ARGYROS-HILOUT-625.dvi


CUBO A Mathematical Journal
Vol.12, No¯ 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

and

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 and 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 and 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 and 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 and 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

F (x) = 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 and also in engineering are solved by

finding the solutions of certain equations. For example, dynamic systems are mathematically mod-

eled by difference or differential equations, and 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 = T (x), for some suitable operator T , where x is the state. Then the equilibrium states are de-

termined 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, and integral equa-

tions), 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 hand. Since all of these methods have the same recursive struc-

ture, they can be introduced and discussed in a general framework.

We consider the Secant method in the form

xn+1 = xn − δF (xn−1, xn)−1 F (xn) (n ≥ 0), (x−1, x0 ∈ D) (1.2)

where δF (x, y) ∈ L(X , Y) (x, y ∈ D) is a consistent approximation of the Fréchet–derivative of F
[5], [13]. Bosarge and Falb [7], Dennis [9], Potra [16], Argyros [1]–[5], Gutiérrez [10] and others

[11], [15], [18], have provided sufficient convergence conditions for the Secant method based on

Lipschitz–type conditions on δF .



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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 vio-

lated. Indeed, let ℓ = 1, η = .18, and 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 optimiza-

tion considerations, and the above observation.

Here, using recurrent functions, Lipschitz–type and center–Lipschitz–type conditions, we pro-

vide a semilocal convergence analysis for (1.2). It turns out that our error bounds are more precise

and 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 ℓ0 > 0, ℓ > 0, c > 0, and η ∈ [0, c] be given parameters.

Assume:

ℓ0 (c + η) < 1. (2.4)

Set

δ0 =
ℓ (c + η)

c (1 − ℓ0 (c + η))
. (2.5)

Moreover, assume: there exists δ ∈
[

max{ℓ, δ0},
1

c

)

, such that estimate

ℓ q + ℓ0

(

2 q1/
√

5

1 − q(p−1)/
√

5
+ q2 + q

)

+ ℓ − δ ≤ 0 (2.6)

is satisfied for q = δ c, and p =
1 +

√
5

2
.

Then, scalar sequence {tn} (n ≥ −1) given by

t−1 = 0, t0 = c, t1 = c + η, tn+2 = tn+1 +
ℓ (tn+1 − tn−1) (tn+1 − tn)

1 − ℓ0 (tn+1 − t0 + tn)
(2.7)



164 Ioannis K. Argyros and Saïd Hilout CUBO
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is non–decreasing, bounded above by

t⋆⋆ =
c

q

∞
∑

n=1

qun , (2.8)

and converges to some t⋆ ∈ [0, t⋆⋆], where {un} is the Fibonacci’s sequence: u−1 = u0 = 1, un+1 =
un + un−1, n ≥ 1.

Moreover, the following a priori estimates hold:

0 ≤ t⋆ − tn ≤
c (q1/

√
5)p

n

q

(

1 − qpn (p−1)/
√

5

) (n ≥ 0). (2.9)

Proof. We shall show using induction on k:

ℓ (tk+1 − tk−1)
1 − ℓ0 (tk+1 − t0 + tk)

≤ δ (tk − tk−1), (2.10)

0 ≤ tk+1 − tk ≤ tk − tk−1, (2.11)

and

tk+1 − tk ≤ quk−1 c. (2.12)

Note that if (2.10) holds, then by (2.7), we have:

tk+2 − tk+1 =
ℓ (tk+1 − tk−1) (tk+1 − tk)

1 − ℓ0 (tk+1 − t0 + tk)
≤ δ (tk − tk−1) (tk+1 − tk). (2.13)

Estimate (2.10) can also be written as:

tk+1 − tk ≤ αk (tk − tk−1), (2.14)

where,

αk =
1

ℓ

(

δ (1 − ℓ0 (tk+1 − t0 + tk)) − ℓ
)

. (2.15)

For k = 0, (2.14) becomes

η ≤ 1
ℓ

(

δ (1 − ℓ0 (c + η)) − ℓ
)

c, or δ ≥ δ0,

which is true by the choice of δ. Estimates (2.11) and (2.12) also hold for k = 0 by the initial

conditions.



