½x; y; F ðx  yÞ ¼ F ðxÞ  F ðyÞ:

*

Corresponding author. E-mail address: [email protected] (J. Chen).

0096-3003/$ - see front matter Ó 2006 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2006.10.014

ð1:3Þ

J. Chen, Z. Shen / Applied Mathematics and Computation 188 (2007) 514–524

515

The convergence of the classical secant method (1.2) has been studied by many authors [1–3,6–8]. Since the finite difference approximation of the Jacobian is not very accurate, and the first iterations of this method are usually worse than Newton’s method, in [2,3], the following modification of secant method was discussed: xnþ1 ¼ xn  ½xn ; xn þ an ðxn1  xn Þ; F 1 F ðxn Þ;

ð1:4Þ

n P 0;

where a 2 [0, 1]. This new iterative method is, in general, a good alternative to Newton’s method, since [xn, xn + an(xn1  xn); F] is always the good approximation to F 0 (xn). Moreover, it is more efficient than classical secant method (1.2), we refer the reader to [1,7] and the references therein. Now we suppose x* is a solution of (1.1), analyze the local convergence of secant type method (1.4) and in particular, estimate the radius of the convergence ball of the secant method. Let x* denote the solution of (1.1), B(x, r) denote an open ball with radius r and center x, and let Bðx; rÞ denote its closure. With the derivative of F satisfying the Lipschitz condition: 1

kF 0 ðx Þ ðF 0 ðxÞ  F 0 ðyÞÞk 6 Lkx  yk;

8x; y 2 Bðx ; rÞ; for some L > 0:

Traub and Wozniakowski [9] and Wang [10] independently gave an exact estimate of radius r ¼ vergence ball of Newton’s method: 1

xnþ1 ¼ xn  F 0 ðxn Þ F ðxÞ;

ð1:5Þ 2 3L

for the con-

n ¼ 0; 1; 2; . . .

Under the hypothesis that F 0 (x) satisfies the following Lipschitz conditions: Z qðxÞ 0  1 0 0 s LðuÞ du; 8x 2 Bðx ; rÞ; kF ðx Þ ðF ðxÞ  F ðx ÞÞk 6

ð1:6Þ

sqðxÞ

0

 1

0

0



kF ðx Þ ðF ðxÞ  F ðx ÞÞk 6

Z

qðxÞ

8x 2 Bðx ; rÞ;

LðuÞ du;

ð1:7Þ

0

where q(x) = kx  x*k, xs = x* + s(x  x*), and L is monotonic function, Wang [11,12] studied the convergence of the Newton’s method. In this paper, we study the convergence of the secant type method (1.4) with the following Lipschitz conditions [13–16] for operator F which are more general than (1.6), (1.7): 1

8x; y 2 Bðx ; rÞ;

ð1:8Þ

1

8x 2 Bðx ; rÞ;

ð1:9Þ

kF 0 ðx Þ ðF 0 ðxÞ  F 0 ðyÞÞk 6 f0 ðkx  ykÞ; kF 0 ðx Þ ðF 0 ðxÞ  F 0 ðx ÞÞk 6 g0 ðkx  x kÞ;

where f0, g0 are all monotonic functions. The radius estimate of the convergence ball is given, and the corresponding error estimate is also established. Especially, with the Lipschitz condition (1.5), we show the convergence of the sequence {xn}(n P 1) generated by the secant type method is quadratic at least. And a similar result under the Ho¨lder condition is also obtained. Finally, some applications and numerical experiments are provided to illuminate the convergence theories of secant type methods. The paper is organized as follows: in Sections 2–4, we give the local convergence theorems, the estimates of radii for the convergence ball of secant type method, the uniqueness of the solution and theirs applications. Section 5 shows the (1 + p)-order convergence of the secant type method. In Section 6, we present some numerical experiments. 2. Convergence analysis under weak Lipschitz conditions The condition on the function F kF ðxÞ  F ðxs Þk 6 Lkx  xs k;

8x 2 Bðx ; rÞ;

ð2:1Þ

where xs = x* + s(x  x*), 0 6 s 6 1, is usually called radius Lipschitz condition in the ball B(x*, r) with constant L. Sometimes if it is only required to satisfy kF ðxÞ  F ðx Þk 6 Lkx  x k;

