




































©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+yn
4

)
− F ′

(
xn+yn
2

)
+ 2F ′

(
xn+3yn
4

)
. 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 2The 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 i f t 6= 0

0 i f 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 = 5K0
6 (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).

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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 ∗ = 3
5K0

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) + 2K0t

3.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 = η
1−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 4and
tk+1 ≤

(1− αk+2)η
1− α < T.

Evidently, (2.4) holds if
4K(1 + α)αk−1η

9(1−K0 1−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)tk−1η + 9K0(1 + t + . . .+ tk−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 + . . .+ tk)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−K0 1−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 + . . .+ tk+1)η − 2. (2.16)

Then, (2.15) holds if
f
(2)
k (t) ≤ 0 at t = α2. (2.17)

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Eur. J. Math. Anal. 10.28924/ada/ma.2.13 5We get
f
(2)
k+1(t) = K(2 + t)(1 + t)tkη + 2K0(1 + t + . . .+ tk+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 = 5K0
3 (sk+tk) < 5K0

3 ( η
1−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 + . . .+ tn)η − 1.

This time we have
f
(1)
n+1(t) = f

(1)
n (t) + g

(1)
n (t)tn−1η, (2.23)

where
g
(1)
n (t) =

4KK0
3

tn+2η +
4KK0

3
tn+1η +

4

3
Kt −

4K

3
(1−K0η).

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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)tn−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 73. 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,

1
K0

).(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

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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−1k 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)

)
,

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Eur. J. Math. Anal. 10.28924/ada/ma.2.13 9so 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

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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)‖ ≤ K
2 (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 =
∫ 1
0 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

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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.

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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, 12), 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,
1
K0

) ∩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 + 1
K0

),

|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
.

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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− 1
K0
, 1 + 1

K0
], then

K =
5%3 + q

9%2
,

where % = 4−q
3−q and K < K1 for all q ∈ (0, 12).

Next, we verify conditions (2.2), (2.3), (2.24) and (2.25).
Then for q = 0.95, 6K0 = 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 − 10K0η3 = 0.8861, and (4K3 + 4
3K0Kη + 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
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https://doi.org/10.1007/s11075-012-9585-7

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

