# Semilocal Convergence for  a Fifth-Order Newton's Method Using Recurrence Relations in Banach Spaces

Authors: Liang Chen; Chuanqing Gu; Yanfang Ma
Journal: Journal of Applied Mathematics
 (2011)

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## Abstract

We study a modified Newton's method with fifth-order convergence for nonlinear equations in Banach spaces. We make an attempt to establish the semilocal convergence of this method by using recurrence relations.
The recurrence relations for the method are derived, and then an existence-uniqueness theorem is given to establish the
R-order of the method to be five and a priori error bounds. Finally, a numerical application is presented to demonstrate
our approach.


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## Body

## 1. Introduction

Many scientific problems can be expressed in the form of a nonlinear equation(1.1)F(x)=0,
where F:Ω⊂X→Y is a nonlinear operator on an open convex subset Ω of a Banach space X with values in a Banach space Y.The Newton’s method [1, 2], which has quadratically order convergence, is one of the most well-known methods for solving nonlinear equations. Recently, numerous variants of Newton’s method with high-order convergence are developed in the literature [3–7]; these methods improve the local order of convergence of Newton’s method by an additional evaluation of the function. In this paper, we consider the semilocal convergence for the method proposed in [7]. We first extend this method to Banach spaces and write it as (1.2)un=xn-ΓnF(xn),zn=xn-[F′(yn)]-1F(xn),JF(xn,yn)=[3F′(yn)-2F′(xn)]-1F′(yn),xn+1=zn-JF(xn,yn)ΓnF(zn),
where I is the identity operator on X, Γn=[F′(xn)]-1, yn=xn+(1/2)(un-xn).In many papers, the convergence of iterative methods for solving nonlinear operator equation in Banach spaces is established from the convergence of majorizing sequences, which is obtained by applying the real function to a polynomial [8–10]. An alternative approach is developed to establish this convergence by using recurrence relations. For example, the recurrence relations are used in establishing the convergence of Newton’s method [2] and some high-order methods [11–17]. In this paper, we consider the semilocal convergence of the method given by (1.2) using the recurrence relations. We construct the system of recurrence relations and prove the convergence of the method, along with an error estimate. Finally, numerical results are presented to demonstrate our approach.