8x 2 Bðx ; rÞ;

ð2:2Þ

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J. Chen, Z. Shen / Applied Mathematics and Computation 188 (2007) 514–524

we call it the center Lipschitz condition in the ball B(x*, r) with constant L. Furthermore, the L in the Lipschitz condition need not be a constant, but a positive integrable function. If this is the case, then (2.1) or (2.2) is replaced by kF ðxÞ  F ðxs Þk 6 f0 ðð1  sÞqðxÞÞ;

8x 2 Bðx ; rÞ; 0 6 s 6 1;

ð2:3Þ

or kF ðxÞ  F ðx Þk 6 g0 ðqðxÞÞ;

8x 2 Bðx ; rÞ;

ð2:4Þ

x*k.

At the same time, the corresponding ‘Lipschitz condition’ is referred to as having the L where q(x) = kx  average. By Banach theorem [5,12], the following result can be obtained directly. Lemma 2.1. Suppose that F has a continuous derivative in B(x*, r), F 0 (x*)1 exists and F 0 (x*)1F 0 satisfies the Lipschitz condition: 1

kF 0 ðx Þ F 0 ðxÞ  Ik 6 g0 ðqðxÞÞ;

8x 2 Bðx ; rÞ;

ð2:5Þ

where g0 is positive integrable function. Let r satisfy g0 ðrÞ 6 1;

ð2:6Þ

then F 0 (x) is invertible in this ball and kF 0 ðxÞ1 F 0 ðx Þk 6

1 : 1  g0 ðqðxÞÞ

ð2:7Þ

Theorem 2.1. Suppose x* satisfies (1.1), F has a continuous derivative in B(x*, r), F 0 (x*)1 exists and F 0 (x*)1F 0 (x) satisfies the Lipschitz condition: kF 1 ðx ÞðF ðxÞ  F ðxs ÞÞk 6 f0 ðð1  sÞqðxÞÞ;

8x 2 Bðx ; rÞ; 0 6 s 6 1;

ð2:8Þ

s

where x = x* + s(x  x*), q(x) = kx  x*k, and f0 is increasing. Let r > 0 satisfy R1 0

f0 ðð1  sÞrÞ ds 6 1: 1  f0 ðrÞ

ð2:9Þ

Then secant type method is convergent for all x0, x1 2 B(x*, r) and R1 f0 ðð1  sÞðð1  an Þqðxn Þ þ an qðxn1 ÞÞÞ ds  kxnþ1  x k 6 kxn  x k; R 10 1  0 f0 ðð1  an ð1  sÞÞqðxn Þ þ an ð1  sÞqðxn1 ÞÞ ds

n ¼ 0; 1; 2; . . . ;

ð2:10Þ

where R1 q¼

1

R 10 0

f0 ðð1  sÞðð1  a0 Þqðx0 Þ þ a0 qðx1 ÞÞÞ ds

f0 ðð1  a0 ð1  sÞÞqðx0 Þ þ a0 ð1  sÞqðx1 ÞÞ ds

ð2:11Þ

is less than 1. Proof. Arbitrarily choosing x0 2 B(x*, r), where r satisfies (2.9), then q determined by (2.11) is less than 1. In fact, by the monotonicity of f0, we have R1 R1 f ðð1  sÞðð1  a0 Þqðx0 Þ þ a0 qðx1 ÞÞÞ ds f0 ðð1  sÞrÞ ds 0 0 q¼ < 0 6 1: R1 R1 1  0 f0 ðð1  a0 ð1  sÞÞqðx0 Þ þ a0 ð1  sÞqðx1 ÞÞ ds 1  0 f0 ðrÞ ds Using the mean-value theorem of integration, the following expression can be obtained: Z 1 F ½x; yðx  yÞ ¼ F ðxÞ  F ðyÞ ¼ F 0 ðsx þ ð1  sÞyÞðx  yÞ ds; 8x; y 2 Bðx ; rÞ; 0

J. Chen, Z. Shen / Applied Mathematics and Computation 188 (2007) 514–524

517

i.e., F ½x; y

Z

1

F 0 ðsx þ ð1  sÞyÞ ds;

8x; y 2 Bðx ; rÞ;

