View of Convergence analysis for the two-step Newton method of order four

Rev. Anal. Num´er. Th´eor. Approx., vol. 43 (2014) no. 1, pp. 33–44 ictp.acad.ro/jnaat

CONVERGENCE ANALYSIS

FOR THE TWO-STEP NEWTON METHOD OF ORDER FOUR

IOANNIS K. ARGYROS^∗and SANJAY K. KHATTRI^†

Abstract. We provide a tighter than before convergence analysis for the two- step Newton method of order four using recurrent functions. Numerical examples are also provided in this study.

MSC 2000. 65H10; 65G99; 65J15; 47H17; 49M15.

Keywords. Two-step Newton method, Newton’s method, Banach space, Kan- torovich hypothesis, majorizing sequence, Lipschitz/center-Lipschitz condition.

1. INTRODUCTION

In this study, we are concerned with the problem of approximating a locally unique solutionx^? of equation

(1.1) F(x) = 0,

where, F is Fr´echet-differentiable operator defined on a convex subset D of a Banach spaceX with values in a Banach space Y.

Many problems in computational mathematics can be brought in the form (1.1). The solutions of these equations are rarely found in closed form. There- fore most solution methods for these equations are iterative. Newton’s method (1.2) xn+1=xn− F⁰(xn)⁻¹F(xn) (n≥0), (x0∈ D)

is undoubtedly the most popular method for generating a sequence{x_n} con- verging quadratically to x^? [5, 13, 15]. Two-step Newton method (TSNM)

yn=xn− F⁰(xn)⁻¹F(xn) (n≥0), (x0 ∈ D), x_n+1=y_n− F⁰(y_n)⁻¹F(y_n),

(1.3)

generates a converging sequence{x_n}tox^?with order four [5, 9]. The following conditions have been used to show the semilocal convergence for the Newton’s

∗Department of Mathematical Sciences, Cameron University, Lawton, Oklahoma 73505- 6377, USA, e-mail: [email protected].

†Department of Engineering, Stord Haugesund University College, Norway, e-mail:

[email protected].

method (1.2) and consequently the semilocal convergence of (TSNM) [5, 13, 15, 17] (C_K):

F⁰(x₀)⁻¹ ∈L(Y,X) for somex₀∈ D;

F⁰(x0)⁻¹F(x0)≤ν

F⁰(x₀)⁻¹F⁰(x)− F⁰(y)≤Lkx−yk for all x, y∈ D;

h_K =Lη≤ ¹₂ (1.4)

and

U(x₀, λ) =x∈ X kx−x₀k ≤λ ⊆ D, for specifiedλ≥0.

Note that (1.4) is the, famous for its simplicity and clarity, Kantorovich sufficient convergence hypothesis for the Newton’s method (1.2). A current survey on Newton-type methods can be found in [5] and the references therein (see also [1–4] and [6–17]). We have shown [5] the quadratic convergence of the Newton’s method (1.2) using the set of conditions (C_AH)

F⁰(x0)⁻¹ ∈L(Y,X) for somex0 ∈ D;

F⁰(x₀)⁻¹F(x₀)≤η

F⁰(x₀)⁻¹F⁰(x)− F⁰(x₀)≤L₀kx−x₀k for allx∈ D;

F⁰(x0)⁻¹F⁰(x)− F⁰(y)≤Lkx−yk for allx, y∈ D;

h_AH =Lη ≤ ¹₂ (1.5)

and

U(x0, λ0)⊆ D, for some specifiedλ₀≥0, where

L= ¹₈L+ 4L₀+^pL²+ 8L₀L. (1.6)

Note that

(1.7) L₀ ≤L

holds in general, and L/L0 can be arbitrarily large [4, 5]. Moreover, the L0

Center-Lipschitz is not an additional condition, sinceL₀ is a special case ofL.

Furthermore, we have by (1.4)-(1.7)

(1.8) h_K≤ ¹₂ =⇒ h_AH ≤ ¹₂

but not necessarily vice versa unless ifL0 =L. The error analysis under (1.5) is also tighter than (1.4). Hence, the applicability of Newton’s method (1.2) has been extended.

In this study, we provide the sufficient convergence conditions for (TSNM) corresponding to (1.4). The paper is organized as follows: §2 contains the

semilocal convergence analysis for (TSNM), whereas the numerical examples are given in§3.

2. SEMILOCAL CONVERGENCE ANALYSIS FOR (TSNM)

We need the following result on majorizing sequence for (TSNM).