## 2. Preliminary Results

Letx0∈Ω, and the nonlinear operator F:Ω⊂X→Y be continuously second-order Fréchet differentiable, where Ω is an open set and X and Y are Banach spaces. We assume that (C1)
∥Γ0F(x0)∥≤η,(C2)
∥Γ0∥≤β,(C3)
∥F′′(x)∥≤M,x∈Ω,(C4)
there exists a positive real numberN such that(2.1)‖F′′(x)-F′′(y)‖≤N‖x-y‖,∀x,y∈Ω.Now, we define the following scalar functions which will be often used in the later developments. Let(2.2)g(t,s)=2-t2-3t[t32(2-t)2+t24-2t+34s]+22-t,h(t,s)=11-tg(t,s),φ(t,s)=t2(1-t)2[t32(2-t)2+t24-2t+34s]2+(7s8(1-t)+t2(2-t)(1-t)+2t22-3t)[t32(2-t)2+t24-2t+34s].
Let p(t)=tg(t,s)-1, and let σ(s) be the smallest positive zero of the scalar function p(t) for s∈(0,5). Then, using MATLAB, we obtain that σ(s) is decreasing and σ(s)>0 for all s∈(0,5) and (2.3)max⁡0<s<5σ(s)≈0.6117085589952558<23.
Some properties of the functions defined above are given in the following lemma.Lemma 2.1.
Let the real functionsg,h, and  φ be given in (2.2). Then  (a)
g(t,s) and h(t,s) are increasing and g(t,s)>1, h(t,s)>1 for all t∈(0,σ(s)) and s∈(0,5),(b)
φ(t,s) is increasing for all t∈(0,σ(s)) and s∈(0,5),(c)
g(θt,θ2s)<g(t,s), h(θt,θ2s)<h(t,s), and φ(θt,θ2s)<θ4φ(t,s) for θ∈(0,1), t∈(0,σ(s)), and s∈(0,5).Assume that the conditions (C1)–(C4) hold. we now denote η0=η, β0=β, a0=Mβ0η0, b0=Nβ0η02, and c0=h(a0,b0)φ(a0,b0). Let a0<σ(s) and h(a0,b0)c0<1, then we can define the following sequences for n≥0:(2.4)ηn+1=cnηn,(2.5)βn+1=h(an,bn)βn,(2.6)an+1=Mβn+1ηn+1,(2.7)bn+1=Nβn+1ηn+12,(2.8)cn+1=h(an+1,bn+1)φ(an+1,bn+1).
From the definition of an+1, bn+1, (2.4) and (2.5), we also have (2.9)an+1=h(an,bn)cnan,(2.10)bn+1=h(an,bn)cn2bn.Nextly we will study some properties of the previous scalar sequences defined in (2.4)–(2.10), and later developments will require the following lemma.Lemma 2.2.
Let the real functionsg,h, and  φ be given in (2.2). If
(2.11)a0<σ(s),h(a0,b0)c0<1,
then one has (a)
h(a0,b0)>1 and cn<1 for n≥0,(b)
the sequences{ηn}, {an}, {bn}, and {cn} are decreasing while {βn} is increasing,(c)
g(an,bn)an<1 and h(an,bn)cn<1 for n≥0.Proof.
By Lemma2.1 and (2.11), h(a0,b0)>1 and cn<1 hold. It follows from the definitions that η1<η0, a1<a0, b1<b0. Moreover, by Lemma 2.1, we have 1<h(a1,b1)<h(a1,b0)<h(a0,b0) and φ(a1,b1)<φ(a1,b0)<φ(a0,b0). This yields c1<c0, β1>β0, and (b) holds. Based on these results we obtain g(a1,b1)a1<g(a0,b0)a0<1 and h(a1,b1)c1<h(a0,b0)c0<1 and (c) holds. By induction we can derive that the items (a), (b), and (c) hold.Lemma 2.3.
Under the assumptions of Lemma2.2 and defining  γ=h(a0,b0)c0, then
(2.12)cn≤λγ5n,n≥0,
where λ=1/h(a0,b0). Also, for n≥0, one has
(2.13)∏i=0nci≤λn+1γ(5n+1-1)/4.Proof.
By the definition ofan+1 and bn+1 given in (2.9)-(2.10), we obtain a1=h(a0,b0)c0a0=γa0, b1=h(a0,b0)c02b0<γ2b0; by Lemma 2.1 we have
(2.14)c1<h(γa0,γ2b0)φ(γa0,γ2b0)<γ4h(a0,b0)φ(a0,b0)=γ51-1c0=λγ51.
Suppose ck≤λγ5k, k≤1. Then, by Lemma 2.2, we have ak+1<ak, bk+1<bk and h(ak,bk)ck<1. Thus,
(2.15)ck+1<h(ak,bk)φ(h(ak,bk)ckak,h(ak,bk)ck2bk)<h(ak,bk)φ(h(ak,bk)ckak,h2(ak,bk)ck2bk)<h5(ak,bk)ck4φ(ak,bk)=h4(ak,bk)ck5<λγ5k+1.
Therefore, it holds that cn≤λγ5n, n≥0.
By (2.12), we get(2.16)∏i=0nci≤∏i=0nλγ5i=λn+1γ∑i=0n5i=λn+1γ(5n+1-1)/4,n≥0.
This shows that (2.13) holds. The proof is completed.Lemma 2.4.
Under the assumptions of Lemma2.2, let  γ=h(a0,b0)c0 and  λ=1/h(a0,b0). The sequence  {ηn} satisfies
(2.17)ηn≤ηλnγ(5n-1)/4,n≥0.
Hence, the sequence {ηn} converges to 0. Moreover, for any n≥0, m≥1, it holds that
(2.18)∑i=nn+mηi≤ηλnγ(5n-1)/41-λm+1γ5n(5m+3)/41-λγ5n.Proof.
From the definition of sequence{ηn} given in (2.4) and (2.13), we have
(2.19)ηn=cn-1ηn-1=cn-1cn-2ηn-2=⋯=η(∏i=0n-1ci)≤ηλnγ(5n-1)/4.
Because λ<1 and γ<1, it follows that ηn→0 as n→∞; hence, the sequence {ηn} converges to 0.
Since(2.20)∑i=nn+mλiγ5i/4≤λnγ5n/4+γ5n(∑i=n+1n+mλiγ5i-1/4)=λnγ5n/4+λγ5n(∑i=nn+m-1λiγ5i/4)=λnγ5n/4+λγ5n(∑i=nn+mλiγ5i/4-λn+mγ5n+m/4),
where n≥0, m≥1, we can obtain
(2.21)∑i=nn+mλiγ5i/4≤λnγ5n/41-λm+1γ5n(5m+3)/41-λγ5n.
Furthermore,
(2.22)∑i=nn+mηi≤η∑i=nn+mλiγ(5i-1)/4=ηγ-1/4∑i=nn+mλiγ5i/4≤ηλnγ(5n-1)/41-λm+1γ5n(5m+3)/41-λγ5n.
Therefore, ∑n=0∞ηn exists. The proof is completed.


