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On the Ostrowski Method for Solving Equations

Ioannis K. Argyros1,∗, Santhosh George2, Christopher I. Argyros3
1Department of Mathematical Sciences, Cameron University, Lawton, OK 73505, USA

iargyros@cameron.edu
2 Department of Mathematical and Computational Sciences,National Institute of Technology Karnataka,

India-575 025
sgeorge@nitk.edu.in

3Department of Computing and Technology, Cameron University, Lawton, OK 73505, USA
christopher.argyros@cameron.edu

∗Correspondence: iargyros@cameron.edu

Abstract. In this paper, we revisited the Ostrowski’s method for solving Banach space valued equa-tions. We developed a technique to determine a subset of the original convergence domain and usingthis new Lipschitz constants derived. These constants are at least as tight as the earlier ones leadingto a finer convergence analysis in both the semi-local and the local convergence case. These tech-niques are very general, so they can be used to extend the applicability of other methods withoutadditional hypotheses. Numerical experiments complete this study.

1. Introduction
One of the most challenging tasks in computational mathematics is the problem of determininga solution x∗ of equation

F (x) = 0, (1.1)
where F : Ω ⊂ B −→ B1 is an operator acting between Banach spaces B and B1 with Ω 6= ∅. Theclosed form derivation of x∗ is possible only in rare cases. This leads practitioners and researchersin developing solution methods that are iterative.In this work, we consider Ostrowski’s method defined for x0 ∈ Ω and each n = 0, 1, 2, . . . by

yn = xn −F ′(xn)−1F (xn)

xk+1 = yn −A−1n F (yn), (1.2)
Received: 7 Nov 2021.
Key words and phrases. Ostrowski’s method; Banach space; convergence criterion.1

https://adac.ee
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Eur. J. Math. Anal. 10.28924/ada/ma.2.3 2
where An = 2[yn,xn; F ] −F ′(xn). The convergence order is four obtained under certain conditionson the initial data (Ω,F,F ′,x0) and Taylor expansion [17, 25]. So, the assumptions on the fourthderivative reduce the applicability of these schemes.For example: Let B = B1 = R, Ω = [−0.5, 1.5]. Define λ on Ω by

λ(t) =

{
t3 log t2 + t5 − t4 if t 6= 0

0 if t = 0.

Then, we get t∗ = 1, and
λ′′′(t) = 6 log t2 + 60t2 − 24t + 22.

Obviously λ′′′(t) is not bounded on Ω. So, the convergence of scheme (1.2) is not guaranteed bythe analyses in [17, 24].We study two types of convergence called local and semi-local. In the first one based on thesolution x∗ we find the radii of the convergence balls. But in the second one based on the starter
x0 we develop criteria that guarantee convergence of sequence {xn}. There is a plethora of thistypes of results [10, 15, 16, 22, 28, 38]. But what all these results have in common is that the regionof accessibility (or convergence region) is limited in general reducing the applicability of Newton’sand other methods [8,20,26,28,31]. Moreover, the error bounds on distances ‖xk+1−xk‖ or ‖xk−x∗‖are pessimistic. The same is true for the uniqueness ball of these methods. These problems becomemore difficult when studying methods of convergence order three or higher [8, 17, 19, 31–33]. Wehave developed different techniques to addres these problems.In technique 1, we determine a subset Ω of Ω also containing the iterates. But in this set Ω theLipschitz-like parameters (or functions) are at least as tight as the original ones, so the resultingconvergence is finer. This technique does not depend on the convergence order of the method. Butwe shall demonstrate it in case of fourth order methods. These methods require the evaluation ofthe second order Fréchet derivative of operator F. Notice that for a system (nonlinear) of i equationswith i unknowns, the first derivative is a matrix with i2 entries (values), whereas the second Fréchetderivative has i3 entries. That is why there is a need for avoiding F ′′.The rest of the paper is organized as follows: In Section 2 we develop the second technique basedon majorizing sequences. The local convergence analysis results appear in Section 3. Numericalexamples can be found in Section 4. The paper ends with some concluding remarks.

