Open Access

The convergence theorem for fourth-order super-Halley method in weaker conditions

Journal of Inequalities and Applications20162016:289

https://doi.org/10.1186/s13660-016-1227-5

Received: 26 August 2016

Accepted: 8 November 2016

Published: 22 November 2016

Abstract

In this paper, we establish the Newton-Kantorovich convergence theorem of a fourth-order super-Halley method under weaker conditions in Banach space, which is used to solve the nonlinear equations. Finally, some examples are provided to show the application of our theorem.

Keywords

nonlinear equations in Banach spacessuper-Halley methodsemilocal convergenceNewton-Kantorovich theoremweaker conditions

MSC

65J1565H1065G9947J2549M15

1 Introduction

For a number of problems arising in scientific and engineering areas one often needs to find the solution of nonlinear equations in Banach spaces
$$ F(x)=0, $$
(1)
where F is a third-order Fréchet-differentiable operator defined on a convex subset Ω of a Banach space X with values in a Banach space Y.
There are kinds of methods to find a solution of equation (1). Generally, iterative methods are often used to solve this problem [1]. The best-known iterative method is Newton’s method
$$ x_{n+1}=x_{n}-F'(x_{n})^{-1}F(x_{n}), $$
(2)
which has quadratic convergence. Recently a lot of research has been carried out to provide improvements. Third-order iterative methods such as Halley’s method, Chebyshev’s method, super-Halley’s method, Chebyshev-like’s method etc. [212] are used to solve equation (1). To improve the convergence order, fourth-order iterative methods are also discussed in [1319].
Kou et al. [20] presented a variant of the super-Halley method which improves the order of the super-Halley method from three to four by using the values of the second derivative at \(( x_{n}-\frac{1}{3}f(x_{n})/f'(x_{n}) )\) instead of \(x_{n}\). Wang et al. [15] established the semilocal convergence of the fourth-order super-Halley method in Banach spaces by using recurrence relations. This method in Banach spaces can be given by
$$ x_{n + 1}= x_{n} - \biggl[I+\frac{1}{2}K_{F}(x_{n}) \bigl[I-K_{F}(x_{n}) \bigr]^{-1} \biggr] \Gamma_{n} F(x_{n}), $$
(3)
where \(\Gamma_{n}=[F'(x_{n})]^{-1}\), \(K_{F}(x_{n})=\Gamma_{n}F''(u_{n})\Gamma_{n}F(x_{n})\), and \(u_{n}=x_{n}-\frac{1}{3}\Gamma_{n}F(x_{n})\).
Let \(x_{0} \in\Omega\) and the nonlinear operator \(F: \Omega\subset X\rightarrow Y\) be continuously third-order Fréchet differentiable where Ω is an open set and X and Y are Banach spaces. Assume that
  1. (C1)

    \(\Vert \Gamma_{0}F(x_{0})\Vert \leq\eta\),

     
  2. (C2)

    \(\Vert \Gamma_{0} \Vert \leq\beta\),

     
  3. (C3)

    \(\Vert F^{\prime\prime}(x) \Vert \leq M, x \in\Omega\),

     
  4. (C4)

    \(\Vert F'''(x)\Vert \leq N, x \in\Omega\),

     
  5. (C5)
    there exists a positive real number L such that
    $$\bigl\Vert F^{\prime\prime\prime}(x)- F^{\prime\prime\prime}(y) \bigr\Vert \leq L \Vert x-y \Vert ,\quad \forall x, y \in\Omega. $$
     

Under the above assumptions, we apply majorizing functions to prove the semilocal convergence of the method (3) to solve nonlinear equations in Banach spaces and establish its convergence theorems in [21]. The main results is as follows.