## 3. Recurrence Relations for the Method

We firstly give an approximation of the operatorF in the following lemma, which will be used in the next derivation.Lemma 3.1.
Assume that the nonlinear operatorF:Ω⊂X→Y is continuously second-order Fréchet differentiable, where  Ω is an open set and  X and  Y are Banach spaces. Then, one has
(3.1)F(zn)=∫01F′′(un+t(zn-un))(1-t)dt(zn-un)2-∫01F′′(yn+t(un-yn))(un-yn)dt[F′(yn)]-1[F′(yn)-F′(xn)](un-xn)+∫01[F′′(xn+t(un-xn))-F′′(xn)](1-t)dt(un-xn)2+12∫01[F′′(xn)-F′′(xn+12t(un-xn))]dt(un-xn)2,(3.2)F(xn+1)=∫01[F′′(xn+t(un-xn))-F′′(xn)](un-xn)dt(xn+1-zn)-32∫01[F′′(xn+12t(un-xn))-F′′(xn)](un-xn)dt(xn+1-zn)-2F′′(xn)(yn-xn)[I+3H(xn,yn)]-1H(xn,yn)ΓnF(zn)-12∫01[F′′(xn+12t(un-xn))-F′′(xn)](un-xn)dtΓnF(zn)+∫01[F′(zn+t(xn+1-zn))-F′(un)]dt(xn+1-zn),
where H(xn,yn)=Γn[F′(yn)-F′(xn)] and JF(xn,yn)=I-2[I+3H(xn,yn)]-1H(xn,yn).Proof.
By the Taylor Expansion, we obtain(3.3)F(zn)=F(un)+F′(un)(zn-un)+∫01F′′(un+t(zn-un))(1-t)dt(zn-un)2,(3.4)F(un)=F(xn)+F′(xn)(un-xn)+12F′′(xn)(un-xn)2+∫01[F′′(xn+t(un-xn))-F′′(xn)](1-t)dt(un-xn)2,(3.5)F′(un)=F′(yn)+∫01F′′(yn+t(un-yn))(un-yn)dt,F′(yn)=F′(xn)+12F′′(xn)(un-xn)+12∫01[F′′(xn+12t(un-xn))-F′′(xn)]dt(un-xn),
and we obtain
(3.6)F′(un)(zn-un)=F′(yn)(zn-un)+∫01F′′(yn+t(un-yn))(un-yn)dt(zn-un),(3.7)[F′(yn)-F′(xn)](un-xn)=12F′′(xn)(un-xn)2+12∫01[F′′(xn+12t(un-xn))-F′′(xn)]dt(un-xn)2.
By the first two steps of method given in (1.2) and (3.4), (3.6), (3.7), we obtain
(3.8)F′(un)(zn-un)=-12F′′(xn)(un-xn)2+12∫01[F′′(xn)-F′′(xn+12t(un-xn))]dt(un-xn)2+∫01F′′(yn+t(un-yn))(un-yn)dt(zn-un).
Substituting (3.4) and (3.8) into (3.3), we obtain (3.1).
We now considerF(xn+1). Since(3.9)[3F′(yn)-2F′(xn)](xn+1-zn)+[F′(yn)-F′(xn)]ΓnF(zn)+F(zn)=0,
using Taylor’s formula, we have
(3.10)F(xn+1)=F(zn)+F′(un)(xn+1-zn)+∫01[F′(zn+t(xn+1-zn))-F′(un)]dt(xn+1-zn)=[F′(un)-F′(yn)](xn+1-zn)-2[F′(yn)-F′(xn)](xn+1-zn)-[F′(yn)-F′(xn)]ΓnF(zn)+∫01[F′(zn+t(xn+1-zn))-F′(un)]dt(xn+1-zn).
Similarly, we obtain
(3.11)F′(un)=F′(xn)+F′′(xn)(un-xn)+∫01[F′′(xn+t(un-xn))-F′′(xn)](un-xn)dt,
It follows that
(3.12)F′(un)-F′(yn)=12F′′(xn)(un-xn)+∫01[F′′(xn+t(un-xn))-F′′(xn)](un-xn)dt-12∫01[F′′(xn+12t(un-xn))-F′′(xn)](un-xn)dt,F′(yn)-F′(xn)=12F′′(xn)(un-xn)+12∫01[F′′(xn+12t(un-xn))-F′′(xn)](un-xn)dt.