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Convergence Conditions for the Secant Method 165

Assume that (2.10) (i.e., (2.13)), (2.11), and (2.12) hold for all k ≤ n.

By the induction hypotheses we have

0 ≤ tk+2 − tk+1 ≤ δ quk−1−1 c (tk+1 − tk) = quk−1 (tk+1 − tk) < tk+1 − tk. (2.16)

That is (2.11) holds for n = k + 1. On the other hand:

0 ≤ tk+2 − tk+1 ≤ quk−1 (tk+1 − tk) ≤ quk−1 quk c = quk+1 c, (2.17)

which shows (2.12) for n = k + 1.

We also have the estimate

0 ≤ tk+2 − t0 ≤ (t1 − t0) + (t2 − t1) + · · · + (tk+2 − tk+1)
≤ c

q
(qu0 + qu1 + · · · + quk ) < t⋆. (2.18)

Clearly, we have:

uk =
1√
5

[(

1 +
√

5

2

)k+1

−
(

1 −
√

5

2

)k+1]

≥ 1√
5

(

1 +
√

5

2

)k

=
pk√

5
, (2.19)

So, for any k ≥ 0, m ≥ 0, we get in turn:

0 ≤ tk+m − tk ≤ (tk − tk+1) + (tk+1 − tk+2) + · · · + (tk+m − tk+m−1)
≤ c

q
(quk + quk+1 + · · · + quk+m−1 )

≤ c
q

(qp
k/

√
5 + qp

k+1/
√

5 + · · · + qp
k+m−1/

√
5).

(2.20)

Using Bernoulli’s inequality, we obtain:

tk+m − tk ≤
c

q
qp

k/
√

5 (1 + q(p
k+1−pk )/

√
5 + q(p

k+2−pk )/
√

5 · · · + q(p
k+m−1−pk )/

√
5)

=
c

q
qp

k/
√

5 (1 + qp
k (p−1)/

√
5 + qp

k (p2−1)/
√

5 · · · + qp
k (pm−1−1)/

√
5)

≤ c
q

qp
k/

√
5 (1 + qp

k (p−1)/
√

5 + qp
k (1+2 (p−1)−1)/

√
5 · · · + qp

k (pm−1−1)/
√

5)

=
c

q
qp

k/
√

5

(

1 + qp
k (p−1)/

√
5 +

(

qp
k (p−1)/

√
5

)2

· · · +
(

qp
k (p−1)/

√
5

)m−1)

=
c

q
qp

k/
√

5 1 − qp
k (p−1) m/

√
5

1 − qpk (p−1)/
√

5.

(2.21)

In particular, from (2.18) we have:



166 Ioannis K. Argyros and Saïd Hilout CUBO
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tk+m ≤
c

q
(qu0 + · · · + quk+m−1 ) + c (2.22)

In order for us to show (2.10), it suffices:

δ (tk − tk−1) ≤ αk+1 (2.23)

or

δ (tk − tk−1) ≤
1

ℓ

(

δ (1 − ℓ0 (tk+2 − t0 + tk+1)) − ℓ
)

or

ℓ δ quk−1−1 + δ ℓ0 (tk+2 + tk+1 − t0) ≤ δ − ℓ

or

ℓ δ quk−1−1 c + δ ℓ0

(

c

q
(qu0 + qu1 + · · · + quk+1 ) + c+

c

q
(qu0 + qu1 + · · · + quk ) + c − c

)

≤ δ − ℓ

or

ℓ δ quk−1−1 c + ℓ0

(

2 (qu0 + qu1 + · · · + quk ) + quk+1 + δ c
)

≤ δ − ℓ

or

ℓ quk−1 + ℓ0

(

2 (qu0 + qu1 + · · · + quk ) + quk+1 + q
)

≤ δ − ℓ (2.24)

or

ℓ q + ℓ0

(

2 q1/
√

5

1 − q(p−1)/
√

5
+ q2 + q

)

+ ℓ − δ ≤ 0,

which is true by the choice of δ and q given by (2.6).

