Convergence Conditions for the Secant Method

CUBO A Mathematical Journal Vol.12, N ō 01, 161 174. March 2010 Convergence Conditions for the Secant Method Ioannis K. Argyros Department of Mathematics Sciences, Lawton, OK 73505, USA email: iargyros@cameron.edu Saïd Hilout Poitiers university, Laboratoire de Mathématiques et Applications, 86962 Futuroscope Chasseneuil Cedex, France email: said.hilout@math.univ poitiers.fr ABSTRACT We provide new sufficient convergence conditions for the convergence of the Secant method to a locally unique solution of a nonlinear equation in a Banach space. Our new idea uses recurrent functions, Lipschitz type center Lipschitz type instead of just Lipschitz type conditions on the divided difference of the operator involved. It turns out that this way our error bounds are more precise than earlier ones under our convergence hypotheses we can cover cases where earlier conditions are violated. Numerical examples are also provided in this study. RESUMEN Son dadas nuevas condiciones suficientes para la convergencia del método de la secante para una solución localmente única de una ecuación no lineal en un espacio de Banach. Estas ideas nuevas usan funciones recurrentes, tipo-lipschitz y tipo centro-lipschitz sobre la diferencia dividida de los operadores envolvidos. Resulta que esta manera las cotas de errores son mas

162 Ioannis K. Argyros Saïd Hilout CUBO 12, 1 2010 precisas que las anteriores y bajo nuestras hipótesis de convergencia nosotros podemos cubrir casos donde las condiciones previas eran violadas. Ejemplos numéricos son dados en este estudio. Key words phrases: Secant method, Banach space, majorizing sequence, divided difference, Fréchet derivative. Math. Subj. Class.: 65H10, 65B05, 65G99, 65N30, 47H17, 49M15. 1 Introduction In this study we are concerned with the problem of approximating a locally unique solution x of equation Fx = 0, 1.1 where F is a Fréchet differentiable operator defined on a convex subset D of a Banach space X with values in a Banach space Y. A large number of problems in applied mathematics also in engineering are solved by finding the solutions of certain equations. For example, dynamic systems are mathematically modeled by difference or differential equations, 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 ẋ = Tx, for some suitable operator T, where x is the state. Then the equilibrium states are determined by solving equation 1.1. Similar equations are used in the case of discrete systems. The unknowns of engineering equations can be functions difference, differential, 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 h. Since all of these methods have the same recursive structure, they can be introduced discussed in a general framework. We consider the Secant method in the form x n+1 = x n δfx n 1, x n 1 Fx n n 0, x 1, x 0 D 1.2 where δfx, y LX, Y x, y D is a consistent approximation of the Fréchet derivative of F [5], [13]. Bosarge Falb [7], Dennis [9], Potra [16], Argyros [1] [5], Gutiérrez [10] others [11], [15], [18], have provided sufficient convergence conditions for the Secant method based on Lipschitz type conditions on δf.

CUBO 12, 1 2010 Convergence Conditions for the Secant Method 163 The sufficient convergence condition for the Secant method used in most references is given l c + 2 l η 1, 1.3 where, l, c, η are non negative parameters to be precised later. This hypothesis is easily violated. Indeed, let l = 1, η =.18, c =.185. Then, 1.3 does not holds, since l c + 2 l η = 1.033528137. Hence, there is not guarantee that an equation using the information l, c, η has a solution that can be found using Secant method 1.2. In this study we are motived by optimization considerations, the above observation. Here, using recurrent functions, Lipschitz type center Lipschitz type conditions, we provide a semilocal convergence analysis for 1.2. It turns out that our error bounds are more precise our convergence conditions hold in cases where the corresponding hypotheses mentioned in earlier references mentioned above are violated. Newton s method is also examined as a special case. Numerical examples are also provided in this study. 2 Semilocal Convergence Analysis of the Secant Method We need the following result on majorizing sequences for the Secant method 1.2. Lemma 2.1. Let l 0 > 0, l > 0, c > 0, η [0, c] be given parameters. Assume: Set δ 0 = Moreover, assume: there exists δ l 0 c + η < 1. 2.4 l c + η c 1 l 0 c + η. 2.5 [ max{l, δ 0 }, 1, such that estimate c 2 q 1/ 5 l q + l 0 1 q + p 1/ 5 q2 + q + l δ 0 2.6 is satisfied for q = δ c, p = 1 + 5. 2 Then, scalar sequence {t n } n 1 given by t 1 = 0, t 0 = c, t 1 = c + η, t n+2 = t n+1 + l t n+1 t n 1 t n+1 t n 1 l 0 t n+1 t 0 + t n 2.7

