THIRD-ORDER FAMILY OF METHODS IN BANACH SPACES

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1 THIRD-ORDER FAMILY OF METHODS IN BANACH SPACES Changbum Chun Department of Mathematics, Sungkyunkwan University Suwon , Republic of Korea; Tel.: , Fax: Pantelimon Stănică Naval Postgraduate School, Department of Applied Mathematics Monterey, CA 93943; Tel.: (83) , Fax: (83) Beny Neta Naval Postgraduate School, Department of Applied Mathematics Monterey, CA 93943; Tel.: (83) , Fax: (83) Abstract. Recently, Parida and Gupta [J. Comp. Appl. Math. 6 (7), ] used Rall s recurrence relations approach (from 96) to approximate roots of nonlinear equations, by developing several methods, the latest of which is free of second derivative and it is of third order. In this paper, we use an idea of Kou and Li [Appl. Math. Comp. 87 (7), 7-3] and modify the approach of Parida and Gupta, obtaining yet another third-order method to approximate a solution of a nonlinear equation in a Banach space. We give several applications to our method.. Introduction Newton s method and its variants are used to solve nonlinear operator equations F (x) = or systems of nonlinear equations. The convergence of these methods was established using Kantorovich theorem (see e.g. [,, 3]). The convergence of the sequences obtained by these methods in Banach spaces is derived from the convergence of majorizing sequences (see [4] and references therein). Rall [5] has suggested a different approach for the convergence of these methods, based on recurrence relations. Parida [6], and Parida and Gupta [7] used this idea for several third-order methods (see also the work of Candela and Marquina [8, 9], Ezquerro and Hernández [], and Gutiérrez and Hernández [, ]). Here we apply the idea to the third-order method free of second derivative proposed by Kou and Li [3]. They developed a family of methods for the solution of Date: January 4,. Mathematics Subject Classification. 4A5, 65D99. Key words and phrases. Newton s method; Iterative methods; Nonlinear equations; Order of convergence; Root-finding methods. Corresponding author: This research was supported by Basic Science Research Program through the National Research Foundation of Korea(NRF) funded by the Ministry of Education, Science and Technology(-7)

2 Report Documentation Page Form Approved OMB No Public reporting burden for the collection of information is estimated to average hour per response, including the time for reviewing instructions, searching existing data sources, gathering and maintaining the data needed, and completing and reviewing the collection of information. Send comments regarding this burden estimate or any other aspect of this collection of information, including suggestions for reducing this burden, to Washington Headquarters Services, Directorate for Information Operations and Reports, 5 Jefferson Davis Highway, Suite 4, Arlington VA -43. Respondents should be aware that notwithstanding any other provision of law, no person shall be subject to a penalty for failing to comply with a collection of information if it does not display a currently valid OMB control number.. REPORT DATE 4 JAN. REPORT TYPE 3. DATES COVERED -- to TITLE AND SUBTITLE Third-Order Family of Methods in Banach Spaces 5a. CONTRACT NUMBER 5b. GRANT NUMBER 5c. PROGRAM ELEMENT NUMBER 6. AUTHOR(S) 5d. PROJECT NUMBER 5e. TASK NUMBER 5f. WORK UNIT NUMBER 7. PERFORMING ORGANIZATION NAME(S) AND ADDRESS(ES) Naval Postgraduate School,Department of Applied Mathematics,Monterey,CA, PERFORMING ORGANIZATION REPORT NUMBER 9. SPONSORING/MONITORING AGENCY NAME(S) AND ADDRESS(ES). SPONSOR/MONITOR S ACRONYM(S). DISTRIBUTION/AVAILABILITY STATEMENT Approved for public release; distribution unlimited 3. SUPPLEMENTARY NOTES Computers & Mathematics with Applications 6 (), SPONSOR/MONITOR S REPORT NUMBER(S) 4. ABSTRACT Recently, Parida and Gupta [J. Comp. Appl. Math. 6 (7) ] used Rall s recurrence relations approach (from 96) to approximate roots of nonlinear equations, by developing several methods, the latest of which is free of second derivative and it is of third order. In this paper, we use an idea of Kou and Li [Appl. Math. Comp. 87 (7), 7-3] and modify the approach of Parida and Gupta, obtaining yet another third-order method to approximate a solution of a nonlinear equation in a Banach space. We give several applications to our method. 5. SUBJECT TERMS 6. SECURITY CLASSIFICATION OF: 7. LIMITATION OF ABSTRACT a. REPORT unclassified b. ABSTRACT unclassified c. THIS PAGE unclassified Same as Report (SAR) 8. NUMBER OF PAGES 4 9a. NAME OF RESPONSIBLE PERSON Standard Form 98 (Rev. 8-98) Prescribed by ANSI Std Z39-8