Theorem 1

[21]

Let X and Y be two Banach spaces and \(F:\Omega\subseteq X \rightarrow Y\) be a third-order Fréchet differentiable on a non-empty open convex subset Ω. Assume that all conditions (C1)-(C5) hold and \(x_{0} \in \Omega\), \(h=K \beta\eta\leq1/2\), \(\overline{B(x_{0}, t^{*})} \subset\Omega \), then the sequence \(\{x_{n}\}\) generated by the method (3) is well defined, \(x_{n} \in \overline{B(x_{0}, t^{*})}\) and converges to the unique solution \(x^{*}\in B(x_{0}, t^{**})\) of \(F(x)\), and \(\Vert x_{n}-x^{*}\Vert \leq t^{*}-t_{n}\), where
$$\begin{aligned} &t^{*} =\frac{1-\sqrt{1-2h}}{h}\eta,\qquad t^{**} = \frac{1+\sqrt{1-2h}}{h} \eta, \\ & K \geq M \biggl[ 1 + \frac{N}{M^{2} \beta} +\frac{35 L}{36 M^{3} \beta^{2}} \biggr]. \end{aligned}$$
(4)
We know the conditions of Theorem 1 cannot be satisfied by some general nonlinear operator equations. For example,
$$ F(x)=\frac{1}{6}x^{3}+\frac{1}{6}x^{2}- \frac{5}{6}x+\frac{1}{3}=0. $$
(5)
Let the initial point \(x_{0}=0\), \(\Omega= [-1, 1]\). Then we know
$$\begin{aligned} &\beta= \bigl\Vert F'(x_{0})^{-1} \bigr\Vert = \frac{6}{5}, \qquad\eta= \bigl\Vert F'(x_{0})^{-1}F(x_{0}) \bigr\Vert =\frac{2}{5}, \\ &M=\frac{4}{3},\qquad N=1,\qquad L=0. \end{aligned}$$
From (4), we can get \(K \geq M\), so
$$h=K \beta\eta\geq M \beta\eta=\frac{16}{25}> \frac{1}{2}. $$
The conditions of Theorem 1 cannot be satisfied. Hence, we cannot know whether the sequence \(\{x_{n}\}\) generated by the method (3) converges to the solution \(x^{*}\).

In this paper, we consider weaker conditions and establish a new Newton-Kantorovich convergence theorem. The paper is organized as follows: in Section 2 the convergence analysis based on weaker conditions is given and in Section 3, a new Newton-Kantorovich convergence theorem is established. In Section 4, some numerical examples are worked out. We finish the work with some conclusions and references.

2 Analysis of convergence

Let \(x_{0} \in\Omega\) and nonlinear operator \(F: \Omega\subset X\rightarrow Y\) be continuously third-order Fréchet differentiable where Ω is an open set and X and Y are Banach spaces. We assume that:
  1. (C6)

    \(\Vert \Gamma_{0}F(x_{0})\Vert \leq\eta\),

     
  2. (C7)

    \(\Vert F'(x_{0})^{-1}F^{\prime\prime}(x_{0}) \Vert \leq\gamma\),

     
  3. (C8)

    \(\Vert F'(x_{0})^{-1}F'''(x)\Vert \leq N, x \in\Omega\),

     
  4. (C9)
    there exists a positive real number L such that
    $$ \bigl\Vert F'(x_{0})^{-1} \bigl[F^{\prime\prime\prime}(x)- F^{\prime\prime\prime}(y) \bigr] \bigr\Vert \leq L \Vert x-y \Vert ,\quad \forall x, y \in\Omega. $$
    (6)
     
Denote
$$ g(t)=\frac{1}{6}Kt^{3}+\frac{1}{2}\gamma t^{2}-t+\eta, $$
(7)
where \(K, \gamma, \eta\) are positive real numbers and
$$ \frac{5L}{12N \eta+ 36\gamma}+N \leq K. $$
(8)

Lemma 1

[19]

Let \(\alpha=\frac{2}{\gamma+\sqrt{\gamma^{2}+2K}}, \beta=\alpha-\frac{1}{6}K \alpha^{3}-\frac{1}{2}\gamma \alpha^{2}=\frac{2 (\gamma+2\sqrt{\gamma^{2}+2K} )}{3 (\gamma+\sqrt{\gamma^{2}+2K} )}\). If \(\eta\leq \beta\), then the polynomial equation \(g(t)\) has two positive real roots \(r_{1}, r_{2}\) (\(r_{1}\leq r_{2}\)) and a negative root \(-r_{0}\) (\(r_{0}>0\)).