ð2:12Þ

0

we have   Z  0  1 1 0   1 1 0  kI  F 0 ðx Þ ½x; y; F k ¼ kF 0 ðx Þ ðF 0 ðx Þ  ½x; y; F Þk ¼  F ðx Þ ðF ðx Þ  F ðsx þ ð1  sÞyÞ dsÞ   Z

0

1

6

kF 0 ðx Þ1 ðF 0 ðx Þ  F 0 ðsx þ ð1  sÞyÞkÞ ds

0

Z

1

f0 ðksðx  x Þ þ ð1  sÞðy  x ÞkÞ ds

6 0

Z

1 



f0 ðksðx  x Þk þ kð1  sÞðy  x ÞkÞ ds <

6 0

Z

1

f0 ðrÞ ds 6 1:

0

And then by Lemma 2.1, we know for all x,y 2 B(x*, r), [x, y; F] is invertible and k½x; y; F 1 F 0 ðx Þk 6

1

R1 0

1 f0 ðksðx  x Þ þ ð1  sÞðy  x ÞkÞ ds

;

8x; y 2 Bðx ; rÞ:

ð2:13Þ

Now if xn 2 B(x*, r), ~xn ¼ xn þ an ðxn1  xn Þ, we have by (1.4) 1

1

xnþ1  x ¼ xn  x  ½xn ; ~xn ; F  F ðxn Þ ¼ ½xn ; ~xn ; F  ðF ðx Þ  F ðxn Þ þ ½xn ; ~xn ; F ðxn  x ÞÞ Z 1 1 0  1 ¼ ½xn ; ~xn ; F  F ðx Þ F 0 ðx Þ ð½xn ; ~xn ; F   F 0 ðxs ÞÞðxn  x Þ ds; 0

1

¼ ½xn ; ~xn ; F  F 0 ðx Þ

Z

1

1

F 0 ðx Þ ðF 0 ðsxn þ ð1  sÞ~xn Þ  F 0 ðxs ÞÞðxn  x Þ ds;

0

where xs = x* + s(xn  x*). Hence, by inequality (2.13) we obtain Z 1 1 1 kF 0 ðx Þ ðF 0 ðsxn þ ð1  sÞ~xn Þ  F 0 ðxs ÞÞk  qðxn Þ ds kxnþ1  x k 6 k½xn ; ~xn ; F  F 0 ðx Þk R1 6 1

R1 0

0

f0 ðksðxn  x Þ þ ð1  sÞð~xn  x ÞkÞ ds

R1 6 1

R 10 0

1

R 10 0

qðxn Þ

f0 ðð1  sÞðð1  an Þqðxn Þ þ an qðxn1 ÞÞÞ ds

f0 ðð1  an ð1  sÞÞqðxn Þ þ an ð1  sÞqðxn1 ÞÞ ds

R1 6

0

f0 ðkð1  sÞð~xn  x ÞkÞ ds

f0 ðð1  sÞðð1  an Þqðxn Þ þ an qðxn1 ÞÞÞ ds

f0 ðð1  an ð1  sÞÞqðxn Þ þ an ð1  sÞqðxn1 ÞÞ ds

qðxn Þ qðxn Þ:

Taking n = 0 above, we obtain kx1  x*k 6 qkx0  x*k < kx0  x*k. Hence, i.e. x1 2 B(x*, r), this shows that (1.4) can be continued an infinite number of times. By mathematical induction, all xn belong to B(x*, r) and q(xn) = kxn  x*k decreases monotonically. Thus (2.10) follows. h Theorem 2.2. Suppose x* satisfies (1.1), F has a continuous derivative in B(x*, r), F 0 (x*)1 exists and F 0 (x*)1F 0 (x) satisfies the Lipschitz condition: kF 1 ðx ÞðF ðxÞ  F ðx ÞÞk 6 g0 ðqðxÞÞ;

8x 2 Bðx ; rÞ;

where q(x) = kx  x*k, and g0 is increasing. Let r > 0 satisfy R1 g0 ðrÞ þ 0 g0 ðsrÞ ds 6 1: 1  g0 ðrÞ