The induction for (2.10)–(2.12) is now complete.

It follows from (2.21) that scalar sequence {tn} is Cauchy in the complete space R, and 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 ℓ0 > 0, ℓ > 0, c > 0, and η ∈ (0, c] be given parameters.

Assume:

ℓ0 (c + η) < 1;



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Convergence Conditions for the Secant Method 167

there exists

δ ∈
(

max{ℓ + 2 ℓ0, δ0, δ1}, min{
1

c
, δ∞}

)

, (2.25)

and,

v1 ≥ δ1, (2.26)

where, δ1 =
q1

c
, is such that q1 is the unique zero in (0, 1) of equation

Q(t) = ℓ0 t
4 + ℓ0 t

2 + ℓ t − ℓ = 0,

δ∞ =
q∞

c
, is such that q∞ is the unique positive zero of equation

f∞(t) = ℓ0

(

2 t1/
√

5

1 − t(p−1)/
√

5
+ t

)

+ ℓ − δ = 0,

and

v1 is the unique positive zero of equation

f0(t) = ℓ0 t
2 + (3 ℓ0 + ℓ) t + 2 ℓ0 + ℓ − δ = 0;

and

ℓ0 t
2 uk + ℓ0 t

uk + ℓ tuk−2 − ℓ ≥ 0 (k ≥ 1) for t ≥ δ1. (2.27)

Then, the conclusions of Lemma 2.1 hold true.

Proof.

We follow the proof of Lemma 2.1 until estimate (2.23). Then, let us define functions fk, gk

by

fk(t) = ℓ t
uk−1 + ℓ0

(

2 (1 + tu1 + · · · + tuk ) + tuk+1 + t
)

+ ℓ − δ, (2.28)

and

gk(t) = gk(t) t
uk−1 , (2.29)

where,

gk(t) = ℓ0 t
2 uk + ℓ0 t

uk + ℓ tuk−2 − ℓ. (2.30)



168 Ioannis K. Argyros and Saïd Hilout CUBO
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We need to find the relationship between two consecutive functions fk:

fk+1(t) = ℓ t
uk + ℓ0

(

2 (t0 + tu1 + · · · + tuk+1 ) + tuk+2 + t
)

+ ℓ − δ

= ℓ tuk−1 − ℓ tuk−1 + ℓ tuk + ℓ0
(

2 (t0 + tu1 + · · · + tuk )+

tuk+1 + tuk+1 + tuk+2 + t

)

+ ℓ − δ

= fk(t) + ℓ (t
uk − tuk−1 ) + ℓ0 (tuk+1 + tuk+2 )

= fk(t) + gk(t).

(2.31)

Using hypothesis 2 ℓ0 + ℓ−δ < 0, and (2.28), we get: f0(0) = 2 ℓ0 + ℓ−δ < 0, fk(0) = δ −ℓ < 0
(k ≥ 1). Moreover for sufficiently large t, we also have fk(t) > 0 for all k ≥ 0. It then follows from
the intermediate value theorem that: for each k ≥ 0, there exists vk ≥ 0, with fk(vk) = 0. Each vk
is the unique positive zero of fk, since f

′
k(t) > 0 (k ≥ 0). That is the graph of function fk crosses

the positive axis only once.

By the definition of fk and vk, there exists a polynomial pk−1 of degree k −1, with pk−1(s) > 0
(s > 0), such that:

fk(t) = (t − vk) pk−1(t).