Lemma 1. Let L₀, L, η be constants. Assume: there exist parameters α and φsuch that

Lη

2(1−L₀η) ≤α, (2.1)

L1η

2(1−L₂η) ≤φ≤φ0

(2.2) and

η ≤η0

(2.3) where,

L₁ =α²L, L₂= (1 +α)L₀, (2.4)

φ1 = ^4L0α

2(L0+L2)α−L+

√

[2(L0+L)α−L]²+8L0Lα, (2.5)

φ₂= ^2L1

L1+√

L²₁+8L1L2

, φ₃= ^2α[1−(L_Lη^0+L2)η], (2.6)

φ₀ = min{φ₁, φ₂, φ₃}, (2.7)

η₁= _L ²

1+2L2(1+φ), η₂= _L ¹

0+L2, (2.8)

η0 = min{η₁, η2}. (2.9)

Then, sequences {s_n}, {t_n} generated by

(2.10) t₀ = 0, s₀ =η, t_n+1 =s_n+^L(s_2(1−L^n−tn)2

0sn), sn+1=tn+1+^L(t_2(1−L^n+1−sn)2

0tn+1), are non-decreasing, bounded from above by

(2.11) t^??=^1+α_1−φη,

and converge to their common least upper bound t^? ∈ [0, t^??]. Moreover, the following estimates holds

0≤t_n+1−s_n ≤α(s_n−t_n), (2.12)

and

0≤s_n+1−t_n+1≤φ(s_n−t_n).

(2.13)

Proof. We shall show using induction onk:

0≤ _2(1−L^L(sk−tk)

0s_k) ≤α, (2.14)

and

0≤ _2(1−L^L1(sk−tk)

0tk+1)≤φ.

(2.15)

Note that estimates (2.12) and (2.13) will then follow from (2.14) and (2.15), respectively. Estimates (2.14) and (2.15) hold by the left hand side hypotheses in (2.1) and (2.2), respectively. It follows from (2.10), (2.14) and (2.15) that estimates (2.12) and (2.13) hold forn= 0. Let us assume estimates (2.14) and (2.15) hold for allk≤n. It then follows that estimates (2.12) and (2.13) hold forn=k. We then have:

0≤s_k−t_k≤φ(sk−1−tk−1)≤φ·φ(sk−2−tk−2)≤ · · · ≤φ^kη, (2.16)

0≤t_k+1−s_k≤α(s_k−t_k)≤αφ^kη, (2.17)

and

t_k+1≤s_k+αφ^kη≤t_k+αφ^kη+φ^kη

≤sk−1+αφ^k−1η+αφ^kη+φ^kη

≤tk−1+φ^k−1η+αφ^k−1η+αφ^kη+φ^kη

=tk−1+ (φ^k−1+φ^k)η+α(φ^k−1+φ^k)η≤ · · ·

≤s0+α(η+φη+· · ·+φ^kη) +α(φη+· · ·+φ^kη)

= (1 +α)(1 +φ+· · ·+φ^kη)≤t^??. (2.18)

In view of (2.16) and (2.18), estimate (2.14) certainly holds if

0≤ ^Lφkη

2[^1−L2(1+φ+···+φ^k−1)η−L0t^k−1η] ≤α, (2.19)

or

Lφ^kη+ 2αL2(1 +φ+· · ·+φ^k−1)η−2α+ 2L0αt^k−1η≤0.

(2.20)

Estimate (2.20) motivates us to introduce functionsf_k on [0,1) by (2.21) f_k(t) =Lηt^k+ 2αL2(1 +t+· · ·+t^k−1)η+ 2L0αt^k−1η−2α.

We need a relationship between two consecutive functionsf_k:

f_k+1(t) =Lt^k+1η+ +2αL₀t^kη+ 2αL₂(1 +t+· · ·+t^k)η−2α−Lt^kη

−2αL₂(1 +t+· · ·+t^k−1)η−2L₀αt^k−1η+ 2α+f_k(t)

=f_k(t) +Lt^k+1η−Lt^kη+ 2αL₂t^kη+ 2L₀αt^kη−2L₀αt^k−1η

=f_k(t) +g(t)t^k−1η, (2.22)

where

g(t) =Lt²+ [2α(L2+L0)−L]t−2L0α.