2. Semi-local Convergence
We base our semi-local convergence analysis on scalar parameters and functions. Let η ≥

0,K0 > 0,K > 0,K1 > 0,K2 > 0,K3 > 0,L0 > 0 with K0 ≤ K, L0 ≤ 2K1 and K4 = K2 + K3.Define polynomials g1 and g2 on the interval [0, 1) by
g1(t) = K1t

5 + (2K1 + K3)t
4 + K1t

3 + (
K

2
−K3)t2 −

K

2
(2.1)

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Eur. J. Math. Anal. 10.28924/ada/ma.2.3 3
and

g2(t) = L0t
4 + (L0 + K3)t

3 + K4t
2 −K3t −K4. (2.2)

We have g1(0) = −K2 < 0,g1(1) = 4K1 > 0,g2(0) = −K4 < 0 and g2(1) = 2L0 > 0. It thenfollows from the intermediate value theorem that polynomials g1 and g2 have at least one root in
(0, 1). Denote by δ1 and δ2 the least such roots, respectively. Moreover, it is convenient to definescalar sequences and parameters

t0 = 0, s0 = η, t1 = s0 +
K0
2

(s0 − t0)2

1 − 2K1s0
,

sn+1 = tn+1 +
(K3(tn+1 − sn) + K4(sn − tn))(tn+1 − sn)

1 −L0tn+1

tn+2 = sn+1 +
K(sn+1 − tn+1)2

2(1 − (K1(sn+1 + tn+1) + K3(sn+1 − tn+1))
, (2.3)

αn =
K(sn − tn)

2(1 − (K1(sn+1 + tn+1) + K3(sn+1 − tn+1))
,

γn =
K3(tn − sn) + K4(sn − tn)

1 −L0tn+1
,

for all n = 0, 1, 2, . . . ,
δn = max{αn,γn},

λ = min{δ1,δ2}

and
µ = max{δ1,δ2}.

Next, we present a convergence result for sequences {tn} and {sn}.
LEMMA 2.1. Suppose: there exists δ satisfying

0 ≤ δ0 ≤ λ ≤ δ ≤ µ < 1 −K1η. (2.4)
Then, sequences {tn},{sn} are well defined nondecreasing, bounded from above bt s∗∗ = η1−δ and
as such they converge to their unique least upper bound s∗ ∈ [η,s∗∗]. Moreover, the following error
estimates hold for all n = 1, 2, . . .

0 ≤ tn+1 − sn ≤ δ(sn − tn) ≤ δ2n+1η, (2.5)
0 ≤ sn − tn ≤ δ(tn − sn−1) ≤ δ2nη (2.6)

and

tn ≤ sn ≤ tn+1. (2.7)

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Eur. J. Math. Anal. 10.28924/ada/ma.2.3 4
Proof. Items (2.5)-(2.7) hold if

0 ≤ αm ≤ δ, (2.8)
0 ≤ γm ≤ δ (2.9)

and
tm ≤ sm ≤ tm+1 (2.10)

are true for all m = 0, 1, 2, . . . . Notice that by the definition of s0,t1 and (2.4), t1 ≥ 0. We alsohave (2.8) and (2.9) hold for m = 0. Suppose (2.8)-(2.10) hold for m = 1, 2, . . . ,n. Then, we canobtain in turn that
sm ≤ tm + δ2mη ≤ sm−1 + δ2m−1η + δ2mη

≤ η + . . . + δη + . . . + δ2mη

=
1 −δ2m+1

1 −δ
η ≤

η

1 −δ
= s∗∗,

and
tm+1 ≤ sm + δ2m+1η ≤ tm + δ2mη + δ2m+1η

≤ η + δη + . . . + δ2m+1η

=
1 −δ2m+2

1 −δ
η ≤

η

1 −δ
.

Hence, by (2.7) and the induction hypotheses, we deduce that sequences {tm} and {sm} arenondecreasing. Evidently, (2.8) holds if
K

2
δ2nη + δK1

(
1 −δ2n+3

1 −δ
η

)
+δK1

1 −δ2n+2

1 −δ
η + K3δ

2(n+1)η −δ ≤ 0. (2.11)
Estimate (2.11) motivates us to introduce recurrent functions h(1)n (t) on the interval [0, 1) by

h
(1)
n )t) =

K

2
t2n−1η + K1(1 + t + . . . + t

2n+2)η

+K1(1 + t + . . . + t
2n+1)η + K3t

2n+3η − 1. (2.12)

