On the convergence of Broyden’s method in Hilbert spaces

On the convergence of Broyden’s method in Hilbert spaces

Applied Mathematics and Computation 242 (2014) 945–951 Contents lists available at ScienceDirect Applied Mathematics and Computation journal homepag...

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Applied Mathematics and Computation 242 (2014) 945–951

Contents lists available at ScienceDirect

Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc

On the convergence of Broyden’s method in Hilbert spaces I.K. Argyros a, Y.J. Cho b,c,⇑, S.K. Khattri d a

Department of Mathematical Sciences, Cameron University, Lawton, OK 73505-6377, USA Department of Mathematics Education and the RINS, Gyeongsang National University, Jinju 660-701, Republic of Korea Department of Mathematics, King Abdulaziz University, Jeddah 21589, Saudi Arabia d Department of Engineering, Stord Haugesund University College, Norway b c

a r t i c l e

i n f o

a b s t r a c t In this paper, we present a new semilocal convergence analysis for an inverse free Broyden’s method in a Hilbert space setting. In the analysis, we apply our new idea of recurrent functions concepts of divided differences of order one and Lipschitz/center–Lipschitz conditions on the operator involved. Our analysis extends the applicability of Broyden’s method in cases not covered before. Finally, we give an example to illustrate the main result in this paper. Ó 2014 Published by Elsevier Inc.

Keywords: Broyden’s method Hilbert space Semilocal convergence Majorizing sequence Divided differences Inverse free method

1. Introduction In this paper, we are concerned with the problem of approximating a locally unique solution xH of the equation

FðxÞ ¼ 0;

ð1:1Þ

where F is a continuous operator defined on an open convex subset D of a Hilbert space X with values in a Banach space Y. The field of computational sciences has seen a considerable development in mathematics, engineering sciences and economic equilibrium theory. For example, dynamic systems are mathematically modeled by difference or differential equations and their solutions usually represent the states of the systems. For the sake of simplicity, assume that a time–invariant system is driven by the equation x_ ¼ T ðxÞ for some suitable operator T , where x is the state. Then the equilibrium states are determined by solving Eq. (1.1). Similar equations are used in the case of discrete systems. The unknowns of engineering equations can be functions (difference, differential and integral equations), vectors (systems of linear or nonlinear algebraic equations) or real or complex numbers (single algebraic equations with single unknowns). Except in special cases, the most commonly used solution methods are iterative – when starting from one or several initial approximations a sequence is constructed that converges to a solution of the equation. Iteration methods are also applied for solving optimization problems. In such cases, the iteration sequences converge to an optimal solution of the problem at hand. Since all of these methods have the same recursive structure, they can be introduced and discussed in a general framework. We note that, in computational sciences, the practice of numerical analysis for finding such solutions is essentially connected to variants of Newton’s method [1–15]. Broyden’s method defined for each n P 0 by

9 yn ¼ Fðxnþ1 Þ  Fðxn Þ; > > = M0 ¼ ½x0 ; x1 j F 1 ðx1 ; x0 2 D; n P 0Þ;   > > ; Mnþ1 ¼ M n I  Fðxnþ1 Þ hyn i

xnþ1 ¼ xn  M n Fðxn Þ;

ð1:2Þ

hyn ;yn i

⇑ Corresponding author at: Department of Mathematics Education and the RINS, Gyeongsang National University, Jinju 660-701, Republic of Korea. E-mail addresses: [email protected] (I.K. Argyros), [email protected] (Y.J. Cho), [email protected] (S.K. Khattri). http://dx.doi.org/10.1016/j.amc.2014.06.001 0096-3003/Ó 2014 Published by Elsevier Inc.

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generates a sequence which approximates xH under certain conditions [11]. Here, M n 2 LðY; X Þ the space of bounded linear operators from Y into X ; h; i denotes the inner product and ½x0 ; x1 j F  is a divided difference of order one. A local as well as semilocal convergence results for Broyden–like methods have been given by several authors [1,3,8,10,11] (see also [4,7], and the references therein). Note that Broyden’s method (1.2) is an ‘‘inversion-free’’ method. That is, no linear sub-problem needs to be solved at each iteration. The study is organized as follows: The Section 2 contains some preliminary results for Lipschitz continuous divided difference of order one. In Section 3, we develop a new semilocal convergence analysis of Broyden’s method (1.2). For this purpose, we use our new idea of recurrent functions in combination with Lipschitz continuous and center-Lipschitz continuous divided differences of order one [4,7,13]. Finally, in Section 4, we provide special cases, as applications, as well as numerical examples. 2. Preliminaries: divided differences In order to make the paper as self-contained as possible, we need some results on divided differences. Definition 2.1. Let G be a nonlinear operator from an open convex subset D of a Banach space X into a Banach space Y. Furthermore, let x; y be two distinct points in D. A linear and bounded operator from X into Y – denoted by ½x; y j G – which satisfied the condition