3 a nonlinear equation f(x) = as follows y n = x n θ f(x n) f (x n ), x n+ = x n f(y n) + (θ + θ )f(x n ) θ f. (x n ) (.) It turns out that this method is of third order when approximating a simple root.. Recurrence relations In this section, we discuss a third-order method for solving nonlinear operator equations F (x) =, (.) where F : Ω X Y is a nonlinear operator on an open convex subset Ω of a Banach space X with values in a Banach space Y. The third-order method [3] is defined as follows: y n = x n θf (x n ) F (x n ), x n+ = x n θ + θ θ F (x n ) F (x n ) θ F (x n ) F (y n ). (.) This family uses two evaluations of F and one evaluation of F. In [7] they discuss a third-order method requiring one evaluation of F and two evaluations of F. Several choices of θ were suggested in [4] and [5]. Let F be a twice Fréchet differentiable in Ω and BL(Y, X) be the set of bounded linear operators from Y into X. Let us assume that Γ = F (x ) BL(Y, X) exists at some x Ω and the following conditions hold: () F (x) F (y) k x y, x, y Ω, () F (x) M, x Ω, (3) Γ β, (4) Γ F (x ) η. Let us also denote a = k βη, α = θ + θ + θ θ, (.3) Now, we define the sequences γ = M βη. a = b =, d = α + γ, b =, a n a n+ =, aa n d n b n+ = a n+ βηc n, d n+ = θ + θ θ b n+ + θ a n+βη [ θ C n + M ] θ b n+, (.4)

4 where with C n = M K n + k θ b n K n + M θ b n, (.5) K n = θ + (θ ) + θ θ b n + M a nβb nη. (.6) Note that we can rewrite d n+ also in the form or, equivalently, as d n+ = αb n+ + γa n+ b n+, (.7) d n+ = d b n+ + γb n+ (a n+ b n+ ). The polynomials C n and K n can be rewritten as C n = ( P + P a n b n + P a nb n) b n, K n = (Q + Q a n b n ) b n. Lemma. Under the previous assumptions, we prove the following: (I n ) Γ n = F (x n ) a n β, (II n ) Γ n F (x n ) b n η, (III n ) x n+ x n d n η, (IV n ) x n+ y n (d n + K n + θb n )η. Proof. We use induction to prove the above claims. Notice that (I ) and (II ) follow immediately from the assumptions. To prove (III ), we start with the first substep of (.), F (y ) = F (y ) θf (x ) F (x )(y x ). (.8) This can be written as F (y ) = ( θ)f (x ) + F (y ) F (x ) F (x )(y x ) = ( θ)f (x ) + F (x + t(y x ))( t)dt(y x ). Now multiply by Γ and use the assumptions, it follows that so that (.9) Γ F (y ) θ η + M βθ η, (.) x x θ + θ θ Γ F (x ) + θ Γ F (y ) ( θ + θ + θ + M ) βη η = d η, (.) θ and (III ) holds. We have (θ + )(θ ) x y = θ Γ F (x ) θ Γ F (y ), (.) so that it follows from (.) that x y θ + (θ ) θ Γ F (x ) + θ Γ F (y ) ( θ + (θ ) + θ + M ) βη η = d η θ (d + K + θb )η, (.3) 3