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Eur. J. Math. Anal. 10.28924/ada/ma.2.3 5
We need a relationship between two consecutive functions f (1)n (t). By this definition, we have inturn that

h
(1)
n+1(t) =

K

2
t2n+1η + K1(1 + t + . . . + t

2n+4)η

+K1(1 + t + . . . + t
2n+3)η + K3t

2n+3η − 1

−
K

2
t2n−1η −K1(1 + t + . . . + t2n+2)η

−K1(1 + t + . . . + t2n+1)η −K3t2n+1η + 1 + h
(1)
n (t)

= h
(1)
n (t) +

K

2
t2n+1η −

K

2
t2n−1η

+K1(t
2n+3 + t2n+4)η + K1(t

2n+2 + t2n+3)η + K3t
2n+3η −K3t2n+1η

= h
(1)
n (t) + g1(t)t

2n−1η. (2.13)
Notice that by the definition of δ1

h
(1)
n+1(δ1) = h

(1)
n (δ1).

By (2.11)-(2.13), estimate (2.11) shall be true if for 4K1 < K
h
(1)
n (δ1) ≤ 0. (2.14)

Let
h(1)∞ (t) = lim

n−→∞
h
(1)
n (t). (2.15)

But then
h(1)∞ (δ) =

2K1η

1 −δ
− 1. (2.16)

Hence, instead of (2.13) we can show
h(1)∞ (δ) ≤ 0, (2.17)

which is true by (2.4). If 4K1 ≥ K then f∞(t) ≥ fn(t), so again f∞(δ) ≤ 0 holds. Similarly, (2.9)holds if
K3δ

2n+1η + K4δ
2nη + δL0

1 −δ2n+2

1 −δ
η −δ ≤ 0 (2.18)

or
h
(2)
n (δ) ≤ 0, (2.19)

where
h
(2)
n (t) = K3t

2nη + K4t
2n−1η + L0(1 + t + . . . + t

2n+1)η − 1. (2.20)

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Eur. J. Math. Anal. 10.28924/ada/ma.2.3 6
This time we have

h
(2)
n+1(t) = K3t

2n+2η + K4t
2n+1η + L0(1 + t + . . . + t

2n+3)η

−1 −K3t2nη −K4t2n−1η −L0(1 + t + . . . + t2n+1)η + 1 + h
(2)
n (t)

= h
(2)
n (t) + K3t

2n+2η −K3t2nη + K4t2n+1 −K4t2n+1η

+L0(t
2n+2 + t2n+3)η

= h
(2)
n (t) + [K3t

3 −K3t + K4t2 −K4 + L0t2 + L− 0t4]t2n−1δ

= h
(2)
n (t) + g2(t)t

2n−1η. (2.21)
By the definition of δ2

h
(2)
n+1(δ2) = h

(2)
n (δ).Let h(2)∞ (t) = limn−→∞h(2)n (t). Then, we get

h(2)∞ (δ) =
L0η

1 −δ
− 1.

Hence, instead of (2.19), we can show
h(2)∞ (δ) ≤ 0,which is true by (2.4). The induction for (2.8)-(2.10) is completed. Therefore, sequences {tn},{sn}are nondecreasing, bounded from above by s∗∗ and as such they converge to s∗.

�The semi-local convergence analysis shall be based on conditions (A).Suppose:
(A1) There exists x0 ∈ Ω, η ≥ 0 such that F ′(x0)−1 ∈ L(E1,E) and

‖F ′(x0)−1F (x0)‖≤ η.

(A2) For each x ∈ Ω
‖F ′(x0)−1(F ′(x) −F ′(x0))‖≤ L0‖x −x0‖.

set Ω0 = U[x0, 1L0 ] ∩ Ω.(A3) For each x,y ∈ Ω0
‖F ′(x0)−1(F ′(y) −F ′(x))‖≤ K‖y −x‖,

‖F ′(x0)−1([y,x; F ] −F ′(x0))‖≤ K1(‖y −x0‖ + ‖x −x0‖),

‖F ′(x0)−1([z,y; F ] − [y,x; F ])‖≤ K2(‖z −y‖ + ‖y −x‖)and
‖F ′(x0)−1([z,y; F ] −F ′(y))‖≤ K3‖z −y‖.(A4) U[x0,s∗] ⊂ Ω and(A5) Conditions of Lemma 2.1 hold.