(2.23)

Using (2.21), we see that (2.20) holds

if fk(φ)≤0 (2.24)

or f₁(φ)≤0, (2.25)

since, g(φ)≤0 and f_k+1(φ) =f_k(φ) +g(φ)φ^kη≤f_k(φ) (2.26)

whereφis chosen as in the right hand side inequality of (2.1). But (2.23) also holds by (2.1). Moreover, define function f∞ on [0,1) by

f∞(t) = lim

k→∞f_k(t).

(2.27)

Then, we have by (2.24)

f∞(φ)≤0.

Hence, (2.12) and (2.14) hold for all k. Similarly, (2.15) holds if L₁φ^kη ≤2φ^h1−L₂(1 +φ+· · ·+φ^k)ηⁱ (2.28)

or

L₁φ^kη+ 2φL₂(1 +φ+· · ·+φ^k)η−2φ≤0.

(2.29)

As in (2.21) we define functions h_k on [0,1) by

(2.30) h_k(t) =L₁t^kη+ 2tL₂(1 +t+· · ·+t^k)η−2φ.

We need a relationship between two consecutive functionsh_k:

h_k+1(t) =L₁t^k+1η+ 2tL₂(1 +t+· · ·+t^k+1)η−2φ−L₁t^kη−

−2tL₂(1 +t+· · ·+t^k)η+ 2φ+h_k(t)

=h_k(t) +L₁t^k+1η−L₁t^kη+ 2L₂t^k+2η

=h_k(t) +g1(t)t^kη (2.31)

where

g₁(t) = 2L₂t²+L₁t−L₁. (2.32)

In view of (2.30), estimate (2.29) holds if

(2.33) if hk(φ)≤0 or h1(φ)≤0

since, g₁(φ)≤0 and h_k+1(φ) =h_k(φ) +g₁(φ)φ^kη ≤h_k(φ) (2.34)

where φ is chosen as in the right hand side of (2.2). Note now that (2.33) holds by (2.3). Furthermore, define functionsh∞ on [0,1) by

(2.35) h∞(t) = lim

k→∞h_k(t).

We then have

(2.36) h∞(φ)≤0.

That completes the induction for (2.13) and (2.15). Finally, in view of (2.12), (2.13) and (2.18), sequences {t_n}, {s_n} converge to t^?. That completes the

proof of the Lemma.

We need an Ostrowski-type relationship between iterates {x_n} and {y_n} [5, 14].

Lemma2. Let us assume iterates{x_n}and{y_n}in (TSNM) are well defined for all n≥0. Then, the following identities hold:

F(x_n+1) = Z 1

0

F⁰(y_n+θ(x_n+1−y_n))− F⁰(y_n)(x_n+1−y_n)dθ, (2.37)

and

F(y_n) = Z 1

0

F⁰(x_n+θ(y_n−x_n))− F⁰(x_n)(y_n−x_n)dθ.

(2.38)

Proof. Identity (2.37) follows from the Taylor’s theorem and the first itera- tion in (TSNM), whereas (2.38) follows from Taylor’s theorem and the second iteration in (TSNM). That completes the proof of the Lemma.

We can show the following semilocal convergence result for (TSNM).

Lemma 3. Let F :D ⊂ X → Y be Fr´echet-differentiable operator. Assume:

there exist x0 ∈ D, L0>0,L >0 and η ≥0 such that for allx, y∈ D:

F⁰(x₀)⁻¹ ∈L(Y,X), (2.39)

F⁰(x₀)⁻¹F(x₀)≤η, (2.40)

F⁰(x₀)⁻¹ F⁰(x)− F⁰(x₀)≤L₀kx−x₀k, (2.41)

F⁰(x0)⁻¹ F⁰(x)− F⁰(y)≤Lkx−yk, (2.42)

U(x0, t^?)⊆ D;

(2.43)

hypotheses of Lemma 2.1 hold, where t^? is given in Lemma 2.1. Then, se- quences {x_n} and {y_n} generated by (TSNM) are well defined, remain in U(x0, t^?) for all n ≥ 0 and converge to a solution x^? ∈ U(x0, t^?) of equa- tion F(x) = 0. Moreover, the following estimates hold

ky_n−xnk ≤sn−tn, (2.44)

kx_n+1−ynk ≤tn+1−sn, (2.45)

kx_n+1−xnk ≤tn+1−tn, (2.46)

ky_n+1−y_nk ≤s_n+1−s_n, (2.47)

kx_n−x^?k ≤t^?−t_n, (2.48)

ky_n−x^?k ≤t^?−sn. (2.49)

