©2022 Ada Academica https://adac.eeEur. J. Math. Anal. 2 (2022) 13doi: 10.28924/ada/ma.2.13 On the Semi-Local Convergence of a Third Order Scheme for Solving Nonlinear Equations Samundra Regmi1, Ioannis K. Argyros2,∗, Santhosh George3, Christopher Argyros4 1Learning Commons, University of North Texas at Dallas, Dallas, TX, USA samundra.regmi@untdallas.edu 2Department of Mathematical Sciences, Cameron University, Lawton, OK 73505, USA iargyros@cameron.edu 3Department of Mathematical and Computational Sciences,National Institute of Technology Karnataka, India-575 025 sgeorge@nitk.edu.in 4Department of Computing and Technology, Cameron University, Lawton, OK 73505, USA christopher.argyros@cameron.edu ∗Correspondence: iargyros@cameron.edu Abstract. The semi-local convergence analysis of a third order scheme for solving nonlinear equationin Banach space has not been given under Lipschitz continuity or other conditions. Our goal isto extend the applicability of the Cordero-Torregrosa scheme in the semi-local convergence underconditions on the first Fréchet derivative of the operator involved. Majorizing sequences are used forproving our results. Numerical experiments testing the convergence criteria are given in this study. 1. Introduction Cordero and Torregrosa in [10] considered the third order scheme, defined for n = 0, 1, 2, . . . , by yn = xn −F ′(xn)−1F (xn) xn+1 = xn − 3M−1n F (xn), (1.1) for solving the nonlinear equation F (x) = 0, (1.2) where Mn = 2F ′(3xn+yn4 )− F ′(xn+yn2 ) + 2F ′(xn+3yn4 ) . Here F : D ⊂ E −→ E1 is an operatoracting between Banach spaces E and E1 with D 6= ∅. In general a closed form solution for (1.2) isnot possible, so iterative schemes are used for approximating a solution x∗ of (1.2) (see [1–27]). Received: 14 Feb 2022. Key words and phrases. semi-local convergence; Cordero-Torregrosa scheme; iterative schemes; Banach space; con-vergence criterion. 1 https://adac.ee https://doi.org/10.28924/ada/ma.2.13 Eur. J. Math. Anal. 10.28924/ada/ma.2.13 2 The local convergence of the this scheme in the special case when E = E1 = R was shown to beof order three using Taylor expansion and assumptions on the fourth order derivative of F, whichis not on these schemes [10]. So, the assumptions on the fourth derivative reduce the applicabilityof these schemes [1–27].For example: Let E = E1 = R, D = [−0.5, 1.5]. Define λ on D by λ(t) = { t3 log t2 + t5 − t4 if t 6= 0 0 if t = 0. Then, we get f (1) = 0, and λ′′′(t) = 6 log t2 + 60t2 − 24t + 22. Obviously λ′′′(t) is not bounded on D. So, the convergence of scheme (1.1) is not guaranteed bythe previous analyses in [1–27].In this study we introduce a majorant sequence and use general continuity conditions to extendthe applicability of scheme (1.1). Our analysis includes error bounds and results on uniqueness of x∗ based on computable Lipschitz constants not given before in [1–27] and in other similar studiesusing Taylor series. Our idea is very general. So, it applies on other schemes too.The rest of the study is set up as follows: In Section 2 we present results on majorizing sequences.Sections 3,4 contain the semi-local and local convergence, respectively, where in Section 4 thenumerical experiments are presented. Concluding remarks are given in the last Section 5. 