ð2:14Þ

ð2:15Þ

518

J. Chen, Z. Shen / Applied Mathematics and Computation 188 (2007) 514–524

Then secant type method is convergent for all x0, x1 2 B(x*, r), n = 0, 1, 2, . . ., and R1



kxnþ1  x k 6

g0 ðð1  an ð1  sÞÞqðxn Þ þ an ð1  sÞqðxn1 ÞÞ ds kxn  x k R1 1  0 g0 ðð1  an ð1  sÞÞqðxn Þ þ an ð1  sÞqðxn1 ÞÞ ds R1 g ðsqðxn ÞÞ ds 0 0 þ kxn  x k; R1 1  0 g0 ðð1  an ð1  sÞÞqðxn Þ þ an ð1  sÞqðxn1 ÞÞ ds 0

ð2:16Þ

where R1 g0 ðð1  an ð1  sÞÞqðx0 Þ þ a0 ð1  sÞqðx1 ÞÞ ds þ 0 g0 ðsqðx0 ÞÞ ds R1 1  0 g0 ðð1  a0 ð1  sÞÞqðx0 Þ þ a0 ð1  sÞqðx1 ÞÞ ds

R1 q¼

0

ð2:17Þ

is less than 1. Proof. Arbitrarily choosing x0 2 B(x*, r), where r satisfies (2.15), then q determined by (2.17) is less than 1. In fact, by the monotonicity of f0, g0, we have R1 R1 g0 ðð1  an ð1  sÞÞqðx0 Þ þ a0 ð1  sÞqðx1 ÞÞ ds þ 0 g0 ðsqðx0 ÞÞ ds g0 ðrÞ þ 0 g0 ðsrÞ ds 6 1: < R1 1  g0 ðrÞ 1  0 g0 ðð1  a0 ð1  sÞÞqðx0 Þ þ a0 ð1  sÞqðx1 ÞÞ ds

R1 q¼

0

By the expression (2.12), we have   Z   0  1 1 0  0  kI  F ðx Þ ½x; y; F k ¼ kF ðx Þ ðF ðx Þ  ½x; y; F Þk ¼ F ðx Þ ðF ðx Þ  F ðsx þ ð1  sÞyÞ dsÞ  0

 1

 1

0

Z

0



0

1

1

kF 0 ðx Þ ðF 0 ðx Þ  F 0 ðsx þ ð1  sÞyÞk dsÞ

6 0

Z

1

g0 ðksðx  x Þ þ ð1  sÞðy  x ÞkÞ ds

6 0

Z 6

1 



g0 ðksðx  x Þk þ kð1  sÞðy  x ÞkÞ ds <

0

Z

1

g0 ðrÞ ds 6 1:

0

And then by Lemma 2.1, we know for all x 2 B(x*, r), [x, y; F] is invertible and 1

k½x; y; F  F 0 ðx Þk 6

1

R1 0

1 g0 ðksðx 

x Þ

þ ð1  sÞðy 

x ÞkÞ ds

;

8x; y 2 Bðx ; rÞ:

ð2:18Þ

Now if xn 2 B(x*, r), ~xn ¼ xn þ an ðxn1  xn Þ, we have by (1.4) 1

1

xnþ1  x ¼ xn  x  ½xn ; ~xn ; F  F ðxn Þ ¼ ½xn ; ~xn ; F  ðF ðx Þ  F ðxn Þ þ ½xn ; ~xn ; F ðxn  x ÞÞ Z 1 1 1 ¼ ½xn ; ~xn ; F  F 0 ðx Þ F 0 ðx Þ ð½xn ; ~xn ; F   F 0 ðxs ÞÞðxn  x Þ ds; 1