Lemma 2

Let \(r_{1}, r_{2}, -r_{0}\) be three roots of \(g(t)\) and \(r_{1} \leq r_{2}, r_{0}>0\). Write \(u=r_{0}+t\), \(a=r_{1}-t\), \(b=r_{2}-t\), and
$$ q(t)=\frac{{b - u}}{{a - u}} \cdot\frac{{(a - u)b^{3} + u(u - a)b^{2} + u^{2} (a - u)b - au^{3} }}{{(b - u)a^{3} + u(u - b)a^{2} + u^{2} (b - u)a - bu^{3} }} . $$
(9)
Then as \(0\leq t\leq r_{1}\), we have
$$ q(0)\leq q(t) \leq q(r_{1})\leq1. $$
(10)

Proof

Since \(g(t)=\frac{K}{6}abu\) and \(g''(t)\geq 0\) (\(t\geq0\)), we have
$$ u-a-b \geq0. $$
(11)
Differentiating q and noticing \(q'(t)\geq0\) (\(0\leq t\leq r_{1}\)), we obtain
$$ q(0)\leq q(t)\leq q(r_{1}). $$
(12)
On the other hand, since
$$q(t)-1 \leq0, $$
the lemma is proved.
Now we consider the majorizing sequences \(\{t_{n}\}\), \(\{ s_{n} \}\) \((n\geq0)\), \(t_{0}=0\),
$$ \textstyle\begin{cases} s_{n} = t_{n} - \frac{{g(t_{n} )}}{{g'(t_{n} )}}, \\ h_{n} = - g'(t_{n} )^{ - 1} g''(r_{n} ) (s_{n} - t_{n} ), \\ t_{n + 1} = t_{n} - [ {1 + \frac{1}{2} \frac{{h_{n} }}{{1 - h_{n} }}} ]\frac{{g(t_{n} )}}{{g'(t_{n} )}} , \end{cases} $$
(13)
where \(r_{n}=t_{n}+1/3(s_{n}-t_{n})\). □

Lemma 3

Let \(g(t)\) be defined by (7) and satisfy the condition \(\eta\leq\beta\), then we have
$$ \frac{{(\sqrt[3]{\lambda_{2}} \theta)^{4^{n} } }}{{\sqrt[3]{\lambda_{2}} - (\sqrt[3]{\lambda_{2}} \theta)^{4^{n} } }}(r_{2} - r_{1} ) \le r_{1} - t_{n} \le\frac{{(\sqrt[3]{\lambda_{1}} \theta)^{4^{n} } }}{{\sqrt[3]{\lambda_{1}} - (\sqrt[3]{\lambda_{1}} \theta)^{4^{n} } }}(r_{2} - r_{1} ),\quad n = 0,1, \ldots, $$
(14)
where \(\theta = \frac{{r_{1} }}{{r_{2} }}, \lambda_{1} = q(r_{1} ), \lambda_{2} = q(0)\).