2. Majorizing Sequences Scalar sequences are developed that majorize scheme (1.1). Let K0 > 0,K > 0 and η > 0 begiven constants. Define sequences {tn},{sn} by t0 = 0, s0 = η tn+1 = sn + 2K(sn − tn)(tn+1 − tn) 9(1 −K0tn)(1 −pn) , sn+1 = tn+1 + K(tn+1 − tn + sn − tn)(tn+1 − tn) 2(1 −K0tn+1) , (2.1) where pn = 5K06 (sn + tn). Notice that tn+1 is given implicitly in the first substep of sequence (2.1).It we solve for tn+1, we get its explicit form tn+1 = 9sn(1 −K0tn)(1 −pn) − 2tnK(sn − tn) 9(1 −K0tn)(1 −pn) − 2K(sn − tn) . But for the convergence analysis in Theorem 3.1 we prefer tn+1 in its implicit form.Next, we present sufficient conditions for the convergence scheme (1.1). https://doi.org/10.28924/ada/ma.2.13 Eur. J. Math. Anal. 10.28924/ada/ma.2.13 3 LEMMA 2.1. Suppose that 5(tn + sn) < 6 K0 . (2.2) for all n = 0, 1, 2, . . . . Then, sequences {tn} is nondecreasing and bounded from above by T∗ = 35K0 and as such it converge to its unique least upper T ∈ [0,T∗]. Proof. It follows from the definition (2.1) of sequences {tn} and (2.2) that this sequence isnondecreasing and bounded from above by T∗, and as such it converges to T. �The next result shows the convergence of sequence {tn}, under stronger but easier to verifyconditions than (2.2). But first we need to introduce some functions and parameters. Definefunctions g1 and g2 on the interval (0, 1) by g1(t) = 4K(1 + t)t − 4K(1 + t) + 9K0t, and g2(t) = K(2 + t)(1 + t)t −K(2 + t)(1 + t) + 2K0t3.Then, we get g1(0) = −4K, g1(1) = 9K0, g2(0) = −2K and g2(1) = 2K0.Hence, functions g1 and g2 have roots in (0, 1). Denote the minimal such roots by α1 and α2, re-spectively. Set a = 2K(t1−t0) 9(1−K0t)(1−p0) , b = K(t1−t0+s0−t0)(t1−t0) 2η(1−K0t1) , c̄ = min{a,b}, c = max{a,b},α3 = min{α1,α2} and α = max{α1,α2}.Then, we can show the second result on majorizing sequences for method (1.2). LEMMA 2.2. Suppose 0 < c̄ ≤ c ≤ α3 ≤ α ≤ 1 − 10 3 K0η. (2.3) Then, sequence {tn} is nondecreasing, bounded from above by T = η1−α and as such it converges to its unique least upper bound t∗ ∈ [0,T ]. Proof. Items 0 ≤ 2K(tk+1 − tk) 9(1 −K0tk)(1 −pk) ≤ α, (2.4) 0 ≤ K(tk+1 − tk + sk − tk)(tk+1 − tk) 2(1 −K0tk+1) ≤ α(sk − tk), (2.5) 0 ≤ 1 1 −pk ≤ 2 (2.6) and tk ≤ sk ≤ tk+1 (2.7)are shown using induction on k. These estimates are true for k = 0 by (2.3). Suppose thesehold for all k smaller than n − 1. By induction hypotheses and (1.2), we have 0 ≤ sk − tk ≤ α(sk−1 − tk−1) ≤ . . . ≤ αkη, tk+1 − tk = (tk+1 − sk) + (sk − tk) ≤ (1 + α)(sk − tk) https://doi.org/10.28924/ada/ma.2.13 Eur. J. Math. Anal. 10.28924/ada/ma.2.13 4 and tk+1 ≤ (1 −αk+2)η 1 −α < T. Evidently, (2.4) holds if 4K(1 + α)αk−1η 9(1 −K01−α k+1 1−α η ≤ α, (2.8) where we used (2.6). Define recurrent polynomials f (1) k on the interval (0, 1) by f (1) n (t) = 4K(1 + t)t