½x; y j Gðx  yÞ ¼ GðxÞ  GðyÞ will be called a divided difference of G at the points x and y. The above condition does not determine uniquely the divided difference with the exception of the case when X is one dimensional. Let us suppose that we have associated to each pair ðx; yÞ of distinct points from D a divided difference of G at these points. We often require that the mapping ðx; yÞ ! ½x; y j G satisfied a Lipschitz condition. We say that G has a Lipschitz continuous divided difference on D if there exists a non-negative constant a such that

k½x; y j G  ½u; v j Gk 6 aðkx  uk þ ky  v kÞ for all x; y; u; v 2 D with x – y and u – v . This condition allows us to extend the mapping ðx; yÞ ! ½x; y j G to the whole Cartesian product D  D. It is well known that if G is Fréchet-differentiable on D, then ½x; x j G ¼ F 0 ðxÞ [13]. 3. Semilocal convergence analysis of Broyden’s method In this section, we present the semilocal convergence analysis of Broyden’s method. Let l0 ; l; c and g be non-negative constants. It is convenient for us to define parameters d0 , d1 and d2 by

8 gÞ > d0 :¼ 1llðcþ ; > 0 ðcþgÞ > > > < 2ð1ððlþ2l0 Þgþl0 cÞÞ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi d1 :¼ ðlþ2l0 Þgþ ððlþ2l0 ÞgÞ2 þ4l0 gð1ððlþ2l0 Þgþl0 cÞÞ  pffiffiffiffiffiffiffiffiffiffiffi > > > > lþ l2 þ4 l0 l > for l0 – 0: : d2 :¼ 12 l0

Notice that d2 is the unique positive root in ð0; 1Þ of the polynomial

f ðtÞ ¼ l0 t3 þ ðl0 þ lÞt2  l ¼ 0:

ð3:1Þ 0

2

Indeed, using (3.1), we obtain f ð0Þ ¼ l < 0; f ð1Þ ¼ 2l0 > 0 and f ðtÞ ¼ 3l0 t þ 2ðl0 þ lÞt > 0 for any t > 0. The existence of d2 follows from the intermediate value theorem applied to the polynomial f on the interval ½0; 1. Accordingly, the function f crosses the positive x-axis only once at the point d2 ð2 ð0; 1). We need the following result on the majorizing sequence for the Broyden’s method (1.2): Lemma 3.1. Let l0 ; l; c be given non-negative constants and g 2 ð0; c. Suppose that

ðl þ 2l0 Þg þ l0 c < 1

ð3:2Þ

d0 6 minfd1 ; d2 g:

ð3:3Þ

and

Then, the scalar sequence ftn g ðn P 1Þ given by

t1 ¼ 0;

t 0 ¼ c;

tnþ2 ¼ t nþ1 þ

t 1 ¼ c þ g;

lðt nþ1  t n1 Þðt nþ1  t n Þ 1  l0 ðtnþ1  t n þ 2ðt n  t 0 Þ þ t 0 Þ

ð3:4Þ

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is non-decreasing, bounded from above by

tHH ¼

g 1d

þc

ð3:5Þ

and converges to its unique least upper bound t H such that

0 6 tH 6 t HH ;

ð3:6Þ

where

 d :¼

d1 ;

d1 6 d2 ;

d2 ;

d2 6 d1 :

Moreover, the following estimate holds: for each n P 0,

0 6 tnþ2  tnþ1 6 dðt nþ1  t n Þ 6 dnþ1 g:

ð3:7Þ

Proof. It follows from (3.2), we have d0 2 ½0; 1Þ and d1 > 0. Next, we show, using mathematical induction on k P 0, that

0 6 tkþ2  tkþ1 6 dðt kþ1  t k Þ:

ð3:8Þ

For k ¼ 0, we must show, by (3.4), that

06

lðt 1  t1 Þ lðc þ gÞ 6 d or 0 6 6 d; 1  l0 t 1 1  l0 ðc þ gÞ

which is true from (3.2) and the choice of d P d0 . Let us assume that (3.8) holds for k P n þ 1. The induction hypothesis gives

tkþ2 6 tkþ1 þ dðt kþ1  t k Þ 6 tk þ dðtk  tk1 Þ þ dðtkþ1  tk Þ 6 t 1 þ dðt 1  t 0 Þ þ    þ dðtkþ1  t k Þ 6 c þ g þ dg þ    þ dkþ1 g ¼ c þ

1  dkþ2 g g< þ c ¼ t HH : 1d 1d

ð3:9Þ

We must show (3.8) holds with k þ 1 replacing k, i.e.

" kþ1

lðt kþ2  t k Þ þ d l0 ððtkþ2  tkþ2 Þ þ 2ðt kþ1  t 0 Þ þ t0 Þ 6 lððt kþ2  t kþ1 Þ þ ðt kþ1  t k ÞÞ þ d l0 d 6 ðdk þ dkþ1 Þg þ

1  dkþ1 þ2 1d

!#

g þ d l0 c

d l0 ð2  dkþ1  dkþ2 Þg þ d l0 c: 1d

ð3:10Þ

By (3.8) and (3.10) we have the estimate

lðdk þ dkþ1 Þg þ

 d l0  2  dkþ1  dkþ2 g þ dl0 c 6 d 1d

or

lðdk1 þ dk Þg þ l0



   1 þ d þ    þ dk þ 1 þ d þ    þ dkþ1 g þ l0 c  1 6 0:

ð3:11Þ

ð3:12Þ

In view of (3.12), we are motivated to define (for d ¼ s) the functions for k P 1

  fk ðsÞ ¼ lðsk1 þ sk Þg þ l0 2 1 þ s þ    þ sk þ skþ1 g þ l0 c  1:

ð3:13Þ

We need a relationship between two consecutive functions fk . From the preceding equation, we obtain

    fkþ1 ðsÞ ¼ lðsk þ skþ1 Þg þ l0 2 1 þ s þ    þ skþ1 þ skþ2 g þ l0 c  1 ¼ fk ðsÞ þ l skþ1 þ sk1 g þ l0 skþ1 þ skþ2 g ¼ f ðsÞsk1 g þ fk ðsÞ:

ð3:14Þ

Note that d1 is the unique positive root of the polynomial f1 . Instead of (3.12), we show that

fk ðdÞ 6 0 for each k P 0. Case 1: d1 6 d2 The estimate (3.15) holds for k ¼ 1, as the equality. Using (3.14), we get in turn

f2 ðdÞ ¼ f1 ðdÞ þ f ðdÞdg 6 0; since f1 ðdÞ ¼ 0 and f ðdÞ 6 0 by (3.1) and d1 P d. Assume that (3.15) holds for m 6 k. Then, again, by (3.14), we obtain

fkþ1 ðdÞ ¼ fk ðdÞ þ f ðdÞdk1 d 6 0;

ð3:15Þ

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which completes the induction for (3.15). Furthermore, define function f1 on ½0; 1Þ by

f1 ðsÞ ¼ lim f k ðsÞ:

ð3:16Þ

k!1

Using (3.15) and (3.16) yields

f1 ðdÞ ¼ lim f k ðdÞ 6 0:

ð3:17Þ

k!1

Hence we showed the sequence ft n g ðn P 1Þ is non-decreasing and bounded from above by t HH , so that the estimate (3.7) holds. It follows that there exists tH 2 ½0; tHH , so that limn!1 tn ¼ tH . Case 2: d2 6 d1 Considering (3.14), we can show

f2 ðdÞ 6 0; since f2 ðdÞ ¼ fk ðdÞ for each k P 2 instead of (3.14). The preceding equation is true by the choice of d. The rest follows as in Case 1. That completes the proof. h From Broyden’s method (1.2), we get