Substituting (3.12) into (3.10), we can obtain (3.2). The proof is completed.We denoteB(x,r)={y∈X:∥y-x∥<r} and B(x,r)¯={y∈X:∥y-x∥≤r} in this paper.In the following, the recurrence relations are derived for the method given by (1.2) under the assumptions mentioned in the previous section.Forn=0, the existence of Γ0 implies the existence of u0,y0. This gives us
(3.13)∥u0-x0∥=∥Γ0F(x0)∥≤η0,
and this means that u0,y0∈B(x0,Rη), where R=g(a0,b0)/(1-c0). By the initial hypotheses, we have
(3.14)∥H(x0,y0)∥≤∥Γ0∥∥F′(x0)-F′(y0)∥≤12M∥Γ0∥∥u0-x0∥≤12a0.
Because of the assumption a0<σ(s)<2/3, by the Banach lemma [2] it follows that (3.15)∥[I+3H(x0,y0)]-1∥≤11-(3/2)a0.
It is followed that (3.16)‖JF(x0,y0)‖≤1+2∥[I+3H(x0,y0)]-1∥∥H(x0,y0)∥≤1+2a02-3a0.Consequentlyz0 is well defined and (3.17)∥z0-x0∥≤∥[F′(y0)]-1F(x0)∥≤∥[F′(y0)]-1F′(x0)∥∥Γ0F(x0)∥≤22-a0η0.
It is similar to obtain (3.18)∥z0-u0∥=∥[F′(y0)-1-Γ0]F(x0)∥=∥F′(y0)-1[F′(x0)-F′(y0)]Γ0F(x0)∥≤∥F′(y0)-1∥∥[F′(x0)-F′(y0)]∥∥Γ0F(x0)∥≤a02-a0η0.
By Lemma 3.1, we can get (3.19)∥F(z0)∥≤12M∥z0-u0∥2+M2∥F′(y0)-1∥∥u0-y0∥∥y0-x0∥∥u0-x0∥+34N∥u0-x0∥3,(3.20)∥F(x1)∥≤12M∥x1-z0∥2+(M∥z0-u0∥+32N∥u0-x0∥2)∥x1-z0∥.
Therefore, we have
(3.21)∥x1-z0∥≤∥JF(x0,y0)∥∥Γ0∥∥F(z0)∥≤2-a02-3a0[a02(a02-a0)2+a022(2-a0)+34b0]η0,(3.22)∥x1-x0∥≤∥x1-z0∥+∥z0-x0∥≤g(a0,b0)η0.
From the assumption c0<1/h(a0,b0)<1, it follows that x1∈B(x0,Rη).Bya0<σ(s) and g(t,s) is increasing in t∈(0,σ(s)) and s∈(0,5), we have (3.23)∥I-Γ0F′(x1)∥≤∥Γ0∥∥F′(x0)-F′(x1)∥≤Mβ0∥x1-x0∥≤a0g(a0,b0)<1,
and it follows by the Banach lemma [2] that Γ1=[F′(x1)]-1 exists and (3.24)∥Γ1∥≤β01-a0g(a0,b0)=h(a0,b0)β0=β1.
Then, from (3.20) and (3.24), we have
(3.25)∥u1-x1∥=∥Γ1F(x1)∥≤∥Γ1∥∥F(x1)∥≤h(a0,b0)φ(a0,b0)η0=c0η0=η1.
Because of g(a0,b0)>1, we obtain (3.26)∥u1-x0∥≤∥u1-x1∥+∥x1-x0∥≤(g(a0,b0)+c0)η0<g(a0,b0)(1+c0)η<Rη,
which shows u1,y1∈B(x0,Rη).In addition, we have(3.27)M∥Γ1∥∥Γ1F(x1)∥≤h(a0,b0)c0a0=a1,N∥Γ1∥∥Γ1F(x1)∥2≤h(a0,b0)c02b0=b1.
Repeating the above derivation, we can obtain the system of recurrence relations given in the next lemma.Lemma 3.2.
Let the assumptions and the conditions  (C1)– (C4) hold. Then, the following items are true for all  n≥0: (i)
there existsΓn=[F′(xn)]-1 and ∥Γn∥≤βn,(ii)
∥ΓnF(xn)∥≤ηn,(iii)
M∥Γn∥∥ΓnF(xn)∥≤an,(iv)
N∥Γn∥∥ΓnF(xn)∥2≤bn,(v)
∥xn+1-xn∥≤g(an,bn)ηn,(vi)
∥xn+1-x0∥≤Rη, where R=g(a0,b0)/(1-c0).Proof.
The proof of (I)–(V) follows by using the above-mentioned way and invoking the induction hypothesis. We only consider (VI). By (V) and Lemma2.4, we obtain
(3.28)∥xn+1-x0∥≤∑i=0n∥xi+1-xi∥≤∑i=0ng(ai,bi)ηi≤g(a0,b0)∑i=0nηi≤g(a0,b0)η1-λn+1γ(5n+3)/41-c0<Rη.
So the lemma is proved.