To show (2.14), we must have fk(t) ≤ 0. That is

q = t ≤ vk (k ≥ 0). (2.32)

In view of (2.28) and (2.31), we get

fk+1(vk) = fk(vk) + gk(vk) = gk(vk) = qk(vk) v
uk−1
k ≥ 0 for vk ≥ δ1 (by (2.27)).

Therefore, if vk ≥ δ1, then vk+1 ≤ vk, and lim
k−→∞

vk = q∞ exists. Note that v1 ≥ δ1 by (2.23).
Then, it follows v2 ≤ v1. Assume vm ≥ δ1 (m ≤ n), then vm+1 ≤ vm. We must also show
vm+1 ≥ δ1. But this is true, since vm+1 ≥ δ∞ ≥ δ1 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, x0) belonging to the class C(ℓ, ℓ0, η, c, δ)
defined as follows:

Definition 2.3. Let ℓ, ℓ0, η, c, δ be non–negative parameters satisfying the hypotheses of Lemma

2.2.

We say that a triplet (F, x−1, x0) belongs to the class C(ℓ, ℓ0, η, c, δ) if:



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Convergence Conditions for the Secant Method 169

(c1) F is a nonlinear operator defined on a convex subset D of a Banach space X with values in
a Banach space Y;

(c2) x−1 and x0 are two points belonging to the interior D0 of D and satisfying the inequality

‖ x0 − x−1 ‖≤ c;

(c3) F is Fréchet–differentiable on D0, and there exists an operator δF : D0 ×D0 → L(X , Y) such
that:

the linear operator A = δF (x−1, x0) is invertible, its inverse A
−1 is bounded and:

‖ A−1 F (x0) ‖ ≤ η;
‖ A [δF (x, y) − F ′(z)] ‖ ≤ ℓ (‖ x − z ‖ + ‖ y − z ‖);

‖ A [δF (x, y) − F ′(x0)] ‖ ≤ ℓ0 (‖ x − x0 ‖ + ‖ y − x0 ‖)

for all x, y, z ∈ D.

(c4) the set Dc = {x ∈ D; F is continuous at x} contains the closed ball U (x0, t⋆) = {x ∈ X |‖
x − x0 ‖≤ 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, x0) ∈ C(ℓ, ℓ0, η, c, δ), then sequence {xn} (n ≥ −1) generated by Secant
method (1.2) is well defined, remains in U (x0, t

⋆) for all n ≥ 0 and converges to a unique solution
x⋆ ∈ U (x0, t⋆) of equation F (x) = 0.

Moreover the following estimates hold for all n ≥ 0

‖ xn+2 − xn+1 ‖≤ tn+2 − tn+1, (2.33)

and

‖ xn − x⋆ ‖≤ t⋆ − tn, (2.34)

where the sequence {tn} (n ≥ 0) given by (2.7).

Furthermore, if

U

(

x0,
1

2
(c +

1

ℓ0
)

)

⊆ D,

t⋆⋆ <
1

2
(c +

1

ℓ0
) (2.35)

the solution x⋆ is unique in U (x0, t
⋆).

Finally, if

t⋆⋆ <
1

ℓ0
− R, U (x0, R) ⊆ D,

where t⋆⋆ is given by (2.8), then the solution x⋆ is unique in U (x0, R).



170 Ioannis K. Argyros and Saïd Hilout CUBO
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Proof. We first show operator L = δF (xk, xk+1) is invertible for xk, xk+1 ∈ U (x0, t⋆). It
follows from (2.7), (2.8), (c2) and (c3) that:

‖ I − A−1 L ‖=‖ A−1 (L − A) ‖ ≤ ‖ A−1(L − F ′(x0)) ‖ + ‖ A−1(F ′(x0) − A) ‖
≤ ℓ0 (‖ xk − x0 ‖ + ‖ xk+1 − x0 ‖ + ‖ x0 − x−1 ‖)
≤ ℓ0 (tk − t0 + tk+1 − t0 + c)
≤ ℓ0 (t⋆ − t0 + t⋆ − t0 + c)

≤ ℓ0
(

2

[

η

1 − δ
+ c

]

− c
)

≤ 1

(2.36)

since δ ≤ δ∞.