164 Ioannis K. Argyros Saïd Hilout CUBO 12, 1 2010 is non decreasing, bounded above by t = c q q un, 2.8 n=1 converges to some t [0, t ], where {u n } is the Fibonacci s sequence: u 1 = u 0 = 1, u n+1 = u n + u n 1, n 1. Moreover, the following a priori estimates hold: 0 t t n c q 1/ 5 pn n 0. 2.9 q 1 q pn p 1/ 5 Proof. We shall show using induction on k: l t k+1 t k 1 1 l 0 t k+1 t 0 + t k δ t k t k 1, 2.10 0 t k+1 t k t k t k 1, 2.11 t k+1 t k q u k 1 c. 2.12 Note that if 2.10 holds, then by 2.7, we have: t k+2 t k+1 = l t k+1 t k 1 t k+1 t k 1 l 0 t k+1 t 0 + t k Estimate 2.10 can also be written as: δ t k t k 1 t k+1 t k. 2.13 t k+1 t k α k t k t k 1, 2.14 where, α k = 1 l δ 1 l 0 t k+1 t 0 + t k l. 2.15 For k = 0, 2.14 becomes η 1 l δ 1 l 0 c + η l c, or δ δ 0, which is true by the choice of δ. Estimates 2.11 2.12 also hold for k = 0 by the initial conditions.

CUBO 12, 1 2010 Convergence Conditions for the Secant Method 165 Assume that 2.10 i.e., 2.13, 2.11, 2.12 hold for all k n. By the induction hypotheses we have 0 t k+2 t k+1 δ q u k 1 1 c t k+1 t k = q u k 1 t k+1 t k < t k+1 t k. 2.16 That is 2.11 holds for n = k + 1. On the other h: 0 t k+2 t k+1 q u k 1 t k+1 t k q u k 1 q u k c = q u k+1 c, 2.17 which shows 2.12 for n = k + 1. We also have the estimate 0 t k+2 t 0 t 1 t 0 + t 2 t 1 + + t k+2 t k+1 c q qu0 + q u1 + + q u k < t. 2.18 Clearly, we have: u k = 1 [ k+1 k+1 ] 1 + 5 1 5 5 2 2 So, for any k 0, m 0, we get in turn: 1 k 1 + 5 = pk, 2.19 5 2 5 0 t k+m t k t k t k+1 + t k+1 t k+2 + + t k+m t k+m 1 c q qu k + q u k+1 + + q u k+m 1 c q qpk / 5 + q pk+1 / 5 + + q pk+m 1 / 5. 2.20 Using Bernoulli s inequality, we obtain: t k+m t k c q qpk / 5 1 + q pk+1 p k / 5 + q pk+2 p k / 5 + q pk+m 1 p k / 5 = c q qpk / 5 1 + q pk p 1/ 5 + q pk p 2 1/ 5 + q pk p m 1 1/ 5 c / 5 q qpk 1 + q pk p 1/ 5 + q pk 1+2 p 1 1/ 5 + q pk p m 1 1/ 5 = c / 5 q qpk 1 + q pk p 1/ 5 + q pk p 1/ 5 2 + q pk p 1/ m 1 5 2.21 5 = c / 5 1 p 1 m/ qpk q qpk 1 q pk p 1/ 5. In particular, from 2.18 we have:

166 Ioannis K. Argyros Saïd Hilout CUBO 12, 1 2010 t k+m c q qu0 + + q u k+m 1 + c 2.22 In order for us to show 2.10, it suffices: δ t k t k 1 α k+1 2.23 or δ t k t k 1 1 l δ 1 l 0 t k+2 t 0 + t k+1 l or or or or or l δ q u k 1 1 + δ l 0 t k+2 + t k+1 t 0 δ l c l δ q uk 1 1 c + δ l 0 q qu0 + q u1 + + q u k+1 + c+ c q qu0 + q u1 + + q u k + c c δ l l δ q uk 1 1 c + l 0 2 q u0 + q u1 + + q u k + q u k+1 + δ c δ l l q u k 1 + l 0 2 q u0 + q u1 + + q u k + q u k+1 + q δ l 2.24 2 q 1/ 5 l q + l 0 1 q + p 1/ 5 q2 + q + l δ 0, which is true by the choice of δ q given by 2.6. The induction for 2.10 2.12 is now complete. It follows from 2.21 that scalar sequence {t n } is Cauchy in the complete space R, as such it converges to some t [0, t ]. By letting m 0 in 2.21, we obtain 2.9. That completes the proof of Lemma 2.1. We shall also provide another result where condition 2.6 is dropped from the hypotheses of Lemma 2.1: Lemma 2.2. Let l 0 > 0, l > 0, c > 0, η 0, c] be given parameters. Assume: l 0 c + η < 1;