## 4. Semilocal Convergence

Lemma 4.1.
LetR=g(a0,b0)/(1-c0). If  h(a0,b0)c0<1, then  R<1/a0.Proof.
Since(4.1)c0<1h(a0,b0)=1-a0g(a0,b0),
we obtain R<1/a0.Now we give a theorem to establish the semilocal convergence of (1.2), the existence and uniqueness of the solution, and the domain in which it is located, along with a priori error bounds, which lead to the R-order of convergence at least five of iteration (1.2).Theorem 4.2.
LetF:Ω⊂X→Y be a nonlinear two times Fréchet differentiable operator in an open convex subset  Ω of a Banach space  X with values in a Banach space  Y. Assume that  x0∈Ω and all conditions  (C1)– (C4) hold. Let  a0=Mβη,  b0=Nβη2, and  c0=h(a0,b0)φ(a0,b0) satisfy  a0<σ(s) and  h(a0,b0)c0<1, where  g,h, and φ are defined by (2.2). Let  B(x0,Rη)¯⊂Ω, where  R=g(a0,b0)/(1-c0), then, starting from  x0, the sequence  {xn} generated by the method (1.2) converges to a solution  x* of  F(x) with  xn,  x* belonging to  B(x0,Rη)¯ and  x* being the unique solution of  F(x) in  B(x0,(2/Mβ)-Rη)∩Ω. Moreover, a priori error estimate is given by
(4.2)∥xn-x*∥≤g(a0,b0)ηλnγ(5n-1)/41-λγ5n,
where γ=h(a0,b0)c0 and λ=1/h(a0,b0).Proof.
By Lemma3.2, the sequence {xn} is well defined in B(x0,Rη)¯. Now we prove that {xn} is a Cauchy sequence. Since
(4.3)∥xn+m-xn∥≤∑i=nn+m-1∥xi+1-xi∥≤∑i=nn+m-1g(ai,bi)ηi≤g(a0,b0)∑i=nn+m-1ηi≤g(a0,b0)ηλnγ(5n-1)/41-λmγ5n(5m-1+3)/41-λγ5n,
it follows that {xn} is a Cauchy sequence, and hence the sequence {xn} is convergent. So there exists an x* such that limn→∞xn=x*.
By lettingn=0, m→∞ in (4.3), we obtain(4.4)∥x*-x0∥≤Rη.
This shows that x*∈B(x0,Rη)¯.
Now we prove thatx* is a solution of F(x)=0. Since(4.5)∥Γ0∥∥F(xn)∥≤∥Γn∥∥F(xn)∥≤ηn,
by letting n→∞ in (4.5), we obtain ∥F(xn)∥→0 since g(an,bn)<g(a0,b0) and ηn→0. Hence, by the continuity of F in Ω, we obtain F(x*)=0.
We prove the uniqueness ofx* in B(x0,(2/Mβ)-Rη)∩Ω. Firstly we can obtain x*∈B(x0,(2/Mβ)-Rη)∩Ω, since it follows by Lemma 3.2 that(4.6)2Mβ-Rη=(2a0-R)η>1a0η>Rη,
and then B(x0,Rη)¯⊂B(x0,(2/Mβ)-Rη)∩Ω. Let x** be another zero of F(x) in B(x0,(2/Mβ)-Rη)∩Ω. By Taylor theorem, we have
(4.7)0=F(x**)-F(x*)=∫01F′((1-t)x*+tx**)dt(x**-x*).
Since
(4.8)∥Γ0∥‖∫01[F′((1-t)x*+tx**)-F′(x0)]dt‖≤Mβ∫01[(1-t)∥x*-x0∥+t∥x**-x0∥]dt<Mβ2[Rη+2Mβ-Rη]=1,
it follows by the Banach lemma that ∫01F′((1-t)x*+tx**)dt is invertible and hence x**=x*.
Finally, by lettingm→∞ in (4.3), we obtain (4.2) and furthermore(4.9)∥xn-x*∥≤g(a0,b0)ηγ1/4(1-c0)(γ1/4)5n.
This means that the method given by (1.2) is of R-order of convergence at least five. This ends the proof.