According to the Banach Lemma on invertible operators [5], [13], and (2.36), L is invertible

and

‖ L−1 A ‖≤
(

1 − ℓ0 (‖ xk − x0 ‖ + ‖ xk+1 − x0 ‖ +c)
)−1

. (2.37)

The second condition in (c3) implies the Lipschitz condition for F
′

‖ A−1 (F ′(u) − F ′(v)) ‖≤ 2 ℓ ‖ u − v ‖, u, v ∈ D0. (2.38)

By the identity,

F (x) − F (y) =
∫ 1

0

F
′(y + t(x − y)) dt (x − y) (2.39)

we get

‖ A−10 [F (x) − F (y) − F ′(u)(x − y)] ‖≤ ℓ (‖ x − u ‖ + ‖ y − u ‖) ‖ x − y ‖ (2.40)

and

‖ A−10 [F (x) − F (y) − δF (u, v) (x − y)] ‖≤ ℓ (‖ x − v ‖ + ‖ y − v ‖ + ‖ u − v ‖) ‖ x − y ‖ (2.41)

for all x, y, u, v ∈ D0. By a continuity argument (2.38)–(2.41) remain valid if x and/or y belong to
Dc.

We first show (2.33). If (2.33) holds for all n ≤ k and if {xn} (n ≥ 0) is well defined for
n = 0, 1, 2, · · · , k then

‖ x0 − xn ‖≤ tn − t0 < t⋆ − t0, n ≤ k. (2.42)

That is (1.2) is well defined for n = k + 1. For n = −1, and n = 0, (2.33) reduces to
‖ x−1 − x0 ‖≤ c, and ‖ x0 − x1 ‖≤ η. Suppose (2.33) holds for n = −1, 0, 1, · · · , k (k ≥ 0). Using
(2.37), (2.41) and

F (xk+1) = F (xk+1) − F (xk) − δF (xk−1, xk) (xk+1 − xk) (2.43)



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Convergence Conditions for the Secant Method 171

we obtain in turn

‖ xk+2 − xk+1 ‖ = ‖ δF (xk, xk+1)−1 F (xk+1) ‖
≤ ‖ δF (xk, xk+1)−1 A ‖ ‖ A−1 F (xk+1) ‖

≤ ℓ (‖ xk+1 − xk ‖ + ‖ xk − xk−1 ‖)
1 − ℓ0 [‖ xk+1 − x0 ‖ + ‖ xk − x0 ‖ +c]

‖ xk+1 − xk ‖

≤ ℓ (tk+1 − tk + tk − tk−1)
1 − ℓ0 [tk+1 − t0 + tk − t0 + t0 − t−1]

(tk+1 − tk)

= tk+2 − tk+1.

(2.44)

The induction for (2.33) is now complete. It follows from (2.33) and Lemma 2.2 that sequence

{xn} (n ≥ −1) is Cauchy in a Banach space X , and as such it converges to some x⋆ ∈ U (x0, t⋆)
(since U (x0, t

⋆) is a closed set). By letting k → ∞ in (2.44), we obtain F (x⋆) = 0.

Estimate (2.34) follows from (2.33) by using standard majoration techniques [1], [5], [13].

We shall first show uniqueness in U (x0, t
⋆). Let y⋆ ∈ U (x0, t⋆) be a solution of equation (1.1).

Set

M =
∫ 1

0

F ′(y⋆ + t (y⋆ − x⋆)) dt.

It then by (c3):

‖ A−1 (A − M) ‖ ≤ ℓ0 (‖ y⋆ − x0 ‖ + ‖ x⋆ − x0 ‖ + ‖ x0 − x−1 ‖)

≤ ℓ0
(

2 (t⋆ − t0) + t0
)

≤ ℓ0 (2 t⋆⋆ − c) < 1,

(2.45)

since δ ≤ δ∞.