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Eur. J. Math. Anal. 10.28924/ada/ma.2.3 7
Next, we present the semi-local convergence of method (1.2).
THEOREM 2.2. Suppose that conditions (A) hold. Then, sequences {yn},{xn} generated by
method (1.2) are well defined in U[x0,s∗], remain in U[x0,s∗] for each n = 0, 1, 2, . . . and converge
to a solution x∗ ∈ U[x0,s∗] of equation F (x) = 0. Moreover, the following assertion holds

‖xn −x∗‖≤ s∗ − tn. (2.22)
Proof. Mathematical induction on m shall be used to show(Im) ‖ym −xm‖≤ sm − tmand(IIm) ‖xm+1 −ym‖≤ tm+1 − sm.By the first substep of method (1.2) we have

‖y0 −x0‖ = ‖F ′(x0)−1F (x0)‖≤ η = s0 − t0 = s0 ≤ s∗.

So (I0) holds and y0 ∈ U[x0,s∗]. By the first substep of method (1.2) we can write
F (y0) = F (y0) −F (x0) −F ′(x0)(y0 −x0). (2.23)

Using (A2) and (2.23), we have
‖F ′(x0)−1F (y0)‖≤

K0
2
‖y0 −x0‖2 ≤

K0
2

(s0 − t0)2. (2.24)
We need to show the invertability of linear operator A. We have by (A3)

‖F ′(x0)−1(A0 −F ′(x0))‖ ≤ 2‖F ′(x0)−1([y0,x0; F ] −F ′(x0))‖

≤ 2K1(‖y0 −x0‖ + ‖x0 −x0‖)

≤ 2K1(s0 + t0) < 1,

so
‖A−10 F

′(x0)‖≤
1

1 − 2K1(s0 + t0)
, (2.25)

by the Banach lemma on linear invertible operators [19, 25]. Then, iterate x1 exists by the secondsubstep of method (1.2), and we can write
x1 −y0 = (A−10 F

′(x0))(F
′(x0)

−1F (y0)). (2.26)
By (2.24)-(2.26), we get

‖x1 −y0‖ ≤ ‖A−10 F
′(x0)‖‖F ′(x0)−1F (y0)‖

≤
K0(s0 − t0)2

2(1 − 2K1(s0 + t0))
= t1 − s0,

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Eur. J. Math. Anal. 10.28924/ada/ma.2.3 8
showing (II0). Moreover, we have

‖x1 −x0‖ ≤ ‖x1 −y0‖ + ‖y0 −x0‖

≤ t1 − s0 + s0 − t0 = t1 ≤ s∗,

so x1 ∈ U[x0,s∗]. Suppose that (Im) and (IIm) hold, ym,xm+1 ∈ U[x0,s∗] and F ′(xm)−1,A−1m existfor each m = 1, 2, . . . ,n. We shall prove they hold for m = n + 1. Using the second substep ofmethod (1.2), we get
‖F (x0)−1F (xn+1)‖ = ‖F ′(x0)−1(F (xn+1) −F (yn)) −An(xn+1 −yn))‖

= ‖F ′(x0)−1([xn+1,yn; F ] −An)(xn+1 −yn)‖

≤ F ′(x0)−1([xn+1,yn; F ] − [yn,xn; F ])‖

+‖F ′(x0)−1([yn,xn; F ] −F ′(xn))‖

≤ (K2(‖xn+1 −yn‖ + ‖yn −xn‖) + K3‖yn −xn‖)‖xn+1 −yn‖

(2.27)
We need to show F ′(xn+1) is invertible. By (A2) and the induction hypotheses we obtain

‖F ′(x0)−1(F ′(xn+1) −F ′(x0))‖ ≤ L0‖xn+1 −x0‖

≤ L0(tn+1 − t0) = L0tn=1 < 1,

so
‖F ′(xn+1)−1F ′(x0)‖≤

1

1 −L0tn+1
. (2.28)

Hence, we get by (2.27), (2.28) and the first substep of method (1.20 that
‖yn+1 −xn+1‖ = ‖F ′(xn+1)−1F (x0)‖‖F ′(x0)−1F (xn+1)‖

≤
(K2(tn+1 − sn) + (sn − tn)) + K3(sn − tn))(tn+1 − sn)

1 −L0tn+1
= sn+1 − tn+1, (2.29)

since K4 = K2 + K3, showing (Im) for m = n + 1. Then, we also have
‖yn+1 −x0‖ ≤ ‖yn+1 −xn+1‖ + ‖xn+1 −x0‖