¼ ½xn ; ~xn ; F  F 0 ðx Þ 1

¼ ½xn ; ~xn ; F  F 0 ðx Þ

Z Z

0 1

1

F 0 ðx Þ ðF 0 ðsxn þ ð1  sÞ~xn Þ  F 0 ðxs ÞÞðxn  x Þ ds 0 1

1

F 0 ðx Þ ðF 0 ðsxn þ ð1  sÞ~xn Þ  F 0 ðx Þ þ F 0 ðx Þ  F 0 ðxs ÞÞðxn  x Þ ds; 0

J. Chen, Z. Shen / Applied Mathematics and Computation 188 (2007) 514–524

519

where xs = x* + s(xn  x*). Hence, by inequality (2.18) we obtain 1

kxnþ1  x k 6 k½xn ; ~xn ; F  F 0 ðx Þk

Z

1

1

ðkF 0 ðx Þ ðF 0 ðsxn þ ð1  sÞ~xn Þ  F 0 ðx ÞÞk

0 1

þ kF 0 ðx Þ ðF ðx Þ  F 0 ðxs ÞÞkÞ  qðxn Þ ds R1 ðg0 ðksðxn  x Þ þ ð1  sÞð~xn  x ÞkÞ þ g0 ðsðxn  x ÞÞÞ ds 6 0 qðxn Þ R1 1  0 g0 ðksðxn  x Þ þ ð1  sÞð~xn  x ÞkÞ ds R1 R1 g0 ðð1  an ð1  sÞÞqðxn Þ þ an ð1  sÞqðxn1 ÞÞ ds þ 0 g0 ðsqðxn ÞÞ ds 6 0 qðxn Þ: R1 1  0 g0 ðð1  an ð1  sÞÞqðxn Þ þ an ð1  sÞqðxn1 ÞÞ ds Taking n = 0 above, we obtain kx1  x*k 6 qkx0  x*k < kx0  x*k. Hence, i.e. x1 2 B(x*, r), this shows that (1.4) can be continued an infinite number of times. By mathematical induction, all xn belong to B(x*, r) and q(xn) = kxn  x*k decreases monotonically. Thus (2.16) follows. h Similar to the proof procedures of Theorems 2.1 and 2.2, the following conclusion can be established: Theorem 2.3. Suppose x* satisfies (1.1), F has a continuous derivative in B(x*, r), F 0 (x*)1 exists and F 0 (x*)1F 0 (x) satisfies Lipschitz conditions (2.8) and (2.14). Let r > 0 satisfy R1 0

f0 ðð1  sÞrÞ ds 6 1: 1  g0 ðrÞ

ð2:19Þ

Then secant type method is convergent for all x0, x1 2 B(x*, r) and R1 

kxnþ1  x k 6

1

R 10 0

f0 ðð1  sÞðð1  an Þqðxn Þ þ an qðxn1 ÞÞÞ ds

g0 ðð1  an ð1  sÞÞqðxn Þ þ an ð1  sÞqðxn1 ÞÞ ds

kxn  x k;

n ¼ 0; 1; 2; . . . ;

ð2:20Þ

where R1 q¼

1

R 10 0

f0 ðð1  sÞðð1  a0 Þqðx0 Þ þ a0 qðx1 ÞÞÞ ds

g0 ðð1  a0 ð1  sÞÞqðx0 Þ þ a0 ð1  sÞqðx1 ÞÞ ds

ð2:21Þ

is less than 1. 3. The uniqueness ball for the solution Theorem 3.1. Suppose x* satisfies (1.1), F has a continuous derivative in B(x*, r), F 0 (x*)1 exists and F 0 (x*)1F 0 (x) satisfies the Lipschitz condition: kF 1 ðx ÞðF ðxÞ  F ðx ÞÞk 6 g0 ðqðxÞÞ;

8x 2 Bðx ; rÞ;

where xs = x* + s(x  x*), q(x) = kx  x*k, and g0 is increasing. Let r > 0 satisfy Rr g ðtÞ dt 0 0 6 1: r Then Eq. (1.1) has a unique solution x* in B(x*, r).

ð3:1Þ

ð3:2Þ

520

J. Chen, Z. Shen / Applied Mathematics and Computation 188 (2007) 514–524

Proof. Suppose x0 2 B(x*, r), x0 5 x* is also a solution of (1.1). Using the integral expressions (2.12), we have 1

1

kI  F 0 ðx Þ ½x0 ; x ; F k ¼ kF 0 ðx Þ ðF 0 ðx Þ  ½x0 ; x ; F Þk   Z  0  1 1 0   0   ¼ F ðx Þ ðF ðx Þ  F ðsx0 þ ð1  sÞx Þ dsÞ  Z

0

1

1

kF 0 ðx Þ ðF 0 ðx Þ  F 0 ðsx0 þ ð1  sÞx Þk dsÞ 6 0 Rr Z 1 g0 ðtÞ dt ¼ 1; g0 ðsrÞ ds ¼ 0 < r 0

6

Z

1

g0 ðksðx0  x ÞkÞ ds 0

and then by Lemma 2.1, we know [x0, x*; F] is invertible and 1

x0  x ¼ ½x0 ; x ; F  ðF ðx0 Þ  F ðx ÞÞ ¼ 0:

ð3:3Þ

This is in contradiction with the assumption. Thus, it follows that Eq. (1.1) has a unique solution x* in B(x*, r). h 4. Applications In the study of the secant type methods, the assumption that the derivative is Lipschitz continuous is considered traditional. In this section, we will apply the obtained results to some concrete cases. By taking f0(s) = L1sp, g0(s) = L2sp, L1, L2, p are constants, the following corollaries are obtained directly. Corollary 4.1. Suppose x* satisfies (1.1), F has a continuous derivative in B(x*, r), F 0 (x*)1 exists and F 0 (x*)1F 0 (x) satisfies the Lipschitz condition: p

kF 1 ðx ÞðF ðxÞ  F ðxs ÞÞk 6 L1 ðð1  sÞqðxÞÞ ; s

x*

x*),

where x = + s(x  q(x) = kx  sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pþ1 p r¼ 6 1: L1 ðp þ 2Þ

x*k.

8x 2 Bðx ; rÞ; 0 6 s 6 1;

ð4:1Þ

Let r > 0 satisfy ð4:2Þ

Then secant type method is convergent for all x0, x1 2 B(x*, r), q¼

L1 ðð1  a0 Þqðx0 Þ þ a0 qðx1 ÞÞp <1 R1 p ðp þ 1Þð1  L1 0 ðð1  a0 ð1  sÞÞqðx0 Þ þ a0 ð1  sÞqðx1 ÞÞ dsÞ

ð4:3Þ

and the inequality (2.10) holds. Corollary 4.2. Suppose x* satisfies (1.1), F has a continuous derivative in B(x*, r), F 0 (x*)1 exists and F 0 (x*)1F 0 (x) satisfies the Lipschitz condition: p

kF 1 ðx ÞðF ðxÞ  F ðx ÞÞk 6 L2 ðqðxÞÞ ;

8x 2 Bðx ; rÞ;

where q(x) = kx  x*k. Let r > 0 satisfy sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pþ1 p 6 1: r¼ L2 ð2p þ 3Þ Then secant type method is convergent for all x0, x1 2 B(x*, r), R1 p p 0Þ L2 0 ðð1  an ð1  sÞÞqðx0 Þ þ a0 ð1  sÞqðx1 ÞÞ ds þ L2 qðx pþ1 q¼ <1 R1 1  L2 0 ðð1  a0 ð1  sÞÞqðx0 Þ þ a0 ð1  sÞqðx1 ÞÞp ds and the inequality (2.16) holds.

ð4:4Þ

ð4:5Þ

ð4:6Þ

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521

Corollary 4.3. Suppose x* satisfies (1.1), F has a continuous derivative in B(x*, r), F 0 (x*)1 exists and F 0 (x*)1F 0 (x) satisfies Lipschitz conditions (4.1) and (4.4). Let r > 0 satisfy sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pþ1 p r¼ 6 1: ð4:7Þ L1 þ L2 ðp þ 1Þ Then secant type method is convergent for all x0, x1 2 B(x*, r), p



L1 ðð1  a0 Þqðx0 Þ þ a0 qðx1 ÞÞ <1 R1 ðp þ 1Þð1  L2 0 ðð1  a0 ð1  sÞÞqðx0 Þ þ a0 ð1  sÞqðx1 ÞÞp dsÞ

ð4:8Þ

and the inequality (2.20) holds. Corollary 4.4. Suppose x* satisfies (1.1), F has a continuous derivative in B(x*, r), F 0 (x*)1 exists and F 0 (x*)1F 0 (x) satisfies the Lipschitz conditions: p

kF 1 ðx ÞðF ðxÞ  F ðx ÞÞk 6 L2 ðqðxÞÞ ;

8x 2 Bðx ; rÞ;

ð4:9Þ

s

where x = x* + s(x  x*), q(x) = kx  x*k, and g0 are increasing. Let r > 0 satisfy sffiffiffiffiffiffiffiffiffiffiffi p p þ 1 r¼ 6 1: L2