It follows from (2.35), and the Banach lemma on invertible operators that M−1 exists on
U (x0, t

⋆).

Using the identity:

F (x⋆) − F (y⋆) = M (x⋆ − y⋆), (2.46)

we deduce x⋆ = y⋆.

Finally, we shall show uniqueness in U (x0, R). As in (2.45), we arrive at

‖ A−1 (A − M) ‖< ℓ0 (t⋆⋆ + R) ≤ 1.
Hence, again we conclude x⋆ = y⋆.

That completes the proof of Theorem 2.4. ♦



172 Ioannis K. Argyros and Saïd Hilout CUBO
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Remark 2.5. Returning back to the example given in the introduction, say ℓ0 = .1, we obtain

δ0 = 2.047714554,
1

c
= 5.405. Condition (2.4) is true, since

ℓ0 (c + η) = .0365 < 1.

Choose δ = 2.5 ∈
(

δ,
1

c

)

to obtain q = .4625. Then condition (2.6) becomes

.423156212 < q.

That is our results can apply, whereas the ones using (1.3) cannot.

Remark 2.6. Let us define the majoring sequence {wn} used in [4], [5] (under condition (1.3)):

w−1 = 0, w0 = c, w1 = c + η, wn+2 = wn+1 +
ℓ (wn+1 − wn−1) (wn+1 − tn)

1 − ℓ (wn+1 − w0 + wn)
. (2.47)

Note that in general

ℓ0 ≤ ℓ (2.48)

holds , and
ℓ

ℓ0
can be arbitrarily large [3], [5]. In the case ℓ0 = ℓ, then tn = wn (n ≥ −1).

Otherwise

tn+1 − tn ≤ wn+1 − wn, (2.49)

0 ≤ t⋆ − tn ≤ w⋆ − wn, w⋆ = lim
n−→∞

wn. (2.50)

Note also that strict inequality holds in (2.49) for n ≥ 1, if ℓ0 < 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 ℓ0 < 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

|x(s)|.

Let θ ∈ [0, 1] be a given parameter. Consider the "Cubic" integral equation

u(s) = u3(s) + λ u(s)

∫ 1

0

q(s, t) u(t) dt + y(s) − θ. (2.51)

Here the kernel q(s, t) is a continuous function of two variables defined on [0, 1] × [0, 1]; the pa-
rameter λ is a real number called the "albedo" for scattering; y(s) is a given continuous function



CUBO
12, 1 (2010)

Convergence Conditions for the Secant Method 173

defined on [0, 1] and x(s) is the unknown function sought in C[0, 1]. Equations of the form (2.51)
arise of gasses [5], [8]. For simplicity, we choose u0(s) = y(s) = 1, and q(s, t) =

s

s + t
, for all

s ∈ [0, 1], and t ∈ [0, 1], with s + t 6= 0. If we let D = U (u0, 1 − θ), and define the operator F on
D by

F (x)(s) = x3(s) + λ x(s)

∫ 1

0

q(s, t) x(t) dt + y(s) − θ, (2.52)

for all s ∈ [0, 1], then every zero of F satisfies equation (2.51). We define the divided difference
δ F (x, y) by

δ F (x, y) =

∫ 1

0

F ′(y + t (x − y)) dt.

We have the estimate

max
0≤s≤1

|
∫

s

s + t
dt| = ln 2.

Therfore, if we set b =‖ F ′(u0)−1 ‖, then, we obtain:

η = b (|λ| ln 2 + 1 − θ),

ℓ = b (|λ| ln 2 + 3 (2 − θ)) and ℓ0 =
1

2
b (2 |λ| ln 2 + 3 (3 − θ)).

Note that ℓ0 < ℓ for all θ ∈ [0, 1].

Received: October, 2008. Revised: January, 2009.

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