k−1η + 9K0(1 + t + . . . + t k−1)η − 9. (2.9) Then, estimate (2.8) holds if f (1) n (t) ≤ 0 at t = α1. (2.10) We need a relationship between two consecutive polynomials f (1) k : f (1) k+1 (t) = 4K(1 + t)tkη + 3K0(1 + t + . . . + t k)η − 9 + f (1) k (t) −4K(1 + t)tk−1 + 3K0(1 + t + . . . + tk−1)η + 9 = f (1) k (t) + g1(t)t k−1η. (2.11) In particular, one gets f (1) k+1 (α1) = f (1) k (α1) since by the definition of α1 and g1, g1(α1) = 0.Define function f (1)∞ (t) = lim k−→∞ f (1) k (t). (2.12) Then, (2.10) holds if f (1)∞ (t) ≤ 0 at t = α1. (2.13)But by (2.9) and (2.12) one gets f (1)∞ (t) = 9K0η 1 − t − 9, (2.14) so (2.13) holds if f (1)∞ (t) ≤ 0 at t = α1 which is true by (2.3).Similarly, (2.5) holds if K(2 + α)(1 + α)αkη 2(1 −K01−α k+2 1−α η) ≤ α. (2.15) Define polynomials f (2) k (t) on the interval (0, 1) by f (2) k (t) = K(2 + t)(1 + t)tk−1η + 2K0(1 + t + . . . + t k+1)η − 2. (2.16) Then, (2.15) holds if f (2) k (t) ≤ 0 at t = α2. (2.17) https://doi.org/10.28924/ada/ma.2.13 Eur. J. Math. Anal. 10.28924/ada/ma.2.13 5 We get f (2) k+1 (t) = K(2 + t)(1 + t)tkη + 2K0(1 + t + . . . + t k+2)η − 2 + f (2) k (t) −K(2 + t)(1 + t)tk−1η − 2K0(1 + t + . . . + tk+1)η + 2 = f (2) k (t) + g2(t)t k−1η, (2.18) and f (2) k+1 (α2) = f (2) k (α2). (2.19) Define function f (2)∞ (t) = lim k−→∞ f (2) k (t). (2.20) Then, (2.17) holds if f (2)∞ (t) ≤ 0 at t = α2. (2.21) By (2.16) and (2.20), we get f (2)∞ (t) = K0η 1 − t − 1, so (2.21) holds by (2.3). Moreover, estimate (2.6) certainly holds if 2pk = 5K03 (sk +tk) < 5K03 ( η1−α + η 1−α) = 10K0η 3(1−α) < 1, which is true by (2.3). Furthermore, estimate (2.7) holds by (2.4)-(2.6) and thedefinition of sequence {tk}. Hence the induction for estimates (2.4)-(2.7) is completed. It followsthat sequence {tk} is nondecreasing and bounded from above by T∗, and such it converges to T. �If one desires iterates to be given explicitly in (2.1), then define instead sequence {tn} as follows t0 = 0, s0 = η tn+1 = sn + 2K(1 + K0tn)(sn − tn)2 3(1 −K0tn)(1 −pn) (2.22) sn+1 = tn+1 + 2K(tn+1 − tn + sn − tn)(tn+1 − tn) 2(1 −K0tn+1) . Moreover, define recurrent polynomial on the interval [0, 1) by f (1) n (t) = 4K 3 tn−1η + 4KK0 3 tn−1(1 + t + . . . + tn)η2 + K0(1 + t + . . . + t n)η − 1. This time we have f (1) n+1(t) = f (1) n (t) + g (1) n (t)t n−1η, (2.23) where g (1) n (t) = 4KK0 3 tn+2η + 4KK0 3 tn+1η + 4 3 Kt − 4K 3 (1 −K0η). https://doi.org/10.28924/ada/ma.2.13 Eur. J. Math. Anal. 10.28924/ada/ma.2.13 6 We get g(1)n (0) = −4K3 (1 −K0η) < 0 for K0η < 1, and g(1)1 (1) = 4KK0η > 0. Denote by rn thesmallest solution of g(1)n (t),respectively. Notice that these solutions are increasing as n increases,since g(1)n (t) ≤ g(1)n−1(t). Hence, it follows by (2.23) that f (1) n+1(t) ≤ f (1) n (t) + g (1) 1 (t)t n−1η. In particular for α1 = r1, we get f (1) n+1(t) ≤ f (1) n (t) at t = α1. Hence, f (1) n (t) ≤ 0 holds if f (1) 1 (t) ≤ 0 at t = α1.But f (1) 1 (t) = 4K 3 η + 4 3 KK0η 2 + K0η − 1. Define b = 2K(s0−t0) 3 . Then, we arrive at the following convergence results for majorizing