≤ sn+1 − tn+1 + tn+1 − s0 = sn+1 ≤ s∗,

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Eur. J. Math. Anal. 10.28924/ada/ma.2.3 9
so yn+1 ∈ U[x0,s∗]. Operator A−1n+1 shall be shown to exist

‖F ′(x0)−1(An+1 −F ′(x0))‖ ≤ ‖F ′(x0)−1([yn+1,xn+1; F ] −F ′(x0))‖

+‖F ′(x0)−1([yn+1; xn+1; F ] −F ′(xn+1))‖

≤ K1(‖yn+1 −x0‖ + ‖xn+1 −x0‖) + K3‖yn+1 −xn+1‖

≤ K1(sn+1 + tn+1) + K3(sn+1 − tn+1) < 1,

so
‖A−1n+1F

′(x0)‖≤
1

1 − (K1(sn+1 + tn+1) + K3(sn+1 − tn+1))
. (2.30)

By the first substep of method (1.2), we can write
F (yn+1) = F (yn+1) −F (xn+1) −F ′(xn+1)(yn+1 −xn+1),

so
‖F ′(x0)−1F (yn+1)‖≤

K

2
‖yn+1 −xn+1‖2 ≤

K

2
(sn+1 − tn+1)2,

so
‖xn+2 −yn+1‖ ≤ ‖A−1n+1F

′(x0)‖‖F ′(x0)−1F (yn+1)‖

≤
K(sn+1 − tn+1)2

2(1 − (K1(sn+1 + tn+1) + K3(sn+1 − tn+1)))
= tn+2 − sn+1,

showing (IIm) for m = n + 1. We can get
‖xn+2 −x0‖ ≤ ‖xn+2 −yn+1‖ + ‖yn+1 −x0‖

≤ tn+2 − sn+1 + sn+1 − t0 = tn+2 ≤ s∗,

so xn+2 ∈ U[x0,s∗]. Furthermore, we obtain
‖xn+1 −xn‖ ≤ ‖xn+1 −yn‖ + ‖yn −xn‖

= tn+1 − sn + sn − tn = tn+1 − tn,

so sequence {xn} is fundamental in a Banach space B, so it converges to some x∗ ∈ U[x0,s∗]. Byletting n −→∞ in (2.27), we obtain
‖F ′(x0)−1F (xk+1)‖ ≤ (K2(tn+1 − sn) + K4(sn − tn))(sn+1 − sn) −→ 0,

so F (x∗) = 0 by the continuity of F.
�A uniqueness of the solution result is given next.

PROPOSITION 2.3. Suppose:

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Eur. J. Math. Anal. 10.28924/ada/ma.2.3 10
(i) There exists a simple solution x∗ of equation F (x) = 0.(ii) There exists s̄ ≥ s∗ such that

L0(s̄ + s∗) < 2.

Set Ω1 = U[x0, s̄] ∩ Ω. Then, the only solution of equation F (x) = 0 in the region Ω1 is x∗.

Proof. Let x̄ ∈ Ω1 with F (x̄) = 0. Let M = ∫10 F ′(x̄ + θ(x∗ − x̄))dθ. Then, in view of (A2) and(ii), we obtain
‖F ′(x0)−1(M −F ′(x0))‖ ≤ L0

∫ 1
0

[(1 −θ)‖x̄ −x0‖ + θ‖x∗ −x0‖]dθ

≤
L0
2

(s̄ + s∗) < 1,

so x̄ = x∗ since M−1 exists and M(x∗ − x̄) = F (x∗) −F (x̄) = 0 − 0 = 0.
�

REMARK 2.4. Notice that s∗∗ given in closed form can repalce s∗ in the conditions of Theorem
2.2.

3. Local Convergence
As in Section 2 we develop some functions and parameters. Let Li, i = 0, 1, 2, 3, 4 be givenparameters. Define function ϕ1 on the interval T = [0, 1L0 ) by

ϕ1(t) =
Lt

2(1 −L0t)
.

Notice that parameter
rA =

2

2L0 + L
<

1

L0
(3.1)

solves equation
ϕ1(t) = 1.

Define functions on the interval T by
q(t) = L0ϕ1(t)t − 1 and p(t) = (2L1(1 + ϕ1(t)) + L)t.