CUBO 12, 1 2010 Convergence Conditions for the Secant Method 167 there exists δ max{l + 2 l 0, δ 0, δ 1 }, min{ 1 c, δ }, 2.25, v 1 δ 1, 2.26 where, δ 1 = q 1 c, is such that q 1 is the unique zero in 0, 1 of equation Qt = l 0 t 4 + l 0 t 2 + l t l = 0, δ = q c, is such that q is the unique positive zero of equation 2 t 1/ 5 f t = l 0 1 t + t + l δ = 0, p 1/ 5 v 1 is the unique positive zero of equation f 0 t = l 0 t 2 + 3 l 0 + l t + 2 l 0 + l δ = 0; l 0 t 2 u k + l 0 t u k + l t u k 2 l 0 k 1 for t δ 1. 2.27 Then, the conclusions of Lemma 2.1 hold true. Proof. by We follow the proof of Lemma 2.1 until estimate 2.23. Then, let us define functions f k, g k f k t = l t u k 1 + l 0 2 1 + t u1 + + t u k + t u k+1 + t + l δ, 2.28 g k t = g k t t u k 1, 2.29 where, g k t = l 0 t 2 u k + l 0 t u k + l t u k 2 l. 2.30

168 Ioannis K. Argyros Saïd Hilout CUBO 12, 1 2010 We need to find the relationship between two consecutive functions f k : f k+1 t = l t u k + l 0 2 t 0 + t u1 + + t u k+1 + t u k+2 + t + l δ = l t u k 1 l t u k 1 + l t u k + l 0 2 t 0 + t u1 + + t u k + t u k+1 + t u k+1 + t u k+2 + t + l δ 2.31 = f k t + l t u k t u k 1 + l 0 t u k+1 + t u k+2 = f k t + g k t. Using hypothesis 2 l 0 +l δ < 0, 2.28, we get: f 0 0 = 2 l 0 +l δ < 0, f k 0 = δ l < 0 k 1. Moreover for sufficiently large t, we also have f k t > 0 for all k 0. It then follows from the intermediate value theorem that: for each k 0, there exists v k 0, with f k v k = 0. Each v k is the unique positive zero of f k, since f k t > 0 k 0. That is the graph of function f k crosses the positive axis only once. By the definition of f k v k, there exists a polynomial p k 1 of degree k 1, with p k 1 s > 0 s > 0, such that: f k t = t v k p k 1 t. To show 2.14, we must have f k t 0. That is In view of 2.28 2.31, we get q = t v k k 0. 2.32 f k+1 v k = f k v k + g k v k = g k v k = q k v k v u k 1 k 0 for v k δ 1 by 2.27. Therefore, if v k δ 1, then v k+1 v k, lim k v k = q exists. Note that v 1 δ 1 by 2.23. Then, it follows v 2 v 1. Assume v m δ 1 m n, then v m+1 v m. We must also show v m+1 δ 1. But this is true, since v m+1 δ δ 1 by 2.23. Then, estimate 2.32 certainly holds if δ δ, which is true by the choice of δ. That completes the proof of Lemma 2.2. We shall study the Secant method 1.2 for triplets F, x 1, x 0 belonging to the class Cl, l 0, η, c, δ defined as follows: Definition 2.3. Let l, l 0, η, c, δ be non negative parameters satisfying the hypotheses of Lemma 2.2. We say that a triplet F, x 1, x 0 belongs to the class Cl, l 0, η, c, δ if:

CUBO 12, 1 2010 Convergence Conditions for the Secant Method 169 c 1 F is a nonlinear operator defined on a convex subset D of a Banach space X with values in a Banach space Y; c 2 x 1 x 0 are two points belonging to the interior D 0 of D satisfying the inequality x 0 x 1 c; c 3 F is Fréchet differentiable on D 0, there exists an operator δf : D 0 D 0 LX, Y such that: the linear operator A = δfx 1, x 0 is invertible, its inverse A 1 is bounded : for all x, y, z D. A 1 Fx 0 η; A [δfx, y F z] l x z + y z ; A [δfx, y F x 0 ] l 0 x x 0 + y x 0 c 4 the set D c = {x D; F is continuous at x} contains the closed ball Ux 0, t = {x X x x 0 t } where t is given in Lemma 2.1. We present the following semilocal convergence theorem for Secant method 1.2. Theorem 2.4. If F, x 1, x 0 Cl, l 0, η, c, δ, then sequence {x n } n 1 generated by Secant method 1.2 is well defined, remains in Ux 0, t for all n 0 converges to a unique solution x Ux 0, t of equation Fx = 0. Moreover the following estimates hold for all n 0 x n+2 x n+1 t n+2 t n+1, 2.33 x n x t t n, 2.34 where the sequence {t n } n 0 given by 2.7. Furthermore, if U x 0, 1 2 c + 1 D, l 0 t < 1 2 c + 1 l 0 2.35 the solution x is unique in Ux 0, t. Finally, if t < 1 l 0 R, Ux 0, R D, where t is given by 2.8, then the solution x is unique in Ux 0, R.

170 Ioannis K. Argyros Saïd Hilout CUBO 12, 1 2010 Proof. We first show operator L = δfx k, x k+1 is invertible for x k, x k+1 Ux 0, t. It follows from 2.7, 2.8, c 2 c 3 that: I A 1 L = A 1 L A A 1 L F x 0 + A 1 F x 0 A l 0 x k x 0 + x k+1 x 0 + x 0 x 1 l 0 t k t 0 + t k+1 t 0 + c l 0 t t 0 + t t 0 + c [ ] η l 0 2 1 δ + c c 1 2.36 since δ δ. According to the Banach Lemma on invertible operators [5], [13], 2.36, L is invertible 1 L 1 A 1 l 0 x k x 0 + x k+1 x 0 +c. 2.37 The second condition in c 3 implies the Lipschitz condition for F A 1 F u F v 2 l u v, u, v D 0. 2.38 By the identity, we get Fx Fy = 1 0 F y + tx y dt x y 2.39 A 1 0 [Fx Fy F ux y] l x u + y u x y 2.40 A 1 0 [Fx Fy δfu, v x y] l x v + y v + u v x y 2.41 for all x, y, u, v D 0. By a continuity argument 2.38 2.41 remain valid if x /or y belong to D c. We first show 2.33. If 2.33 holds for all n k if {x n } n 0 is well defined for n = 0, 1, 2,, k then x 0 x n t n t 0 < t t 0, n k. 2.42 That is 1.2 is well defined for n = k + 1. For n = 1, n = 0, 2.33 reduces to x 1 x 0 c, x 0 x 1 η. Suppose 2.33 holds for n = 1, 0, 1,, k k 0. Using 2.37, 2.41 Fx k+1 = Fx k+1 Fx k δfx k 1, x k x k+1 x k 2.43

CUBO 12, 1 2010 Convergence Conditions for the Secant Method 171 we obtain in turn x k+2 x k+1 = δfx k, x k+1 1 Fx k+1 δfx k, x k+1 1 A A 1 Fx k+1 l x k+1 x k + x k x k 1 1 l 0 [ x k+1 x 0 + x k x 0 +c] x k+1 x k l t k+1 t k + t k t k 1 1 l 0 [t k+1 t 0 + t k t 0 + t 0 t 1 ] t k+1 t k = t k+2 t k+1. 2.44 The induction for 2.33 is now complete. It follows from 2.33 Lemma 2.2 that sequence {x n } n 1 is Cauchy in a Banach space X, as such it converges to some x Ux 0, t since Ux 0, t is a closed set. By letting k in 2.44, we obtain Fx = 0. Estimate 2.34 follows from 2.33 by using stard majoration techniques [1], [5], [13]. We shall first show uniqueness in Ux 0, t. Let y Ux 0, t be a solution of equation 1.1. Set It then by c 3 : M = 1 0 F y + t y x dt. A 1 A M l 0 y x 0 + x x 0 + x 0 x 1 l 0 2 t t 0 + t 0 2.45 since δ δ. l 0 2 t c < 1, It follows from 2.35, the Banach lemma on invertible operators that M 1 exists on Ux 0, t. Using the identity: we deduce x = y. Fx Fy = M x y, 2.46 Finally, we shall show uniqueness in Ux 0, R. As in 2.45, we arrive at Hence, again we conclude x = y. That completes the proof of Theorem 2.4. A 1 A M < l 0 t + R 1.