sequence(2.2). LEMMA 2.3. Suppose 5(tn + sn) < 6 K0 , where {tn} is the sequence defined by (2.22). Then, the conclusions of Lemma 2.2 hold for this sequence. LEMMA 2.4. Suppose 0 < c̄ ≤ c ≤ α3 ≤ α ≤ 1 − 10K0 3 η (2.24) and ( 4K 3 + 4 3 K0Kη + K0 ) η ≤ 1. (2.25) Then, the conclusions of Lemma 2.2 hold for sequence {tn} given by (2.22). REMARK 2.5. The solutions α1 and α2 in Lemma 2.2 depend only on K0 and K. Similarly α2 in Lemma 2.4 depends on K0 and K1. But α1 depends K0,K and η. To avoid this dependence pick any γ ∈ (0, 1] and set γ = K0η. Define functions ḡ (1) n (t) on [0, 1) by ḡ (1) n (t) = 4Kγ 3 tn+1 + 4Kγ 3 tn=1 + 4 3 Kt − 4K 3 (1 −γ). Then, according to the proof of Lemma 2.2 we can set α1 = r̄1, where r̄1 is the smallest solution in (0, 1) of equation ḡ(1)1 (t) = 0 assured also to exist. Finally, notice that the first condition shows implicitly and the second explicitly the smallness of η. https://doi.org/10.28924/ada/ma.2.13 Eur. J. Math. Anal. 10.28924/ada/ma.2.13 7 3. Semi-Local Convergence The following sufficient convergence criteria (A) are used. Suppose: (A1) There exist x0 ∈ D and η > 0 such that F ′(x0)−1 exists and ‖F ′(x0)−1F (x0)‖≤ η. (A2) ‖F ′(x0)−1(F ′(w) −F ′(x0))‖≤ K0‖w −x0‖for all w ∈ D. Set D0 = D∩U(x0, 1K0 ).(A3) ‖F ′(x0)−1(F ′(w) −F ′(v)‖≤ K‖w −v‖for all v ∈ D0 and w = v −F ′(v)−1F (v). Denote by L the constant, if (A3) holds for all u,v ∈ D0, and by L1 the constant for all u,v ∈ D. It follows that K ≤ L ≤ L1. In practicewe shall use whichever of K or L is easier to compute (see also the numerical section).(A4) Hypotheses of Lemma 2.1 or Lemma 2.2 hold.and(A5) U[x0,t∗] ⊂ D (or U[x0,T ] ⊆ D).Next, the semi-local convergence of scheme (1.1) is developed based on conditions (A) and theaforementioned notation. THEOREM 3.1. Suppose conditions (A) hold. Then, the following items hold {xn}∈ U(x0,t∗) (3.1) and ‖x∗ −xn‖≤ t∗ − tn, (3.2) where x∗ = limn−→∞xn ∈ U[x0,t∗] and F (x∗) = 0. Proof. Mathematical induction is used to show ‖yk −xk‖≤ sk − tk (3.3) and ‖xk+1 −yk‖≤ tk+1 − sk. (3.4)It follows from (A1) and (1.1) that ‖y0 −x0‖ = ‖F ′(x0)−1F (x0) ≤ η =≤ s0 − t0 = η ≤ T, (3.5) so y0 ∈ U(x0,t∗) and (3.3) hold for k = 0. Let z ∈ U(x0,t∗). In view of (A2), one has ‖F ′(x0)−1(F ′(z) −F ′(x0)) ≤ K0‖z −x0‖≤ K0t∗ < 1, so F ′(z)−1 ∈ L(E1,E) and https://doi.org/10.28924/ada/ma.2.13 Eur. J. Math. Anal. 10.28924/ada/ma.2.13 8 ‖F ′(z)−1F ′(x0)‖≤ 1 1 −K0‖z −x0‖ . (3.6) by a result due to Banach [14] on linear invertible operators. Operator Mk can be shown to beinvertible. Indeed, by the definition of operator Mk, (2.2) and (A2) we obtain ‖(3F ′(x0))−1(Mk − 3F ′(x0))‖ ≤ 1 3 [2‖F ′(x0)−1 ( F ′ ( 3xk + yk 4 ) −F ′(x0))‖ +‖F ′(x0)−1 ( F ′ ( xk + yk 2 ) −F ′(x0) ) ‖ +2‖F ′(x0)−1 ( F ′ ( xk + 3yk 4 ) −F ′(x0) ) ≤ 1 3 (2K0‖ 3xk + yk 4 −x0‖ + K0‖ xk + yk 2 −x0‖ +2K0‖ xk + 3yk 4 −x0‖) ≤ 1 3 (2K0 3tk + sk 4 + K0 sk + tk 2 + 2K0 tk + 3sk 4 ) = 5K0 6 (tk + sk) = pk < 1, so Mk is invertible and ‖M−1 k F ′(x0)‖≤ 1 3(1 −pk) , (3.7) and xk+1 is well defined by the second substep of method (1.1). Then, we can write by method(1.1) that xk+1 = xk −F ′(xk)−1F (xk) + (F ′(xk)−1 − 3M−1k )F (xk) = yk − 1 3 F ′(xk) −1(Mk − 3F ′(xk))M−1k (xk+1 −xk). (3.8) We need the estimate, Mk − 3F ′(xk) = 2F ′ ( 3xk + yk 4 ) −F ′ ( xk + yk 2 ) +2F ′ ( xk + 3yk 4 ) − 3F ′(xk) = ( F ′ ( 3xk + yk 4 ) −F ′ ( xk + yk 2 )) + ( F ′ ( 3xk + yk 4 ) −F ′(xk) ) + 2 ( F ′ ( xk + 3yk 4 ) −F ′(xk) ) , https://doi.org/10.28924/ada/ma.2.13 Eur. J. Math. Anal. 10.28924/ada/ma.2.13 9 so by (A3) ‖F ′(x0)−1(Mk − 3F ′(xk))‖ ≤ K‖ 3xk + yk 4 − 2xk + 2yk 4 ‖ K‖ 3xk + yk 4 − 4xk 4 ‖ + 2K‖ xk + 3yk 4 − 4xk 4 ‖ = 2K‖yk −xk‖≤ 2K(sk − tk). (3.9) Using (1.1), (3.6) (for z = xk) and (3.7)-(3.9) ‖xk+1 −yk‖ ≤ 2K(sk − tk)(tk+1 − tk) 9(1 −K0tk)(1 −pk) = tk+1 − sk. (3.10) We also have ‖xk+1 −x0‖≤‖xk+1 −yk‖ + ‖yk −x0‖≤ tk+1 − sk + sk − t0 = tk+1 ≤ t∗, (3.11) so xk+1 ∈ U(x0,t∗). We can write by method (1.1) F (xk+1) = F (xk+1) −F (xk) − 1 3 Mk(xk+1 −xk) = ∫ 1 0 (F ′(xk + θ(xk+1 −xk))dθ− 1 3 Mk)(xk+1 −xk). (3.12) One can obtain the estimate∫ 1 0 (F ′(xk + θ(xk+1 −xk))dθ− 2 3 F ′ ( 3xk + yk 4 ) + 1 3 F ′ ( xk + yk 2 ) − 2 3 F ′ ( xk + 4yk 4 ) = ∫ 1 0 F ′(xk + θ(xk+1 −xk))dθ−F ′(xk)) + 2 3 (F ′(xk) −F ′ ( 3xk + yk 4 ) ) + 1 3 (F ′(xk) −F ′ ( xk + 3yk 4 ) + 1 3 (F ′ ( xk + yk 2 ) −F ′ ( xk + 3yk 4 ) ), (3.13) so ‖F ′(x0)−1 ∫ 1 0 (F ′(xk + θ(xk+1 −xk))dθ− 1 3 Mk)‖ ≤ K [ ‖xk+1 −xk‖ 2 + ‖yk −xk‖ 6 + ‖yk −xk‖ 4 + ‖yk −xk‖ 12 ] ≤ K( tk+1 − tk 2 + sk − tk 6 + sk − tk 4 + sk − tk 12 ) = K 2 (tk+1 − tk + sk − tk). (3.14) It follows from method (1.1), (3.6) (for z = xk+1), (3.11) and (2.10) that https://doi.org/10.28924/ada/ma.2.13 Eur. J. Math. Anal. 10.28924/ada/ma.2.13 10 ‖yk+1 −xk+1‖ ≤ ‖(F ′(xk+1)−1F ′(x0)F ′(x0)−1F (xk+1)‖ ≤ K(tk+1 − tk + sk − tk)(tk+1 − tk) 2(1 −K0tk+1) = sk+1 − tk+1, (3.15) showing (3.3). Moreover, we get ‖yk+1 −x0‖ ≤ ‖yk+1 −xk+1‖ + ‖xk+1 −x0‖ ≤ sk+1 − tk+1 + tk+1 − t0 = sk+1 ≤ t∗, (3.16) so yk+1 ∈ U(x0,t∗). The induction for (3.3) and (3.6) is completed. It follows from(3.3), (3.6), (3.10)and (3.16) that sequence {xn} is fundamental in Banach space E, and as such it converges to x∗ ∈ U[x0,t∗]. Using (3.9) and letting k −→ ∞ in ‖F ′(x0)−1F (xk+1)‖ ≤ K2 (tk+1 − tk + sk − tk),we obtain F (x∗) = 0. �Next, a uniqueness of the solution x∗ result is presented. PROPOSITION 3.2. Suppose: (1) The element x∗ ∈ U(x∗,s∗) is a simple solution of (1.2), and (A2) holds. (2) There exists δ ≥ s∗ so that K0(s ∗ + δ) < 2. (3.17) Set D1 = D∩U[x∗,δ]. Then, x∗ is the unique solution of equation (1.2) in the domain D1. Proof. Let q ∈ D1 with F (q) = 0. Define S = ∫10 F ′(q + θ(x∗ −q))dθ. Using (H2) and (3.17)one obtains ‖F ′(x0)−1(S −F ′(x0))‖ ≤ K0 ∫ 1 0 ((1 −θ)‖q −x0‖ + θ‖x∗ −x0‖)dθ ≤ K0 2 (s∗ + δ) < 1, so q = x∗, follows from the invertability of S and the identity S(q−x∗) = F (q)−F (x∗) = 0−0 = 0. � REMARK 3.3. (i) Point T given in closed form can repalce t∗ in Theorem 3.1. (ii) We used majorizing sequence {tn} given by (2.1) and Lemma 2.2 to prove Theorem 3.1. But we can also use majorizing sequence {tn} given by (2.22) and Lemma 2.3 