Suppose that these functions have smallest zeros rq and rp in (0, 1L0 ), respectively. Let r1 =
min{rq, rp} and T0 = [0, r1). Define function ϕ2 on T0 by

ϕ2(t) =

[
Lϕ1(t)

2(1 −L0ϕ1(t)t)

+
L4(L2 + L3)(1 + ϕ1(t))ϕ1(t)

(1 −L0ϕ1(t)t)(1 −p(t))

]
t.

Suppose that function
ϕ2(t) − 1

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Eur. J. Math. Anal. 10.28924/ada/ma.2.3 11
has smallest zero r2 ∈ (0, r1). We shall show that parameter

r = min{rA, r2} (3.2)
is a convergence radius for method (1.2). Let T1 = [0, r). Then, it follows by these definitions thatfor each t ∈ T1

L0t < 1 (3.3)
0 ≤ ϕ1(t) < 1, (3.4)
0 ≤ ϕ1(t)t < 1 (3.5)
0 ≤ p(t) < 1 (3.6)

and
0 ≤ ϕ2(t)t < 1 (3.7)

hold.The conditions (H) to be used in the local convergence of method (1.2) are as follows.Suppose:
(H1) There exists a simple solution x∗ ∈ Ω of equation F (x) = 0.(H2) For each x ∈ Ω

‖F ′(x)−1(F ′(x) −F ′(x∗))‖≤ L0‖x −x∗‖.

Set Ω0 = U[x∗, 1L0 ] ∩ Ω.(H3) For each x,y ∈ Ω0
‖F ′(x∗)−1(F ′(y) −F ′(x))‖≤ L‖y −x‖,

‖F ′(x∗)−1(F ′(y) −F ′(x∗))‖≤ L1(‖y −x∗‖ + ‖x −x∗‖),

‖F ′(x∗)−1([y,x; F ] −F ′(x))‖≤ L2‖y −x‖,

‖F ′(x∗)−1([y,x; F ] −F ′(y))‖≤ L3‖y −x‖

and
‖F ′(x∗)−1F ′(x)‖≤ L4‖x −x∗‖.

and(H4) U[x∗, r] ⊂ Ω.In view of conditions (H) and the developed notation we can show the local convergence result formethod (1.2).
THEOREM 3.1. Under the conditions (H), further suppose that x0 ∈ U(x∗, r) −{x∗}. Then, se-
quence {xk},{yn} generated by method (1.2) is well defined in U(x∗, r), remains in U(x∗, r) for
each k = 0, 1, 2, . . . and converges to x∗.

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Eur. J. Math. Anal. 10.28924/ada/ma.2.3 12
Proof. Let u ∈ U(x∗, r) −{x∗}. By (H1) and (H2), we get in turn that

‖F ′(x∗)−1(F ′(u) −F ′(x∗))‖≤ L0‖u −x∗‖≤ L0r < 1,

so F ′(u) is invertibale and
‖F ′(u)−1F ′(x∗)‖≤

1

1 −L0‖u −x∗‖
. (3.8)

Iterate y0 is well defined by the first substep of method (1.2) and (3.8) for u = x0. Then, we canwrite
y0 −x∗ = x0 −x∗ −F ′(x0)−1F (x0)

= (F ′(x0)
−1F ′(x∗))

×(
∫ 1
0

F ′(x∗)
−1(F ′(x∗ + θ(x0 −x∗)) −F ′(x0))dθ(x0 −x∗).

(3.9)
By (3.2), (3.4), (H3), (3.8) and (3.9), we have in turn that

‖y0 −x∗‖ ≤
L0‖x0 −x∗‖2

2(1 −L0‖x0 −x∗‖

≤
L‖x0 −x∗‖2

2(1 −L0‖x0 −x∗‖)
≤ ϕ1(‖x0 −x∗‖)‖x0 −x∗‖≤‖x0 −x∗‖ < r (3.10)

so y0 ∈ U(x∗, r). Next, we show linear operator A) is invertible. Indeed, using (3.2), (3.6), (H3) and(3.10), we get in turn that
‖F ′(x∗)−1(A0 −F ′(x∗))‖ ≤ ‖F ′(x∗)−1([y0,x0; F ] −F ′(x∗))‖