Furthermore, if there exists R≥t^? such that U(x0, R)⊆ D (2.50)

and

L0(t^?+R)<2, (2.51)

then,x^? is the only solution ofF(x) = 0 in U(x0, R)

Proof. We shall show using induction onkthat (TSNM) is well defined, the iterates remain in U(x0, t^?) for alln≥0 and estimates (2.44) and (2.45) hold for all n≥ 0. Iterate y₀ is well defined by the first equation in (TSNM) for n= 0 and (2.39). We also have by (2.6) and (2.40)

ky₀−x₀k=

F⁰(x₀)⁻¹F(x₀)

≤η=s₀ =s₀−t₀≤t^?.

That is (2.44) holds for n= 0 and y0 ∈U(x0, t^?). Using (TSNM) for n= 0, we see that x1 is well defined. Let w∈ U(x0, t^?). Then, we have by Lemma 2.1 and (2.41):

(2.52) F⁰(x₀)⁻¹F⁰(w)− F⁰(x₀)≤L₀kw−x₀k ≤L₀t^? <1.

It follows from (2.52) and the Banach lemma on invertible operators [5, 13, 15]

that F⁰(w)⁻¹ exists and

F⁰(w)⁻¹F⁰(x0)≤ _1−L ¹

0kw−x0k. (2.53)

In particular, for x1 ∈U(x0, t^?), we have

F⁰(x₁)⁻¹F⁰(x₀)≤ _1−L ¹

0kx₁−x₀k ≤ _1−L ¹

0(t1−t₀) = _1−L¹

0t1. (2.54)

Moreover, in view of (2.38) forn= 0, (TSNM), (2.6) and (2.40)-(2.42), we get

kx₁−y₀k= (2.55)

=

Z 1 0

hF⁰(y0)⁻¹F⁰(x0)ⁱF⁰(x0)⁻¹F⁰(x0+θ(y0−x0))−F⁰(x0)dθ(y0−x0)

≤ _1−L ^L0

0ky0−x0k

Z 1 0

θky₀−x0k²dθ

= _2(1−L^L0

0ky₀−x₀k)ky₀−x₀k²

≤ _2(1−L^L0

0s0)(s₀−t₀)²=t₁−s₀,

which shows (2.45) forn= 0. We also have

kx₁−x₀k ≤ kx₁−y₀k+ky₀−x₀k ≤t₁−s₀+s₀−t₀ =t₁−t₀ ≤t^?, which implies (2.46) holds for n= 0 and x₁ ∈U(x₀, t^?).

Using (TSNM), (2.6), (2.37) (forn= 0) and (2.54), we get ky₁−x₁k=^hF⁰(x₁)⁻¹F⁰(x₀)^{i h}F⁰(x₀)⁻¹F(x₁)ⁱ

≤F⁰(x1)⁻¹F⁰(x0)F⁰(x0)⁻¹F(x1)

≤ _1−L¹

0t1

Z 1 0

F⁰(x₀)⁻¹F⁰(y₀+θ(x₁−y₀))− F⁰(y₀)dθ(x₁−y₀)

≤ _1−L^L0

0t1

Z 1 0

θkx₁−y₀kdθkx₁−y₀k

≤ _1−L^L

0t1

1

2(t₁−s₀)(t₁−s₀) =s₁−t₁, which implies (2.44) for n= 1. We then have:

ky₁−y₀k ≤ ky₁−x₁k+kx₁−y₀k ≤s₁−t₁+t₁−s₀ =s₁−s₀, ky₁−x0k ≤ ky₁−y0k+ky₀−x0k ≤s1−s0+s0−t0 =s1≤t^?, which imply (2.47) for n= 0 and y₁ ∈ U(x₀, t^?). Let us now assume (2.44)- (2.47), y_n, x_k ∈ U(x₀, t^?) for all n≤ k. Using (TSNM), (2.6), (2.37), (2.38), (2.42), (2.53) and the induction hypotheses, we have in turn:

kx_k+1−x₀k ≤ kx_k+1−x_kk+kx_k−xk−1k+· · ·+kx₁−x₀k

≤tk+1−tk+tk−tk−1+· · ·+t1−t0 =tk+1≤t^?, (2.56)

ky_k−x0k ≤ ky_k−x_kk+kx_k−x0k (2.57)

≤s_k−t_k+t_k−t₀

=s_k≤t^?