5 and (IV ) also holds. Following an inductive procedure and assuming x n Ω and aa n d n <, if x n+ Ω, we have I Γ n F (x n+ ) Γ n F (x n ) F (x n+ ) aa n d n <. (.4) Now, we note that Then Γ n+ is defined and Γ n+ Γ n+ [I (F (x n ) F (x n+ )) Γ n ] = Γ n. (.5) Γ n Γ n F (x n ) F (x n+ ) a n β = a n+ β. (.6) aa n d n Hence, by induction (.6) holds for all n. This proves condition (I n ). Using the first step of (.), we have F (y n ) = F (y n ) θf (x n ) F (x n )(y n x n ) = ( θ)f (x n ) + F (y n ) F (x n ) F (x n )(y n x n ) = ( θ)f (x n ) + F (x n + t(y n x n ))( t)dt(y n x n ). Now subtract the first step of (.) from the second to get (.7) F (x n )(x n+ y n ) = θ3 θ θ + θ F (x n ) θ F (y n). (.8) Using (.7) on the identity and F (x n+ ) = F (x n )(x n+ y n ) + F (y n ) + [F (y n ) F (x n )](x n+ y n ) +F (x n+ ) F (y n ) F (y n )(x n+ y n ) (.9) F (x n+ ) F (y n ) F (y n )(x n+ y n ) = we have F (x n+ ) = θ θ F (x n ) θ F (y n) + + F (y n +t(x n+ y n ))( t)dt(x n+ y n ), F (x n + t(y n x n ))( t)dt(y n x n ) F (y n + t(x n+ y n ))( t)dt(x n+ y n ) (.) (.) or F (x n+ ) = θ +[F (y n ) F (x n )](x n+ y n ), + θ F (x n + t(y n x n ))( t)dt(y n x n ) F (y n + t(x n+ y n ))( t)dt(x n+ y n ) +[F (y n ) F (x n )](x n+ y n ). 4 (.)

6 Hence, From (.7) we get F (x n+ ) M θ θ y n x n + M x n+ y n +k y n x n x n+ y n. (.3) Γ n F (y n ) θ b n η + M a nβ y n x n, (.4) so that since y n x n = θγ n F (x n ) θ b n η, and combining (.4) with (.8), we have x n+ y n θ3 θ θ + + θ θ b n η + M θ a nβ y n x n Hence, where C n is given by (.5). Therefore θ + (θ ) + θ θ b n η + M a nβb nη = K n η. (.5) F (x n+ ) C n η, (.6) Γ n+ F (x n+ ) Γ n+ F (x n+ ) a n+ βc n η = b n+ η, (.7) So, by induction (.7) holds for all n. This proves condition (II n ). Using (.7) with x n and y n replaced by x n+ and y n+ respectively, so that F (y n+ ) θ C n η + M θ b n+η, (.8) Γ n+ F (y n+ ) Γ n+ F (y n+ ) a n+ βη [ θ C n + M θ b n+ Using (.7) and (.9), we therefore have x n+ x n+ θ + θ θ Γ n+ F (x n+ + θ Γ n+f (y n+ ) θ + θ θ b n+ η + θ a n+βη ( θ + θ = b n+ + θ a n+βη θ [ θ C n + M θ b n+ ]. (.9) [ θ C n + M ]) θ b n+ η = d n+ η, (.3) Hence, by induction, this inequality holds for all n. This proves condition (III n ). Since (θ + )(θ ) x n+ y n+ = y n x n+ + θ Γ n F (x n ) (.3) we have from (.4), (.5),(.7) that θ Γ nf (y n ) + θγ n+ F (x n+ ), (.3) x n+ y n+ (K n + θb n+ )η. (.33) 5 ]

7 Hence we have x n+ y n+ x n+ x n+ + x n+ y n+ (.34) d n+ η + (K n + θb n+ )η (.35) = (d n+ + K n + θb n+ )η. (.36) Hence, by induction, condition (IV n ) holds for all n. This proves condition (IV n ). 3. Convergence Analysis In this section, we shall establish the convergence of our third-order method (.). To this end, we have to prove the convergence of the sequence x n defined in a Banach space or, which is the same, to prove that d n is a Cauchy sequence and that the following assumptions hold: () x n Ω, () aa n d n <, n N. The next two lemmas will show the Cauchy property for the sequence d n. Lemma. Assume that x is chosen so as to satisfy < d < /a, that is, a(α + γ) <, where α and γ are given by (.3). Then, the sequence a n > is increasing, as n increases. Proof. We show now that all the involved sequences are positive. Under the imposed conditions, we see that a, b, d, C, K are all positive, and also that aa d >. Assume, now, that all a i, b i, d i, C i, K i, and aa i d i are positive, for i =,,..., n. Since C n > and b n+ = a n+ βηc n, it follows that a n+, b n+ have the same sign, and so a n+ b n+ >. Further, from d n+ = b n+ (α + γa n+ b n+ ), we get that d n+ has the same sign as b n+, and so, all three terms a n+, b n+, d n+ share the same sign. By absurd, we suppose that the implied sign is negative. Then d n + d n+ < d n, and so, aa n (d n + d n+ ) > aa n d n, which renders aa n+ d n+ = aa n(d n + d n+ ) >, aa n d n which implies aa n+ d n+ <, but that is impossible since a n+, d n+ have the same sign and a >. a n Next, since a n+ =, then aa n d n d n = ( ), a a n a n+ and so, by telescoping, we get This will render n d i = a i= ( ), where a =. a a n a n =. n a d i 6 i=