+‖F ′(x∗)−1([y0,x0; F ] −F ′(x∗))‖

+‖F ′(x∗)−1(F ′(x0) −F ′(x∗))‖

≤ 2L1(‖y0 −x∗‖ + ‖x0 −x∗‖) + L0‖x0 −x∗‖

≤ 2L1(1 + ϕ1(‖x0 −x∗‖))‖x0 −x∗‖ + L‖x0 −x∗‖

≤ p(‖x0 −x∗‖) ≤ p(r) < 1,

so
‖A−10 F

′(x∗)‖≤
1

1 −p(‖x0 −x∗‖)
(3.11)

and iterate x1 is well defined by the second substep of method (1.2) for n = 0. Then, we can write
x1 −x∗ = (y0 −x∗ −F ′(y0)−1F (y0)) + F ′(y0)−1(A) −F

′(y0))A
−1
0 F (y0). (3.12)

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Eur. J. Math. Anal. 10.28924/ada/ma.2.3 13
Using (3.2), (3.7), (H3), (3.8) (for u = x0,y0) and (3.10)–(3.13), we obtain in turn that
‖x1 −x∗‖ ≤ ‖y0 −x∗ −F ′(y0)−1F (y0)‖

+‖F ′(y0)−1F ′(x∗)‖‖F ′(x∗)−1(A0 −F ′(y0))‖‖A−10 F
′(x∗)‖‖F ′(x∗)−1F (y0)‖

≤
L‖y0 −x∗‖2

2(1 −L0‖x0 −x∗‖
+

(L2 + L3)‖y0 −x0‖L4‖y0 −x∗‖
(1 −L0‖y0 −x∗‖)(1 −p(‖x0 −x∗‖))

≤ ϕ2(‖x0 −x∗‖)‖x0 −x∗‖≤‖x0 −x∗‖ < r,

so x1 ∈ U(x∗, r), where we also used
‖F ′(x∗)−1(A0 −F ′(y0))‖ ≤ ‖F ′(x∗)−1([y0,x0; F ] −F ′(x0))‖

+‖F ′(x∗)−1([y0,x0; F ] −F ′(y0))‖

≤ (L2 + L3)‖y0 −x0‖≤ (L2 + L3)(‖y0 −x∗‖ + ‖x0 −x∗‖)

≤ (L2 + L3)(1 + ϕ1(‖x0 −x∗‖))‖x0 −x∗‖,

and
‖F ′(x∗)−1F (y0)‖ = ‖

∫ 1
0

F ′(x∗)
−1F ′(x∗ + θ(y0 −x∗))dθ(y0 −x∗)‖

≤ L4‖y0 −x∗‖2.

So, far showed
‖y0 −x∗‖≤ ϕ1(‖x0 −x∗‖)‖x0 −x∗‖ < r

and
‖x1 −x∗‖≤ ϕ2(‖x0 −x∗‖)‖x0 −x∗‖ < r.

By simply replacing x0,y0,x1 by xm,ym,xm+1 in the preceding calculations, we get
‖ym −x∗‖≤ ϕ1(‖xm −x∗‖)‖xm −x∗‖ < r

and
‖xm+1 −x∗‖≤ ϕ2(‖xm −x∗‖)‖xm −x∗‖ < r.

Then, from the estimation
‖xm+1 −x∗‖≤ α‖xm −x∗‖ < r, (3.13)

where α = ϕ2(‖x0 −x∗‖) ∈ [0, 1), limm−→∞xm = x∗ and ym,xm+1 ∈ U(x∗, r).
�

REMARK 3.2. By the definition of r, we see that

r ≤ rA. (3.14)

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Eur. J. Math. Anal. 10.28924/ada/ma.2.3 14
Parameter rA was shown in [4] to be a convergence radius for Newton’s method. Notice the radius
of convergence for Newton’s method given independently by Traub [35] and Rheinbold [29] is

rTR =
2

3M1
,

where L1 is the Lipschitz constant on Ω. So, we have

rTR ≤ rA,

since L ≤ L1 and L0 ≤ M1.

4. Numerical Experiments
We provide some examples, showing that the old convergence criteria are not verified but oursare.

EXAMPLE 4.1. Define function

f (t) = θ0t + θ1 + θ2 sin θ3t, t0 = 0,

where θj, j = 0, 1, 2, 3 are parameters. Then, clearly for θ3 large and θ2 small, L0K can be small
(arbitrarily).

EXAMPLE 4.2. Let B = B1 = U[0, 1] the domain of functions given on [0, 1] which are continuous.
We consider the max-norm. Choose Ω = B(0,d), d > 1. Define F on Ω be

F (x)(s) = x(s) −w(s) −ξ
∫ 1
0

K(s,t)x3(t)dt, (4.1)
x ∈ B,s ∈ [0, 1],w ∈ B is given, ξ is a parameter and K is the Green’s kernel given by

K(s2,s1) =

{
(1 − s2)s1, s1 ≤ s2
s2(1 − s1), s2 ≤ s1.