## 5. Numerical Example

LetX=C[0,1] be the space of continuous functions defined on the interval [0,1], with the max-norm, and consider the integral equation F(x)=0, where (5.1)F(x)(s)=x(s)-1-12∫01scos⁡(x(t))dt,
with s∈[0,1], x∈Ω=B(0,2)⊂X. Integral equations of this kind (called Chandrasekhar equations) arise in elasticity or neutron transport problems.It is easy to obtain the derivatives ofF as (5.2)F′(x)y(s)=y(s)-12∫01ssin⁡(x(t))y(t)dt,y∈Ω,F′′(x)yz(s)=-12∫01scos⁡(x(t))y(t)z(t)dt,y,z∈Ω,
The derivative F′′ satisfies (5.3)∥F′′(x)∥≤12=M,x∈Ω,
and the Lipschitz condition with N=1/2(5.4)∥F′′(x)-F′′(y)∥≤12∥x-y∥,x,y∈Ω,
since the norm is taken as max-norm.Starting atx0(t)=4/3, we have (5.5)∥F(x0)∥=‖12∫01scos⁡(x0(t))dt‖≤12cos⁡43,
since (5.6)∥I-F′(x0)∥=‖12∫01ssin⁡(x0(t))dt‖≤12sin⁡43,
and by the Banach lemma that Γ0 exists and (5.7)∥Γ0∥≤22-sin⁡⁡(4/3)=β.
It follow that (5.8)∥Γ0F(x0)∥≤cos⁡⁡(4/3)2-sin⁡⁡(4/3)=η.
Consequently, we obtain (5.9)a0=Mβη=cos⁡⁡(4/3)(2-sin⁡⁡(4/3))2,b0=Nβη2=cos⁡2(4/3)(2-sin⁡⁡(4/3))3,c0=h(a0,b0)φ(a0,b0),
which satisfy (5.10)a0=0.22257070108520<max⁡⁡bnσ(bn)≈0.6034780649041629<23,c0h(a0,b0)=0.01818885859745<1.
This means that the hypothesis of Theorem 4.2 is satisfied. Hence, the recurrence relations for the method given by (1.2) are demonstrated in Table 1. Besides, the solution x* belongs to B(x0,Rη)¯=B(4/3,0.27761986977716)¯⊂Ω, and it is unique in B(4/3,1.77850432749622)∩Ω.Table 1
Results of recurrence relations.
nηnβnanbncn02.2882e-0011.94542.2257e-0015.0928e-0021.3343e-00213.0530e-0032.65204.0483e-0031.2360e-0054.7865e-01021.4613e-0122.66281.9456e-0122.8432e-0242.5141e-04733.6740e-0592.66284.8916e-0591.7972e-1171.0045e-23343.6904e-2922.66284.9134e-29200


## 6. Conclusions

A family of recurrence relations is developed for establishing the semilocal convergence of a modified Newton’s method (1.2) used for solving F(x)=0 in Banach spaces. Based on these recurrence relations, an existence uniqueness theorem is established to show the R-order convergence of the method to be five. Also a priori error bounds are given. A numerical example is worked out to demonstrate our approach and show that our method can be of practical interest.



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*Source: 786306-2011-11-30.xml*