8 n Certainly, since a >, d i >, for all i, then a d i increases as n increases, n and so, a d i decreases as n increases, which implies that the reciprocal, i= namely, a n is an increasing sequence, and consequently, a n a =. We define the sequence c n = a n b n. Then the sequences {a n }, {b n }, {c n }, {d n } can be rewritten as a n a n a n+ = = aa n d n a (αc n + γc n), b n+ = a n+ βηc n = βηb ( ) nc n P + P c n + P c n a (αc n + γc, ( n) c n+ = a n+ b n+ = βηc n P + P c n + P cn) [ a (αc n + γc n)], d n+ = αb n+ + γa n+ b n+ = βηb ( ) nc n P + P c n + P c n a (αc n + γc (α + γc n+ ). n) That the sequence {c n } is a decreasing sequence under the assumption that a b < can be proved by using the mathematical induction. It is obvious that c = a b < = c. Assuming that c n < c n for some n >, we have ( c n+ = βηc n P + P c n + P cn) [ a ((αc n + γc n)] < βηc n P + P c n + P cn ) [ ( a αcn + γcn )] = c n. Therefore the sequence {c n } becomes a decreasing sequence with c n < for all n. If < s < and c n sc n, then ( c n+ = βηc n P + P c n + P cn) [ a (αc ( n + γc n)] s βηc n P + P sc n + P s cn ) [ ( a αscn + γs cn )] ( s βηc n P + P c n + P cn ) [ ( a αcn + γcn )] = s c n. i= Let ζ = c c = c = a b, then we have < ζ < and c ζc = ζ, so that c ζc, c ζ c, c 3 ζ c, c 4 ζ 3 c 3,. c n+ ζ (n + n + + +) c = ζ n+ ζ,. 7

9 On the other hand, with the sequence {d n } under the assumption that a b < we have d n = aandn aa n = ( ) αc n + γc n a n ( αc n + γcn) a = αc n + γc n (α + γ)c n (α + γ)ζ n ζ since {a n } is an increasing sequence, and a. We have thus proved the following estimates. Lemma 3. We assume that a b <. Then the sequence {c n } is a decreasing sequence and for all n N we have the following estimates where < ζ = a b <. c n+ ζ n+ ζ, d n (α + γ)ζ n ζ Lemma 4. The sequence d n > is a convergent sequence and its limit is. Proof. Since a n is increasing, then /a n is a decreasing sequence and further /a n. Therefore, a n is convergent (it is monotonic in a compact set) to a limit L. Since d n = ( ), then d n is convergent to the limit a a n a n+ a (L L) =. Remark. A similar approach would work for some of the lemmas in the paper [7], as well. Some of their results, like Lemmas 4 7 can be simplified using a similar approach: for instance, in Lemma 7 of [7], it is claimed that d i <, but that n ( is immediate, since d i = lim d i = lim ) = ( L), where L n n a a i= i= n a would be the (finite) limit of /a n. i= Now, we state the semilocal convergence of the method defined by (.). Theorem 5. Let X, Y be Banach spaces and F be a twice Fréchet differentiable in an open convex domain Ω of a Banach space X and BL(Y, X) be the set of bounded linear operators from Y into X. Let us assume that Γ = F (x ) BL(Y, X) exists at some x Ω and the following conditions hold: () F (x) F (y) k x y, x, y Ω () F (x) M, x Ω, (3) Γ β, (4) Γ F (x ) η. Let us denote a = k βη. Suppose that x is chosen so as to satisfy a(α + γ) < and a b <, where α and γ are given by (.3). Then, if B(x, rη) Ω, where r = n= d n, then the sequence {x n } defined by (.) and starting at x converges to a solution x of the equation F (x) =. In this case, the solution x and the iterates x n belong to B(x, rη), and x is the only solution of F (x) = in B(x, /(k β) rη) Ω. 8