ky_k+1−x_k+1k= (2.58)

=F⁰(xk+1)⁻¹F⁰(x0)F⁰(x0)⁻¹F(xk+1)

≤F⁰(x_k+1)⁻¹F⁰(x₀)F⁰(x₀)⁻¹F(x_k+1)

≤ _1−L ¹

0kxk+1−x0k

Z ₁

0

F⁰(x₀)⁻¹F⁰(y_k+θ(x_k+1−y_k))− F⁰(y_k)dθ(x_k+1−y_k)

≤ _1−L^L

0t_k+1

Z 1 0

θkx_k+1−ykk²dθ

≤ _1−L^L

0tk+1

1

2(t_k+1−s_k)²

=sk+1−tk+1,

kx_k+2−y_k+1k=F⁰(y_k+1)⁻¹F⁰(x₀)F⁰(x₀)⁻¹F(y_k+1)≤ (2.59)

≤ _1−L¹

0s_k+1

Z 1 0

F⁰(x0)⁻¹F⁰(xk+1+θ(yk+1−x_k+1))−F⁰(xk+1)dθ(yk+1−x_k+1)

≤ _1−L^L

0sk+1

Z 1 0

θky_k+1−x_k+1k²dθ

≤ _2(1−L^L

0sk+1)(s_k+1−t_k+1)²=t_k+2−s_k+1,

ky_k+2−yk+1k ≤ ky_k+2−xk+2k+kx_k+2−yk+1k

≤sk+2−tk+2+tk+2−sk+1 =sk+2−sk+1, (2.60)

kx_k+2−x_k+1k ≤ kx_k+2−y_k+1k+ky_k+1−x_k+1k

≤t_k+2−s_k+1+s_k+1−t_k+1 =t_k+2−t_k+1 (2.61)

which show (2.44)-(2.47) hold for alln≥0. Estimates (2.48) and (2.49) follow from (2.46) and (2.47), respectively by using standard majorization technique [5, 13, 15]. It follows from Lemma 2.1 and (2.44)-(2.48) that (TSNM) is Cauchy in a Banach space X and as such it converges to somex^? ∈U(x₀, t^?) (since U(x₀, t^?) is a closed set). Moreover, we have by (2.58)

F⁰(x0)⁻¹F(xk+1)≤ ^L₂kx_k+1−ykkkx_k+1−ykk →0, as k→ ∞.

(2.62)

That isF(x^?) = 0.Finally to show uniqueness, lety^?∈U(x₀, R) be a solution of equation F(x) = 0.Let us define linear operatorM by

(2.63) M =

Z 1 0

F⁰(y^?+θ(x^?−y^?))dθ.

Then using (2.41), (2.50) and (2.51), we get in turn

F⁰(x0)M− F⁰(x0)≤L0

Z 1 0

ky^?+θ(x^?−y^?)−x0kdθ

≤L₀ Z 1

0

[(1−θ)ky^?−x₀k+θkx^?−x₀k] dθ

≤ ^L₂⁰(R+t^?)<1.

(2.64)

It follows from (2.60) and the Banach Lemma on invertible operators that M⁻¹ exists. Then, in view of the identity

(2.65) 0 =F(x^?)− F(y^?) =M(x^?−y^?),

we conclude that x^? =y^?. That completes the proof of the Theorem.

Remark 4. 1) Limit point t^? can be replaced by t^??, given in closed form by (2.7), in hypotheses (2.40) and (2.48).

2) The verification of conditions (2.1)-(2.3) require simple algebra (see also Example 3.1).

3) If L0 =L, then scalar sequences {s_n},{t_n} given by (2.6) reduce essen- tially to the ones used in [9]. In particular, we have in this case

(2.66) t₀ = 0, s₀ =η, t_n+1 =s_n+^L(s_2(1−Ls^n−tn)2

n), sn+1=tn+1+^L(t_2(1−Lt^n+1−sn)2

n+1)

IfL0 < Literation (2.6) is tighter than (2.62). Moreover, in view of the proof of the Theorem 2.3, we note that sequence

(2.67)

t₀= 0, s₀ =η, t_n+1 =s_n+^L?(sn−tn)2

2(1−L0sn), s_n+1 =t_n+1+^{L?(tn+1−sn)2}

2(1−L0tn+1), is also majorizing for (TSNM), where

L^?=

L₀, if n= 0 L, if n >0.

