Applied Mathematics and Computation 242 (2014) 945–951
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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
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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
l¼
1 ; 2
l0 ¼
1 ; 4
c¼
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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