Mnþ1 ðyn Þ ¼ Mn ðyn Þ  Mn ðFðxnþ1 ÞÞ ¼ Mn Fðxn Þ ¼ xnþ1  xn ; which shows that Mnþ1 ¼ ½xnþ1 ; xn j F 1 . That is why we choose M0 ¼ ½x0 ; x1 j F 1 in Broyden’s method (1.2). Now, we study Broyden’s method (1.2) for the triplets ðF; x1 ; x0 Þ belonging to the class C ¼ Cðl0 ; l; g; c; c; d1 ; d2 ; dÞ defined as follows: Definition 3.2. Let l0 ; l; g; c; c; d1 ; d2 ; d be non-negative parameters satisfying the hypotheses of Lemma 3.1. We say that a triplet ðF; x1 ; x0 Þ belongs to the class C if: (h1 ) F is a non-linear operator defined on a convex subset D of a Hilbert space X with values in a Hilbert space Y. (h2 ) x1 and x0 are two points belonging in D and satisfying the inequality:

kx0  x1 k 6 c: (h3 ) There exists a divided difference, in the sense of Definition 2.1, such that  M 0 ¼ ½x0 ; x1 j F 1 2 LðY; X Þ; kM 0 Fðx0 Þk 6 g;  Lipschitz continuity condition

kM 0 ð½x; y j F   ½u; v j F Þk 6 lðkx  uk þ ky  v kÞ holds for all x; y; u; v 2 D. Note that, in view of the Lipschitz condition, there exists l0 such that

kM 0 ð½x; y j F   ½x0 ; x1 j F Þk 6 l0 ðkx  x0 k þ ky  x1 kÞ holds for all x; y 2 D. Clearly, l0 6 l holds in general and l=l0 can be arbitrarily large [2,4,7].

(h4 ) Uðx0 ; t H Þ :¼ x 2 X j kx  x0 k 6 t H # D. Here, t H is given in Lemma 3.1. We present the following semilocal convergence theorem for Broyden’s method (1.2): Theorem 3.3. If ðF; x1 ; x0 Þ 2 Cðl0 ; l; g; c; c; d1 ; d2 ; dÞ, then the sequence fxn g ðn P 1Þ generated by Broyden’s method (1.2) is well-defined, remains in Uðx0 ; tH Þ for all n P 0 and converge to a unique solution xH 2 Uðx0 ; t H Þ of the equation FðxÞ ¼ 0. Additionally, the following estimates hold: for all n P 0,

kxn  xn1 k

6 t n  t n1

ð3:18Þ

and

kxn  xH k

6 tH  tn ;

ð3:19Þ H

where the sequence ftn g is given by (3.4). Furthermore, x is the unique solution of the equation FðxÞ ¼ 0 in Uðx0 ; RÞ provided that

R P tH ; Uðx0 ; RÞ # D and

lðt H þ R  t 0 Þ þ l0 ð2t H  t 0 Þ < 1:

I.K. Argyros et al. / Applied Mathematics and Computation 242 (2014) 945–951

949

Proof. In view of (1.2) and (3.4), the estimate (3.18) holds for n ¼ 0; 1. We also have

6 kx2  x1 k þ kx1  x0 k 6 t2  t 1 þ t 1  t 0 ¼ t 2  t 0 6 tH  t0 6 t H :

kx2  x0 k H

ð3:20Þ

H

Thus x2 2 Uðx0 ; t Þ. Let us assume (3.18) and xkþ1 2 Uðx0 ; t Þ hold for all n 6 k. Using the center-Lipschitz condition in (h3 ) we get in turn that

kM 0 ð½xnþ1 ; xn j F   ½x0 ; x1 j F Þk 6 l0 ðkxnþ1  x0 k þ kxn  x1 kÞ 6 l0 ðkxnþ1  x0 k þ kxn  x0 k þ kx0  x1 kÞ 6 l0 ðtnþ1  t0 þ tn  t 0 þ t 0 Þ¼ l0 ðtnþ1 þ t n  t0 Þ ¼ l0 ½tnþ1  t n þ 2ðt n  t 0 Þ þ t0  < 1 ð3:21Þ by the proof in Lemma 3.1. From (3.21) and the Banach Lemma on invertible operators [7,13,24], we obtain

kM nþ1 M1 0 k 6

1 : l0 ½tnþ1  t n þ 2ðt n  t 0 Þ þ t 0 

ð3:22Þ

Besides, it follows from Broyden’s method (1.2) that

Fðxnþ1 Þ ¼ Fðxnþ1 Þ  Fðxn Þ þ Fðxn Þ ¼ ½xnþ1 ; xn j F ðxnþ1  xn Þ  M1 n ðxnþ1  xn Þ ¼ ð½xnþ1 ; xn j F   ½xn ; xn1 j F Þðxnþ1  xn Þ:

ð3:23Þ

It follows from ðh3 Þ and (3.23) that

kM 0 Fðxnþ1 Þk

6 kM 0 ð½xnþ1 ; xn j F   ½xn ; xn1 j F Þk kxnþ1  xn k 6 lðkxnþ1  xn k þ kxn  xn1 kÞ kxnþ1  xn k 6 ðtnþ1  tn þ tn  t n1 Þðt nþ1  t n Þ:

ð3:24Þ

Consequently, by (1.2), (3.4), (3.22) and (3.24), we get

kxnþ2  xnþ1 k ¼ kMnþ1 Fðxnþ1 Þk 6 Mnþ1 M 1 0 kM 0 Fðxnþ1 Þk 6

lðt nþ1  t n1 Þðt nþ1  t n Þ ¼ t nþ2  t nþ1 1  l0 ððt nþ1  t n Þ þ 2ðt n  t0 Þ þ t 0 Þ

ð3:25Þ

and

kxnþ2  x0 k 6 kxnþ2  xnþ1 k þ kxnþ1  x0 k 6 t nþ2  t nþ1 þ t nþ1  t 0 ¼ tnþ2  t0 6 tH ;

ð3:26Þ

which completes the induction for (3.18) and xn 2 Uðx0 ; tH Þ for all n P 0. It follows from (3.18) and Lemma 3.1 that fxn g is a Cauchy sequence in a Hilbert space X and so it converges to some xH 2 Uðx0 ; t H Þ since Uðx0 ; tH Þ is a closed set. By letting n ! 1 in (3.24), we obtain FðxH Þ ¼ 0. The estimate (3.19) follows from (3.18) by using the standard majorization techniques ([3,4,7,13]). Finally, we show the uniqueness part. Let yH be a solution of the equation FðxÞ ¼ 0 in Uðx0 ; tH Þ. Then, by (1.2) and (3.22) and ðh3 Þ, we have

 xnþ1  yH ¼ xn  yH  M n Fðxn Þ ¼ M n ½xn ; xn1 j F ðxn  yH Þ  Fðxn Þ þ FðyH Þ 

 ¼ M n ½xn ; xn1 j F   xn ; yH j F ðxn  yH Þ

ð3:27Þ

and

kxnþ1  yH k 6

lkxn1  yH k kxn  yH k : 1  l0 ð2t H  t0 Þ

ð3:28Þ

But

kxn1  yH k 6 kxn1  xn2 k þ kxn2  yH k 6 kxn1  xn2 k þ kxn2  xn3 k þ    þ kx2  x1 k þ kx1  x0 k þ kx0  yH k 6 ðt n1  tn2 Þ þ ðt n2  t n3 Þ þ    þ ðt 2  t1 Þ þ ðt1  t0 Þ þ kx0  yH k 6 tn2  t 0 þ R 6 t H þ R  t 0 :

ð3:29Þ

In view of (3.29) and the uniqueness hypotheses, it follows from (3.28) that

kxnþ1  yH k < kxn  yH k; which implies limn!1 xn ¼ yH . But we have shown that limn!1 xn ¼ xH . Hence we deduce xH ¼ yH . This completes the proof. h 4. Some remarks and numerical examples Remark 4.1. If we impose the condition ð5l0 þ 2lÞg þ l0 c > 1, then d 2 ð0; 1Þ. Moreover, if

f ðdÞ 6 0; then

d 6 d2 (see also the Remark 4.2 that follows).

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Remark 4.2. Let us define d1 by

d1 ¼

1  l0 ðc þ 2gÞ : 1  l0 c

The corresponding condition in [6] to (3.3) is given by

d0 6 d2 6 d1 : Remark 4.3. Note that iteration ft n g is the majorizing sequence for the Secant method. If l0 ¼ l, then the sufficient convergence condition [13] corresponding to (3.3) is given by

lc þ 2

qffiffiffiffiffi l g 6 1:

If l0 ¼ l, let us denote the sequence ftn g by fun g. Then a simple inductive argument shows that

tn < un ;

t nþ1  t n < unþ1  un ;

t H 6 lim un ¼ uH ; n!1

if l0 < l. Hence our error estimates are tighter. Remark 4.4. The point tHH given in closed form by (3.5) can replace t H in Theorem 3.3. In practice, we test all sufficient convergence condition given in this study to see which one applied (if any). Next, we comment on the majorizing sequences for Broyden’s method (1.2). We choose the constants as follows:

g ¼ 0:1; l0 ¼ 0:3; l ¼ 0:6; c ¼ 0:2; so that the inequality (3.2) is satisfied. Furthermore, we obtain

d0 ¼ 1:978022  1001 ;

d1 ¼ 3:597619  10þ00 ;

d2 ¼ 7:320508  1001 :

Here, d2 is the positive solution of Eq. (3.1). We notice that the inequality (3.3)

d0 6 minfd1 ; d2 g is also true. Furthermore, we have d ¼ d2 since d ¼ minfd1 ; d2 g. The results for the sequence ftn g (given by (3.4)) are presented in Table 1. In Table 1, we observe that  The sequence ft n g is decreasing.  The sequence ft n g is bounded from above by

tHH ¼

g 1d

¼ 5:732051  1001 :

 The estimate (3.7) holds.

Example 4.5. Let X ¼ Y ¼ R and define function F on D by

FðxÞ ¼ x3  a:

ð4:1Þ

Table 1 The scalar sequence ft n g is given by Eq. (3.4). n 1 0 1 2 3 4 5 6 7 8

tn

t nþ2  t nþ1 +00

0.000000  10 2.000000  1001 3.000000  1001 3.197802  1001 3.214066  1001 3.214307  1001 3.214307  1001 3.214307  1001 3.214307  1001 3.214307  1001

tnþ1  t n 01

1.000000  10 1.978022  1002 1.626385  1003 2.407580  1005 2.749435  1008 4.585442  1013 8.723549  1021 2.767806  1033 1.670668  1053 3.199535  1086

dnþ1 g 01

2.000000  10 1.000000  1001 1.978022  1002 1.626385  1003 2.407580  1005 2.749435  1008 4.585442  1013 8.723549  1021 2.767806  1033 1.670668  1053

1.000000  1001 7.320508  1002 5.358984  1002 3.923048  1002 2.871871  1002 2.102355  1002 1.539031  1002 1.126649  1002 8.247642  1003 6.037693  1003

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I.K. Argyros et al. / Applied Mathematics and Computation 242 (2014) 945–951 Table 2 The scalar sequence ftn g is given by Eq. (3.4). n

kxn  xn1 k

kxn  xH k

tn  tn1

tH  t n

1 2 3 4 5 6 7 8 9

5.00  1001 2.86  1001 6.09  1002 2.05  1002 2.06  1003 4.73  1005 1.20  1007 7.18  1012 1.09  1018

2.06  1001 7.94  1002 1.85  1002 2.01  1003 4.75  1005 1.20  1007 7.18  1012 1.09  1018 9.84  1030

5.00000  1001 3.00000  1001 1.77778  1001 7.64444  1002 1.97499  1002 2.02994  1003 4.77958  1005 1.07478  1007 5.57232  1012

5.76050  1001 2.76050  1001 9.82722  1002 2.18277  1002 2.07784  1003 4.79032  1005 1.07484  1007 5.57232  1012 6.48231  1019

F is Fréchet differentiable. The divided difference can be written as

½x; y j F ¼

Z

1

F 0 ðy þ tðx  yÞÞdt ¼ ðy  xÞ2 þ 3 yðy  xÞ þ 3 y2 :

0

Let a ¼ 0:5. Through simple calculations – for x0 ¼ 1 and x1 ¼ 1=2 – we have



1 ; 2

l0 ¼

1 ; 4



1 2

and g ¼

3 : 10

Then the inequality (3.1) holds. To verify the inequalities (3.18) and (3.19) of Theorem 3.3, we apply the Broyden’s method (1.2) for solving Eq. (4.1) and use the majorizing sequence (3.4). Table 2 reports our numerical work. In the Table 2, we observe that the inequalities (3.18) and (3.19) – of Theorem 3.3 – hold. It is worth noticing that the sufficient convergence condition for the Secant method

ynþ1 ¼ yn  ½yn ; yn1 j F 1 Fðyn Þ is given by [11,12]

lc þ 2

qffiffiffiffiffi lg 6 1:

However, the preceding estimate is violated, since lc þ 2 method converges.

pffiffiffiffiffi lg ¼ 1:024596669 > 1. Hence, there is no guarantee that Secant

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