ð4:10Þ

Then Eq. (1.1) has a unique solution x* in B(x*, r). It follows from Corollaries 4.1, 4.2 and 4.3, 4.4 that the radius estimates of convergence balls ffi of secant type qffiffiffiffiffiffiffiffiffiffiffi

methods (1.4) agrees with the results of method in [9,10,12,4] r ¼ 3L2 1 ; r ¼  Newton’s  1 larger than the known results in [3,8] r ¼ 3L1 .

p

pþ1 L1 ðpþ2Þ

, and are also

5. Convergence analysis under Ho¨lder condition In Sections 2–4, we established the local convergence theorems for the secant type method (1.4), and under the weak Lipschitz conditions, we also proved that the sequence {xn} generated by the secant type method (1.4) converges to a solution x* of (1.1) linearly at least. In this section, under the Ho¨lder condition (0 < p 6 1) and taking an = O(kxn  x*k), we will show that the convergence order can be improved to 1 + p at least. Specially, if the Lipschitz condition (1.5) holds, the convergence order is quadratic at least. Theorem 5.1. Suppose x* satisfies (1.1), F has a continuous derivative in B(x*, r), F 0 (x*)1 exists and F 0 (x*)1F 0 (x) satisfies the Ho¨lder condition (4.1) with 0 < p 6 1. Let r > 0 satisfy (4.2). Then secant type method is convergent for all x0, x1 2 B(x*, r), n = 0, 1, 2, . . ., kxnþ1  x k 6

ðp þ 1Þð1  L1

R1 0

L1 ð1  an þ Oð1Þqðxn1 ÞÞp

pþ1

p

kxn  x k

p

kxn  x k

p

kxn  x kpþ1 :

ðð1  a0 ð1  sÞÞqðxn Þ þ an ð1  sÞqðxn1 ÞÞ dsÞ

:

ð5:1Þ

Proof. By Theorem 2.1, it is easy to see kxnþ1  x k 6 ¼

L1 ðð1  an Þqðxn Þ þ an qðxn1 ÞÞ

ðp þ 1Þð1  L1 ðp þ 1Þð1  L1

R1 0

R1 0

p

ðð1  an ð1  sÞÞqðxn Þ þ an ð1  sÞqðxn1 ÞÞ dsÞ L1 ð1  an þ Oð1Þqðxn1 ÞÞ

p

ðð1  a0 ð1  sÞÞqðxn Þ þ an ð1  sÞqðxn1 ÞÞ dsÞ

So (5.1) holds, and that shows the convergence order of the sequence {xn}(n P 0) generated by the secant type method (1.4) is at least 1 + p. Specially, if the Lipschitz condition (1.5) holds, the convergence order is quadratic at least. The proof of the theorem is complete. h

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Similar to the proof procedure of Theorem 5.1, we can obtain the following (1 + p)-order convergence results of secant type method under other Ho¨lder conditions. Theorem 5.2. Suppose x* satisfies (1.1), F has a continuous derivative in B(x*, r), F 0 (x*)1 exists and F 0 (x*)1F 0 (x) satisfies the Ho¨lder condition (4.4) with 0 < p 6 1. Let r > 0 satisfy (4.5). Then secant type method is convergent for all x0, x1 2 B(x*, r), n = 0, 1, 2, . . ., R1 p L2 L ð1  an ð1  sÞ þ Oð1Þð1  sÞqðxn1 ÞÞ ds þ pþ1 0 1 1þp  kxnþ1  x k 6 ð5:2Þ kxn  x k : R1 p 1  L2 0 ðð1  an ð1  sÞÞqðxn Þ þ an ð1  sÞqðxn1 ÞÞ ds Theorem 5.3. Suppose x* satisfies (1.1), F has a continuous derivative in B(x*, r), F 0 (x*)1 exists and F 0 (x*)1F 0 (x) satisfies the Ho¨lder conditions (4.1) and (4.4) with 0 < p 6 1. Let r > 0 satisfy (4.7). Then secant type method is convergent for all x0, x1 2 B(x*, r), n = 0, 1, 2, . . . , p

kxnþ1  x k 6

ðp þ 1Þð1  L2

R1

L1 ðð1  an Þ þ an qðxn1 ÞÞ

ðð1  an ð1  sÞÞqðxn Þ þ an ð1  sÞqðxn1 ÞÞp dsÞ 0

kxn  x k

1þp

:

ð5:3Þ

6. Numerical experiments In this section, we apply the convergence ball result given in Section 2 to solve the following nonlinear equations [13,14]. Example 6.1. Let us consider F ðxÞ ¼ ex  1;

x 2 ð1; 1Þ:

In this case, we have x* = 1 and F 0 (x*) = 1, and by simple computations, we can obtain kF 0 ðx ÞðF 0 ðxÞ  F 0 ðyÞÞk ¼ kex  ey k 6 ekx  yk; 0



0

0



8x; y 2 ð1; 1Þ;

kF ðx ÞðF ðxÞ  F ðx ÞÞk ¼ ke  e k 6 ðe  1Þkx  x k; x

0

8x; y 2 ð1; 1Þ:

2  0:324947. Then the radius of convergence ball for secant type method from Theorem 2.3: r ¼ 3e2 102 In Table 1, we consider the secant type method (1.4) with an ¼ kxn1 xn k. In order to obtain a good approximation, secant method (1.2) needs more than 42 function evaluations and 41 inversions. Whereas, secant type method needs only 16 function evaluations and 8 inversions to arrive at the exact solution.

Table 1 Error estimates in max-norm, x1 = 0.4, x0 = 0.3 Iteration

Secant (1.2)

Secant type (1.4)

2 4 6 8 10 14 18 22 26 30 34 38 40 41

4.29150e002 7.67916e003 1.34777e003 2.35696e004 4.11919e005 1.25784e006 3.84086e008 1.17282e009 3.58122e011 1.09357e012 1.39888e014 4.44089e016 2.2204e016 0.00000e000

2.88623e004 7.39838e009 1.84297e013 0.00000e000

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Example 6.2. Consider the system of nonlinear equations ( 2x1  19 x21  x2 ¼ 0; x1 þ 2x2  19 x22 ¼ 0: The associated nonlinear operator F : R2 ! R2 is given by   F 1 ðx1 ; x2 Þ; F ðx1 ; x2 Þ ¼ ; F 2 ðx1 ; x2 Þ where F 1 ðx1 ; x2 Þ ¼ 2x1  19 x21  x2 and F 2 ðx1 ; x2 Þ ¼ x1 þ 2x2  19 x22 . We use the infinity norm kxk = kxk1 = max(kx1k, kx2k) and the induced matrix norms, and take the divided difference of F, [u, v;F] for u,v 2 R2 as: F i ðu1 ; v1 Þ  F i ðv1 ; v2 Þ ; u1  v 1 F i ðu1 ; v2 Þ  F i ðu1 ; v2 Þ ½u; v; F i2 ¼ u1  v 1 ½u; v; F i1 ¼

where i = 1, 2. Therefore 0 ½u; v; F  ¼ @

2  19

u21 w21 v1 w1

1

1

1 2  19

u22 w22 v2 w2

A;

Now taking x* = (9, 9), then we can easily verify that x* is a solution of the problem. Furthermore, !   2 x 1 2  0 1 1 9 0 0  F ðxÞ ¼ : ; F ðx Þ ¼ 1 2  29 x2 1 0 By simple computations, we have 2 kF 0 ðx ÞðF 0 ðxÞ  F 0 ðyÞÞk ¼ kx  yk; 8x; y 2 R2 ; 9 2 0  0 0  kF ðx ÞðF ðxÞ  F ðx ÞÞk ¼ kx  x k; 8x; y 2 R2 : 9 Then the radius of convergence ball for secant type method from Theorem 2.3: r = 3.00.

Table 2 Error estimates in max-norm, x1 = (11.5, 11.5), x0 = (11.0, 11.0) Iteration

Secant (1.2)

Secant type (1.4)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

4.48980e001 5.67443e002 5.89451e002 5.91877e004 5.92122e005 5.92147e006 5.92150e007 5.92150e008 5.92150e009 5.92150e010 5.92149e011 5.92237e012 6.03961e014 5.32907e015 0.00000e000

3.19602e001 1.02492e002 2.29426e005 2.55217e008 2.83276e011 3.19744e014 0.00000e000

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J. Chen, Z. Shen / Applied Mathematics and Computation 188 (2007) 514–524

We consider the same choice for parameters an with the previous example. As shown in Table 2, the secant type method (1.4) has better numerical behaviors than classical secant method (1.2). References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16]

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