By (4.1), we have

(F ′(x)(z))(s) = z(s) − 3ξ
∫ 1
0

K(s,t)x2(t)z(t)dt,

t ∈ Bs ∈ [0, 1]. Consider x0(s) = w(s) = 1 and |ξ| < 83. We get

‖I −F ′(x0)‖ <
3

8
|ξ|, F ′(x0)−1 ∈ L(B1B),

‖F ′(x0)−1‖≤
8

8 − 3|ξ|
, η =

|ξ|
8 − 3|ξ|

, L0 =
12|ξ|

8 − 3|ξ|
,

K =
6d|ξ|
8−3|ξ|,K1 =

L0
2

and K2 = K2 = K3.

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Eur. J. Math. Anal. 10.28924/ada/ma.2.3 15
EXAMPLE 4.3. Let B = B1 = R3 and Ω be as in the Example 4.2. It is well known that the
boundary value problem [16]

ψ(0) = 0,ψ(1) = 1,

ψ′′ = −ψ −τψ2

can be given as a Hammerstein-like nonlinear integral equation

ψ(s) = s +

∫ 1
0

K(s,t)(ψ3(t) + τψ2(t))dt

where τ is a parameter. Then, define F : Ω −→ T2 by

[F (x)](s) = x(s) − s −
∫ 1
0

K(s,t)(x3(t) + τx2(t))dt.

Choose x0(s) = s and Ω = U(x0, r0). Then, clearly U(x0, r0) ⊂ U(0, r0 + 1), since ‖x0‖ = 1.
Suppose 2τ < 5. Then, by conditions (A) are satisfied for L0 = 2τ+3r0+68 , K =

τ+6r0+3
4

, K1 =
L0
2

and K2 = K2 = K3. and η =
1+τ
5−2τ . Notice that L0 < K.

The rest of the examples are given for the local convergence study of Newton’s method.
EXAMPLE 4.4. Let B = B1 = R3, Ω = U[0, 1] and x∗ = (0, 0, 0)tr. Define mapping E on Ω for
λ = (λ1,λ2,λ3)

tr as

E(λ) = (eλ1 − 1,
e − 1

2
λ22 + λ1,λ3)

tr.

Then, conditions (H) hold provided that L0 = e − 1,L = e
1
L0 and M1 = e, since F ′(x∗)−1 =

F ′(x∗) = diag{1, 1, 1,}. Notice that
L0 < L < M1,

L1 =
L0
2
, L4 = L, L2 = L3 =

L

2
.

rTR = 0.2453 < rA = 0.3827, r = 0.2124.

Hence, our radius of convergence is larger.

EXAMPLE 4.5. Let B = B1 and Ω be as in Example 4.2. Define F on Ω as

F (ϕ1)(x) = ϕ1(x) −
∫ 1
0

xϕ1(j)
3dj.

Then, we obtain

F ′(ϕ1(ψ1))(x) = ψ1(x) − 3
∫ 1
0

xjϕ1(j)
2ψ1(j)dj

for all ψ1 ∈ Ω. So, we can choose L0 = 1.5,L = M1 = 3.

L1 =
L0
2
, L4 = L, L2 = L3 =

L

2
.

But then, we get again
rTR = 0.2222 < rA = 0.3333, r = 0.2663.

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Eur. J. Math. Anal. 10.28924/ada/ma.2.3 16
5. Conclusion

Ostrowski’s method was revisited and its applicability was extended in both the semi-localand local convergence case. In particular, the benefits in the semi-local convergence case include:weaker sufficient convergence criteria (i.e. more starters x0 become available); tighter upper boundson ‖xk+1 −xk‖, ‖xk −x∗‖ (i.e., fewer iterates are computed to reach a predecided error accuracy)and the information on the location of x∗ is more precise.The results are based on generalized continuity which is more general than Lipschitz continuityused before. Our two techniques are very general and can be used to extend the applicability ofother methods.
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https://doi.org/10.1080/01630568708816254

	1. Introduction
	2.  Semi-local Convergence
	3. Local Convergence
	4. Numerical Experiments
	5. Conclusion
	References