to arrive at the conclusions of the Theorem 3.1. Simply notice that in the proof of this theorem we got using the second substep of https://doi.org/10.28924/ada/ma.2.13 Eur. J. Math. Anal. 10.28924/ada/ma.2.13 11 scheme (1.1), (3.8) and (3.9) estimate (3.10) leading to the definition of the first substep of sequence (2.1). But we can use the first substep of scheme (1.1) to write instead of (3.8) that xk+1 = yk −F ′(xk)−1(Mk − 3F ′(xk))M−1k F (xk)(yk −xk) leading to ‖xk+1 −yk‖≤ 2K(1 + K0tk)(sk − tk)2 3(1 −K0tk)(1 −pk) = tk+1 − sk, where, we also used ‖F ′(x0)−1F (xk)‖ = ‖F ′(x0)−1((F ′(xk) −F (x0)) + F ′(x0))‖ ≤ 1 + K0‖xk −x0‖≤ 1 + K0tk. Hence, we arrive at the second semi-local convergence rsult for scheme (1.1). THEOREM 3.4. Suppose:conditions (A) hold with (A4) replaced by (A4)’ Hypotheses of Lemma 2.3 or Lemma 2.4 hold. Then, the conclusions of Theorem 3.1 hold with (2.22) replacing (2.1). In practice we shall use the theorem providing the best results. 4. Numerical Experiments Lipschitz parameters are determinded and convegence criteria are tested for some numericalexperiments. EXAMPLE 4.1. Define scalar function ζ(t) = ξ0t + ξ1 + ξ2 sin ξ3t, x0 = 0, where ξj, j = 0, 1, 2, 3 are parameters. Then, clearly for ξ3 large and ξ2 small, K0L1 can be small (arbitrarily). In particular, notice that K L1 −→ 0. EXAMPLE 4.2. Let E = E1 = C[0, 1] and D = U[0, 1]. It is well known that the boundary value problem [12]. ς(0) = 0,ς(1) = 1, ς′′ = −ς −σς2 can be given as a Hammerstein-like nonlinear integral equation ς(s) = s + ∫ 1 0 Q(s,t)(ς3(t) + σς2(t))dt where σ is a parameter. Then, define F : D −→ E1 by [F (x)](s) = x(s) − s − ∫ 1 0 Q(s,t)(x3(t) + σx2(t))dt. https://doi.org/10.28924/ada/ma.2.13 Eur. J. Math. Anal. 10.28924/ada/ma.2.13 12 Choose ς0(s) = s and D = U(ς0,ρ0). Then, clearly U(ς0,ρ0) ⊂ U(0,ρ0 + 1), since ‖ς0‖ = 1. Suppose 2σ < 5. Then, conditions (A) are satisfied for K0 = 2σ + 3ρ0 + 6 8 , L = σ + 6ρ0 + 3 4 , and η = 1+σ 5−2σ. Notice that K0 < L. EXAMPLE 4.3. Let us consider a scalar function ψ defined on the set D = U[x0, 1 − q] for q ∈ (0, 1 2 ), by ψ(x) = x3 −q. Choose x0 = 1. Then, we obtain the estiamtes |ψ′(x0)−1(ψ′(x) −ψ′(x0))| = |x2 −x20 | ≤ |x + x0||x −x0| ≤ (|x −x0| + 2|x0|)|x −x0| = (1 −q + 2)|x −x0| = (3 −q)|x −x0|, for all x ∈ D, so K0 = 3 −q, D0 = U(x0, 1K0 ) ∩D = U(x0, 1 K0 ), |ψ′(x0)−1(ψ′(y) −ψ′(x)| = |y2 −x2| ≤ |y + x||y −x| ≤ (|y −x0 + x −x0 + 2x0)||y −x| = (|y −x0| + |x −x0| + 2|x0|)|y −x| ≤ ( 1 K0 + 1 K0 + 2)|y −x| = 2(1 + 1 K0 )|y −x|, for all x,y ∈ D0, so L = 2(1 + 1K0 ), |ψ′(x0)−1(ψ′(y) −ψ′(x)| = (|y −x0| + |x −x0| + 2|x0|)|y −x| ≤ (1 −q + 1 −q + 2)|y −x| = 2(2 −q)|y −x|, for all x,y ∈ D and L1 = 2(2 −q). Notice that for all q ∈ (0, 12) K0 < L < L1. Next, set y = x −ψ′(x)−1ψ(x), x ∈ D. Then, we have y + x = x −ψ′(x)−1ψ(x) + x = 5x3 + q 3x2 . Define fundtion ψ̄ on the interval D = [q, 2 −q] by ψ̄(x) = 5x3 + q 3x2 . https://doi.org/10.28924/ada/ma.2.13 Eur. J. Math. Anal. 10.28924/ada/ma.2.13 13 Then, we get by this definition that ψ̄′(x) = 15x4 − 6xq 9x4 = 5(x −q)(x2 + xq + q2) 3x3 , where p = 3 √ 2q 5 is the critical point of function ψ̄. Notice that q < p < 2 − q. It follows that this function is decreasing on the interval (q,p) and increasing on the interval (q, 2 − q), since x2 + xq + q2 > 0 and x3 > 0. So, we can set K1 = 5(2 −q)2 + q 9(2 −q)2 ,η = 1 −q 3 and K1 < K0. But if x ∈ D0 = [1 − 1K0 , 1 + 1 K0 ], then K = 5%3 + q 9%2 , where % = 4−q 3−q and K < K1 for all q ∈ (0, 1 2 ). Next, we verify conditions (2.2), (2.3), (2.24) and (2.25). Then for q = 0.95, 6 K0 = 2.9268 and n 1 2 3 4 5 tn 0.1683 0.1694 0.1694 0.1694 0.1694 α1 = 0.1643 = α3, α2 = 0.6588 = α,a = 0.0030 = c̄,b = 0.0136 = c, 1 − 10K0η 3 = 0.8861, and (4K 3 + 4 3 K0Kη + K0)η = 0.0521 < 1. Hence, conditions (2.2),(2.3), (2.24) and (2.25) hold. 5. Conclusion The semi-local convergence of scheme (1.1) with order three is extended using general conditionson F ′ and recurrent majorizing sequences. References [1] I.K. Argyros, On the Newton - Kantorovich hypothesis for solving equations, J. Comput. Math. 169 (2004) 315-332. https://doi.org/10.1016/j.cam.2004.01.029[2] I.K. Argyros, Computational theory of iterative schemes. Series: Studies in Computational Mathematics, 15, Editors:C.K.Chui and L. Wuytack, Elsevier Publ. Co. New York, U.S.A, 2007.[3] I.K. Argyros, Convergence and applications of Newton-type iterations, Springer Verlag, Berlin, Germany, (2008).[4] I.K. Argyros, S. Hilout, Weaker conditions for the convergence of Newton’s scheme, J. Complex. 28 (2012) 364–387. https://doi.org/10.1016/j.jco.2011.12.003.[5] I.K. Argyros, S. Hilout, On an improved convergence analysis of Newton’s scheme, Appl. Math. Comput. 225 (2013)372-386. https://doi.org/10.1016/j.amc.2013.09.049 https://doi.org/10.28924/ada/ma.2.13 https://doi.org/10.1016/j.cam.2004.01.029 https://doi.org/10.1016/j.jco.2011.12.003 https://doi.org/10.1016/j.amc.2013.09.049 Eur. J. Math. Anal. 10.28924/ada/ma.2.13 14 [6] I.K. Argyros, A.A. Magréñan, Iterative schemes and their dynamics with applications, CRC Press, New York, USA,2017.[7] I.K. Argyros, A.A. Magréñan, A contemporary study of iterative schemes, Elsevier (Academic Press), New York,2018.[8] R. Behl, P. Maroju, E. Martinez, S. Singh, A study of the local convergence of a fifth order iterative scheme, IndianJ. Pure Appl. Math. 51 (2020) 439-455. https://doi.org/10.1007/s13226-020-0409-5.[9] E. Cătinaş, The inexact, inexact perturbed, and quasi-Newton schemes are equivalent models, Math. Comput. 74(2005) 291–301. https://doi.org/10.1090/S0025-5718-04-01646-1.[10] A. Cordero, J.R. Torregrosa, Variants of Newton’s Method using fifth-order quadrature formulas, Appl. Math. Comput.190 (2007) 686–698. https://doi.org/10.1016/j.amc.2007.01.062.[11] J.A. Ezquerro, J.M. Gutiérrez, M.A. Hernández, N. Romero, M.J. Rubio, The Newton scheme: From Newton toKantorovich (Spanish), Gac. R. Soc. Mat. Esp. 13 (2010) 53-76.[12] J.A. Ezquerro, M.A. Hernandez, Newton’s scheme: An updated approach of Kantorovich’s theory, Cham. Switzerland,(2018).