Proof

Let \(a_{n}=r_{1}-t_{n}, b_{n}=r_{2}-t_{n}, u_{n}=r_{0}+t_{n}\), then
$$\begin{aligned} & g(t_{n} ) = \frac{K}{6}a_{n} b_{n} u_{n}, \end{aligned}$$
(15)
$$\begin{aligned} &g'(t_{n} ) = -{\frac{K}{6}} [ {a_{n} u_{n} + b_{n} u_{n} - a_{n} b_{n} } ], \end{aligned}$$
(16)
$$\begin{aligned} &g''(t_{n} ) = { \frac{K}{3}} [ {u_{n} - b_{n} - a_{n} } ]. \end{aligned}$$
(17)
Write \(\varphi(t_{n})= a_{n} u_{n} + b_{n} u_{n} - a_{n} b_{n} \), then we have
$$\begin{aligned} & a_{n + 1} = a_{n} + \biggl[ {1 + \frac{1}{2}\frac{{h_{n} }}{{1 - h_{n} }}} \biggr]\frac{{g(t_{n} )}}{{g'(t_{n} )}} \\ &\phantom{a_{n + 1}} = \frac{{a_{n}^{4} (b_{n} - u_{n} ) [ {(a_{n} - u_{n} )b_{n}^{3} + u_{n} (u_{n} - a_{n} )b_{n}^{2} + u_{n}^{2} (a_{n} - u_{n} )b_{n} - a_{n} u_{n}^{3} } ]}}{\varphi^{2} (t_{n} )(a_{n}^{2} u_{n}^{2} + b_{n}^{2} u_{n}^{2} + a_{n}^{2} b_{n}^{2} ) - 2a_{n}^{2} b_{n}^{2} \varphi(t_{n} )}, \end{aligned}$$
(18)
$$\begin{aligned} & b_{n + 1}= b_{n} + \biggl[ {1 + \frac{1}{2}\frac{{h_{n} }}{{1 - h_{n} }}} \biggr]\frac{{g(t_{n} )}}{{g'(t_{n} )}} \\ &\phantom{b_{n + 1}} = \frac{{b_{n}^{4} (a_{n} - u_{n} ) [ {(b_{n} - u_{n} )a_{n}^{3} + u_{n} (u_{n} - b_{n} )b_{n}^{2} + u_{n}^{2} (b_{n} - u_{n} )a_{n} - b_{n} u_{n}^{3} } ]}}{\varphi^{2} (t_{n} )(a_{n}^{2} u_{n}^{2} + b_{n}^{2} u_{n}^{2} + a_{n}^{2} b_{n}^{2} ) - 2a_{n}^{2} b_{n}^{2} \varphi(t_{n} )}. \end{aligned}$$
(19)
We can obtain
$$ \frac{{a_{n + 1} }}{{b_{n + 1} }} = \frac{{b_{n} - u_{n} }}{{a_{n} - u_{n} }} \cdot\frac{{(a_{n} - u_{n} )b_{n}^{3} + u_{n} (u_{n} - a_{n} )b_{n}^{2} + u_{n}^{2} (a_{n} - u_{n} )b_{n} - a_{n} u_{n}^{3} }}{{(b_{n} - u_{n} )a_{n}^{3} + u_{n} (u_{n} - b_{n} )a_{n}^{2} + u_{n}^{2} (b_{n} - u_{n} )a_{n} - b_{n} u_{n}^{3} }} \cdot \biggl( {\frac{{a_{n} }}{{b_{n} }}} \biggr)^{4}. $$
(20)
From Lemma 2, we have \(\lambda_{2}\leq q(t)\leq\lambda_{1}\). Thus
$$ \frac{{a_{n} }}{{b_{n} }} \le\lambda_{1} \biggl( { \frac{{a_{n - 1} }}{{b_{n - 1} }}} \biggr)^{4} \le \cdots \le ( {\lambda_{1}} )^{1 + 4 + \cdots + 4^{n - 1} } \biggl( {\frac{{a_{0} }}{{b_{0} }}} \biggr)^{4^{n} } = \frac{1}{{\sqrt[3]{\lambda_{1}} }} \bigl( {\sqrt[3]{\lambda_{1}} \theta} \bigr)^{4^{n} }. $$
(21)
In a similar way,
$$ \frac{{a_{n} }}{{b_{n} }} \ge\frac{1}{{\sqrt[3]{\lambda_{2}} }} \bigl( {\sqrt[3]{ \lambda_{2}} \theta} \bigr)^{4^{n} }. $$
(22)
That completes the proof of the lemma. □

Lemma 4

Suppose \(t_{n}, s_{n}\) are generated by (13). If \(\eta< \beta\), then the sequences \(\{t_{n}\}, \{s_{n}\}\) increase and converge to \(r_{1}\), and
$$ 0\leq t_{n} \leq s_{n} \leq t_{n+1} < r_{1}. $$
(23)