10 Furthermore, the error bound on x depends on the sequence {d n } given by x n+ x (α + γ)η d k η ζ k, ζ = a b. (3.) ζ k=n+ k=n+ Proof. It is easy to see that the sequence {x n } is convergent. Hence, there exists a limit x such that lim n x n = x. The sequence {a n } is bounded above since a n =. n a d i a d i i= Since lim n d n =, by (.7), we have lim n b n =. This indicates that lim n C n =. Thus, by (.6) and by the continuity of F, we proved that Also, F (x ) =. x n+ x x n+ x n + x n x n + + x x n d k η rη, (3.) k= where r = n= d n. We conclude that x n lies in B(x, rη) and taking limit as n we have x B(x, rη). To show the uniqueness of the solution, let y B(x, /(k β) rη) be another solution of F (x) =. Then = F (y ) F (x ) = i= F (x + t(y x ))dt(y x ). (3.3) To show that y = x, we have to show that the operator F (x + t(y x ))dt is invertible. Now, for Γ k β k β k β F (x + t(y x )) F (x ) dt x + t(y x ) x dt (( t) x x + t y x )dt ( rη + ) k β rη =, (3.4) it follows from Banach s Theorem [] that the operator F (x + t(y x ))dt has an inverse. Therefore, y = x. For every m n +, we have x m x n+ x m x m + x m x m + + x n+ x n+ m k=n+ d k η rη. (3.5) 9

11 By taking m, we get and from Lemma 3 x n+ x x n+ x k=n+ d k η k=n+ (α + γ)η ζ d k η < rη, (3.6) k=n+ ζ k, < ζ <, (3.7) which shows that {x n } converges and proves (3.). This completes the proof. 4. Examples In this section, we give some examples to illustrate the previous convergence result. Example 4.. [7] Let X = C[, ] be the space of continuous functions on [, ] and consider the integral equation F (x) =, where s F (x)(s) = + x(s) + λx(s) x(t)dt, (4.) s + t where s [, ], x C[, ] and < λ. The norm is taken as sup-norm. We easily find and Since F (x)u(s) = u(s) + λu(s) F (x)(uv)(s) = λu(s) s s x(t)dt + λx(s) u(t)dt, u Ω, (4.) s + t s + t s s v(t)dt + λv(s) u(t)dt, u, v Ω. (4.3) s + t s + t [F (x) F (y)](u) λ ln u x y, (4.4) we get k = λ ln. Since F s (x) λ max s s + t dt = λ ln, (4.5) we get M = λ ln. We also have from [7] that β = λ ln x and η = x + λ ln x. λ ln x In our case, we have a = k βη = λ ln ( x + λ ln x ) ( λ ln x ). If x is chosen so that a(α + γ) <, where α = θ + θ + θ θ, γ = λ ln ( x + λ ln x ) ( λ ln x ), the sequence {x n } defined by (.) and starting at such an x, converges to a solution x of the equation F (x) =.

12 Example 4.. Consider the solution of the nonlinear equation F (x) = x 3 + x on [, 3]. We let θ =. Now the initial condition is x =.7, and it is easy to show that F (x ) = 3x + = 9.67, α =, β = F (x ) =.34663, η = F (x ) F (x ) = (.34663)( 3.387) = and F (x) = 6x 8 = M. Now F (x) F (y) = 3(x y)(x+y) and therefore k = 8. Therefore a = k βη = and γ = M βη = 9(.34663)( ) = Then the condition holds: a(α + γ) = <. As a result, the convergence of this nonlinear equation can be studied by Theorem 5. Remark. One can take a larger interval, i.e. [, A] for A > 3 and still satisfy the condition. Suppose, we let x = A > 3, then F (A) = 3A +, α =, β = F (x ) = 3A +, η = F (x ) F (x ) = A3 +A 3A + and F (x) = 6x 6A = M. Now F (x) F (y) = 3(x y)(x + y) and therefore k = 6A. Therefore a = k βη = 6A A3 +A (3A +) and γ = M βη = 3A A3 +A (3A +). Then the condition holds: ] a(α + γ) = 6A A3 +A + 3A A3 +A (3A +) 8 9 < as A. Clearly the number of iterations (say, n) required for convergence depends on how close A to x =. Experimenting with various values of x yields the following results: (3A +) [ x =.7 n = 3 x = 3. n = 3 x = 5. n = 4 x =. n = 4 Example 4.3. Let us consider the system of three nonlinear equations F (x, y, z) = where F : Ω R 3 where Ω = [a, b ] [a, b ] [a 3, b 3 ] is a domain of F containing a solution of this system, and F (x, y, z) = (x +y +z 4, x +y +z x y, x +y +z 4y z 4). (4.6) Then we have F (x, y, z) = x y z x y z x y 4 z, (4.7) and F (x, y, z) = ( y + z + x) y + z + y z z z + x x z x y + x x + y. (4.8) We use the Frobenius norm in R 3 : X = (x + y + z ) / for X = (x, y, z) R 3. ( 3 The corresponding norm on A R 3 R 3 ) 3 /. is A = i= j= a ij