[13] M. Grau-Sánchez, À. Grau, M. Noguera, Ostrowski type methods for solving systems of nonlinear equations, Appl.Math. Comput. 218 (2011) 2377–2385. https://doi.org/10.1016/j.amc.2011.08.011.[14] L.V. Kantorovich, G.P. Akilov, Functional analysis, Pergamon Press, Oxford, (1982).[15] A.A. Magréñan, I.K. Argyros, J.J. Rainer, J.A. Sicilia, Ball convergence of a sixth-order Newton-like schemebased on means under weak conditions, J. Math. Chem. 56 (2018) 2117-2131. https://doi.org/10.1007/ s10910-018-0856-y.[16] A.A. Magréñan, J.M. Gutiérrez, Real dynamics for damped Newton’s scheme applied to cubic polynomials, J. Comput.Appl. Math. 275 (2015) 527–538. https://doi.org/10.1016/j.cam.2013.11.019.[17] L.M. Ortega, W.C. Rheinboldt, Iterative solution of nonlinear equations in several variables, Academic Press, NewYork, (1970).[18] A.M. Ostrowski, Solution of equations in Euclidean and Banach spaces, Elsevier, Amsterdam, 1973.[19] F.A. Potra, V. Pták, Nondiscrete induction and iterative processes, Research Notes in Mathematics, 103. Pitman(Advanced Publishing Program), Boston, MA. (1984).[20] P.D. Proinov, General local convergence theory for a class of iterative processes and its applications to Newton’sscheme, J. Complex. 25 (2009) 38-62. https://doi.org/10.1016/j.jco.2008.05.006[21] P.D. Proinov, New general convergence theory for iterative processes and its applications to Newton-Kantorovichtype theorems, J. Complex. 26 (2010) 3-42. https://doi.org/10.1016/j.jco.2009.05.001[22] W.C. Rheinboldt, An adaptive continuation process of solving systems of nonlinear equations, Banach Center Publ.3 (1978) 129-142.[23] S.M. Shakhno, O.P. Gnatyshyn, On an iterative algorithm of order 1.839. . . for solving the nonlinear least squaresproblems, Appl. Math. Comput. 161 (2005) 253–264. https://doi.org/10.1016/j.amc.2003.12.025.[24] S.M. Shakhno, R.P. Iakymchuk, H.P. Yarmola, Convergence analysis of a two step scheme for the nonlinear squaresproblem with decomposition of operator, J. Numer. Appl. Math. 128 (2018) 82-95.[25] J.R. Sharma, R.K. Guha, R. Sharma, An efficient fourth order weighted - Newton scheme for systems of nonlinearequations, Numer. Algorithms, 62 (2013) 307–323, https://doi.org/10.1007/s11075-012-9585-7.[26] J.F. Traub, Iterative schemes for the solution of equations, Prentice Hall, New Jersey, U.S.A. (1964).[27] R. Verma, New trends in fractional programming, Nova Science Publisher, New York, USA, (2019). https://doi.org/10.28924/ada/ma.2.13 https://doi.org/10.1007/s13226-020-0409-5 https://doi.org/10.1090/S0025-5718-04-01646-1 https://doi.org/10.1016/j.amc.2007.01.062 https://doi.org/10.1016/j.amc.2011.08.011 https://doi.org/10.1007/ s10910-018-0856-y https://doi.org/10.1007/ s10910-018-0856-y https://doi.org/10.1016/j.cam.2013.11.019 https://doi.org/10.1016/j.jco.2008.05.006 https://doi.org/10.1016/j.jco.2009.05.001 https://doi.org/10.1016/j.amc.2003.12.025 https://doi.org/10.1007/s11075-012-9585-7 1. Introduction 2. Majorizing Sequences 3. Semi-Local Convergence 4. Numerical Experiments 5. Conclusion References