172 Ioannis K. Argyros Saïd Hilout CUBO 12, 1 2010 Remark 2.5. Returning back to the example given in the introduction, say l 0 =.1, we obtain δ 0 = 2.047714554, 1 = 5.405. Condition 2.4 is true, since c Choose δ = 2.5 l 0 c + η =.0365 < 1. δ, 1 to obtain q =.4625. Then condition 2.6 becomes c.423156212 < q. That is our results can apply, whereas the ones using 1.3 cannot. Remark 2.6. Let us define the majoring sequence {w n } used in [4], [5] under condition 1.3: w 1 = 0, w 0 = c, w 1 = c + η, w n+2 = w n+1 + l w n+1 w n 1 w n+1 t n. 2.47 1 l w n+1 w 0 + w n Note that in general l 0 l 2.48 holds, l l 0 can be arbitrarily large [3], [5]. In the case l 0 = l, then t n = w n n 1. Otherwise t n+1 t n w n+1 w n, 2.49 0 t t n w w n, w = lim n w n. 2.50 Note also that strict inequality holds in 2.49 for n 1, if l 0 < l. The proof of 2.49, 2.50 can be found in [5]. Note that the only difference in the proofs is that the conditions of Lemma 2.2 are used here, instead of the ones in [4]. However this makes no difference between in the proofs. We complete this study with an example example to show that l 0 < l. Example 2.7. Let X = Y = C[0, 1] be the space of real valued continuous functions defined on the interval [0, 1] with norm x = max 0 s 1 xs. Let θ [0, 1] be a given parameter. Consider the "Cubic" integral equation us = u 3 s + λus 1 0 qs, tutdt + ys θ. 2.51 Here the kernel qs, t is a continuous function of two variables defined on [0, 1] [0, 1]; the parameter λ is a real number called the "albedo" for scattering; ys is a given continuous function

CUBO 12, 1 2010 Convergence Conditions for the Secant Method 173 defined on [0, 1] xs is the unknown function sought in C[0, 1]. Equations of the form 2.51 arise of gasses [5], [8]. For simplicity, we choose u 0 s = ys = 1, qs, t = s, for all s + t s [0, 1], t [0, 1], with s + t 0. If we let D = Uu 0, 1 θ, define the operator F on D by Fxs = x 3 s + λxs 1 0 qs, txtdt + ys θ, 2.52 for all s [0, 1], then every zero of F satisfies equation 2.51. We define the divided difference δ Fx, y by We have the estimate δ Fx, y = 1 0 max 0 s 1 Therfore, if we set b = F u 0 1, then, we obtain: Note that l 0 < l for all θ [0, 1]. F y + t x y dt. s dt = ln 2. s + t η = b λ ln 2 + 1 θ, l = b λ ln 2 + 3 2 θ l 0 = 1 b 2 λ ln 2 + 3 3 θ. 2 Received: October, 2008. Revised: January, 2009. References [1] Argyros, I.K., The theory application of abstract polynomial equations, St.Lucie/CRC/ Lewis Publ. Mathematics series, 1998, Boca Raton, Florida, U.S.A. [2] Argyros, I.K., On the Newton-Kantorovich hypothesis for solving equations, J. Comput. Appl. Math., 169 2004, 315 332. [3] Argyros, I.K., A unifying local-semilocal convergence analysis applications for two-point Newton-like methods in Banach space, J. Math. Anal. Appl., 298 2004, 374 397. [4] Argyros, I.K., New sufficient convergence conditions for the Secant method, Chechoslovak Math. J., 55, 2005, 175 187. [5] Argyros, I.K., Convergence applications of Newton-type iterations, Springer-Verlag Publ., New-York, 2008. [6] Argyros, I.K. Hilout, S., Efficient methods for solving equations variational inequalities, Polimetrica Publ. Co., Milano, Italy, 2009.

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