In case L₀ < L, (2.26) is even a tighter majorizing sequence than (2.62).

Furthermore,L, L1 can be replaced byL0, L^?₁ =α²L0 at the left hand sides of (2.1) and (2.2), respectively.

4) Ifα= 0, defineL₁ =L, then it is simple algebra to show that conditions of Lemma 2.1 reduce to (1.5). Moreover, ifL0=L, these conditions reduce to (1.4). That is we have Newton’s method (1.2) and iteration (2.6) reduces to (2.68) t₀ = 0, t₁=η, t_n+2=t_n+1+^L(t_2(1−L^n+1−tn)2

0tn+1).

In the case of Newton’s method for L0 = L, we have the well-known Kan- torovich majorizing sequence.

(2.69) ν₀= 0, ν₁ =η, ν_n+2 =ν_n+1+^L(ν_2(1−L^n+1−νn)2

0νn+1).

Note that if L0 < L,{t_n} is a tighter majorizing sequence than {ν_n} for the

Newton’s method [5, 13, 15].

3. NUMERICAL EXAMPLES

Let X = Y = R² be equipped with the max-norm, x₀ = (1,1)^T, D = U(x₀,1−p),p∈[0,1) and defineF on Dby

(3.1) F(x) =ξ₁³−p, ξ₂³−p^T , x= (ξ1, ξ2)^T . Using (2.35)-(2.37), we get

η = ^1−p₃ , L0 = 3−p and L= 2(2−p)> L0. Letp= 0.7. Then, we get

η = 0.1, L0 = 2.3 and L= 2.6.

The Newton-Kantorovich hypothesis (1.4) is satisfied, since

2

3(1−p)(2−p) = 0.26<1 for all p∈[0,1/2).

Using Lemma 2.1, forα = 0.17 and φ= 0.0052, we get

L₁ = 0.07514, L₂= 2.691 φ = 0.756703694, φ2 = 0.111383518, φ3= 0.666923077, φ0 = φ2

η₁ = 0.364622409, η₂= 0.200360649, η₀ = η₂,

Lη/[2(1−L₀η)] = 0.168831169 and L₁η/[2(1−L₂η)] = 0.005140238.

Hence, the hypotheses of Lemma 2.1 are satisfied. Moreover, we have by (2.11) that

t^??= 0.11761158<1−p= 0.3.

Furthermore, using (2.48) (fort^? replaced byt^??), we get t^??< R < _L²

0 −t^??= 0.751953637.

So, we can choose R = 0.3. Hence, hypotheses of Theorem 2.3 hold, and (TSNM) converges to

x^?=√³ 0.7,√³

0.7^T = (0.887904002,0.887904002)^T . We compare (2.6) to (2.62).

n sn−tn tn+1−sn sn−tn tn+1−sn sn−tn tn+1−sn

0 1.00·10⁻⁰¹ 1.69·10⁻⁰² 1.00·10⁻⁰¹ 1.76·10⁻⁰² 1.00·10⁻⁰¹ 1.49·10⁻⁰² 1 5.07·10⁻⁰⁴ 4.57·10⁻⁰⁷ 5.78·10⁻⁰⁴ 6.27·10⁻⁰⁷ 3.49·10⁻⁰⁴ 2.15·10⁻⁰⁷ 2 3.73·10⁻¹³ 2.47·10⁻²⁵ 5.37·10⁻¹³ 1.02·10⁻²⁴ 8.19·10⁻¹⁴ 1.19·10⁻²⁶ 3 1.09·10⁻⁴⁹ 2.11·10⁻⁹⁸ 1.94·10⁻⁴⁸ 1.09·10⁻⁹⁶ 2.49·10⁻⁵² 1.09·10⁻¹⁰³ 4 7.91·10⁻¹⁹⁶ 1.11·10⁻³⁹⁰ 9.44·10⁻¹⁹¹ 1.67·10⁻³⁸⁰ 2.11·10⁻²⁰⁶ 7.88·10⁻⁴¹² 5 2.21·10⁻⁷⁸⁰ 8.70·10⁻¹⁵⁶⁰ 5.24·10⁻⁷⁶⁰ 5.15·10⁻¹⁵¹⁹ 1.09·10⁻⁸²² 2.13·10⁻¹⁶⁴⁴

Table 1. Comparison among (2.6), (2.66) and (2.67)

As expected from the theoretical results iteration (2.6) is faster than (2.66).

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Received by the editors: September 12, 2012.