13 The second derivative is a bilinear operator on R 3 given by f xx f xy f xz f yx f yy f yz f zx f zy f zz f xx f xy f xz F (x, y, z) = f yx f yy f yz f zx f zy f zz f 3xx f 3xy f 3xz f 3yx f 3yy f 3yz f 3zx f 3zy f 3zz =. For a bilinear operator B : R 3 R 3 R 3, given two vectors X, Y R 3, we have B y B(X, Y ) := (x, x, x 3 ) B y B 3 y 3 = b x + b x + b 3 x 3 b x + b x + b 3 x 3 b 3 x + b 3 x + b 33 x 3 b x + b x + b 3 x 3 b x + b x + b 3 x 3 b 3 x + b 3 x + b 33 x 3 b 3 x + b 3 x + b 3 3 x 3 b 3 x + b 3 x + b 3 3 x 3 b 3 3 x + b 3 3 x + b 33 3 x 3 where B = B B B 3 = b b b 3 b b b 3 b 3 b 3 b 33 b b b 3 b b b 3 b 3 b 3 b 33 b 3 b 3 b 3 3 b 3 b 3 b 3 3 b 3 3 b 3 3 b (4.) y y y 3 (4.9),

14 We consider the norm of a bilinear operator B on R 3 by ( ) B = sup b jk i u k, (4.) u = where u = (u, u, u 3 ). In our case, for any triple (x, y, z), Now, since we have i= j= M = F (x, y, z) = 3 sup { u + u + u 3 } / = 3. (4.) u = F (x, y, z) F (u, v, w) = k= x u y v z w x u y v z w x u y v z w, (4.3) F (x, y, z) F (u, v, w) = 3 [ (x u) + (y v) + (z w) ] / = 3 (x, y, z) (u, v, w), and therefore k = 3. We let θ =, and so, α =. If we choose x =., y =., z =., then β = F (x, y, z ) = , η = F (x, y, z ) F (x, y, z ) =.85. Therefore a = k βη = and γ = M βη = Then the left side of the condition holds a(α + γ) = <. As a result, the convergence of this system of equations can be studied by Theorem Acknowledgement The authors would like to thank the referees for their useful comments and constructive suggestions which substantially improved the quality of this paper. References [] L. V. Kantorovich, G. P. Akilov, Functional Analysis, Pergamon Press, Oxford, 98. [] J. M. Ortega, W. C. Rheinboldt, Iterative Solution of Nonlinear Equations in Several Variables, Academic Press, Reading, MA, 97. [3] W. C. Rheinboldt, Methods for Solving Systems of Nonlinear Equations, SIAM, Philadelphia, PA, 966. [4] Q. Wu, Y. Zhao, Third-order convergence theorem by using majorizing function for a modified Newton method in Banach space, Appl. Math. Comput. 75 (6), [5] L. B. Rall, Computational Solution of Nonlinear Operator Equations, Robert E. Krieger, New York, 979. [6] P. K. Parida, Study of Some Third Order Methods for Nonlinear Equations in Banach Spaces, Ph.D. Dissertation, Indian Institute of Technology, Department of Mathematics, Kharagpur, India, 7. [7] P. K. Parida, D. K. Gupta, Recurrence relations for a Newton-like method in Banach spaces, J. Comp. Appl. Math. 6 (7), [8] V. Candella, A. Marquina, Recurrence relations for rational cubic methods I: the Halley method, Computing 44 (99), [9] V. Candella, A. Marquina, Recurrence relations for rational cubic methods II: the Chebyshev method, Computing 45 (99),

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