Proof

Let
$$\begin{aligned} &U(t) = t - g'(t)^{ - 1} g(t), \\ &H(t) = \bigl( {g'(t)^{ - 1} } \bigr)^{2} g''(T)g(t), \\ &V(t) = t + \biggl[ {1 + \frac{1}{2}\frac{{H(t)}}{{1 - H(t)}}} \biggr] \bigl( {U(t) - t} \bigr), \end{aligned}$$
(24)
where \(T= (2t+U(t) ) /3\).
When \(0 \leq t \leq r_{1}\), we can obtain \(g(t)\geq0\), \(g'(t)< 0\), \(g''(t)> 0\). Hence
$$ U(t)=t-\frac{g(t)}{g'(t)} \geq t \geq0. $$
(25)
So \(\forall t \in[0,r_{1}]\), we always have \(U(t)\geq t\).
Since \(T=\frac{2t+U(t)}{3}\geq t \geq0\), we have
$$ H(t)= \frac{g''(T)g(t)}{g'(t)^{2}} \geq0. $$
(26)
On the other hand \(g''(T)g(t)-g'(t)^{2}>0\), then
$$ 0\leq H(t)= \frac{g''(T)g(t)}{g'(t)^{2}}< 1. $$
(27)
Thus
$$V(t)=U(t)+\frac{1}{2}\frac{{H(t)}}{{1 - H(t)}} \bigl( {U(t) - t} \bigr) \geq0, $$
and \(\forall t \in[0, r_{1}]\), we always have \(V(t)\geq U(t)\).
Since
$$\begin{aligned} V'(t)={}& \frac{g(t) [ 3g'(t)^{2} ( {g''(T) - g''(t)} ) ( {g''(T)g(t) - 2g'(t)^{2} } )- Kg(t)^{2} g'(t)g''(t) }{{6g'(t)^{2} [ {g''(T)g(t) - g'(t)^{2} } ]^{2} }} \\ &{} +\frac{ - Kg(t)^{2} g''(t)g'(t) + 3g(t)^{2} g''(t)g''(T) ]}{{6g'(t)^{2} [ {g''(T)g(t) - g'(t)^{2} } ]^{2} }} \\ ={}& \frac{{g(t) [ { - Kg(t)^{2} g'(t)g''(t) - Kg(t)^{2} g''(t)g'(t) + 3g(t)^{2} g''(t)g''(T)} ]}}{{6g'(t)^{2} [ {g''(T)g(t) - g'(t)^{2} } ]^{2} }}, \end{aligned}$$
(28)
we know \(V'(t) > 0\) for \(0\leq t \leq r_{1}\). That is to say that \(V(t)\) is monotonically increasing. By this we will inductively prove that
$$ 0\leq t_{n}\leq s_{n} \leq t_{n+1} < V(r_{1})=r_{1}. $$
(29)
In fact, (29) is obviously true for \(n=0\). Assume (29) holds until some n. Since \(t_{n+1}< r_{1}\), \(s_{n+1}, t_{n+2}\) are well defined and \(t_{n+1}\leq s_{n+1}\leq t_{n+2}\). On the other hand, by the monotonicity of \(V(t)\), we also have
$$ t_{n+2}=V(t_{n+1})< V(r_{1})=r_{1}. $$
Thus, (29) also holds for \(n+1\).

From Lemma 3, we can see that \(\{t_{n}\}\) converges to \(r_{1}\). That completes the proof of the lemma. □

Lemma 5

Assume F satisfies the conditions (C6)-(C9), then \(\forall x \in\overline{B(x_{0}, r_{1})}\), \(F'(x)^{-1}\) exists and satisfies the inequality
  1. (I)

    \(\Vert F'(x_{0})^{-1}F''(x)\Vert \leq g''(\Vert x-x_{0}\Vert )\),

     
  2. (II)

    \(\Vert F'(x)^{-1}F'(x_{0})\Vert \leq -g'(\Vert x-x_{0}\Vert )^{-1}\).

     

Proof

(I) From the above assumptions, we have
$$\begin{aligned} \bigl\Vert F'(x_{0})^{-1}F''(x) \bigr\Vert &= \bigl\Vert F'(x_{0})^{-1}F''(x_{0})+F'(x_{0})^{-1} \bigl[F''(x)-F''(x_{0}) \bigr] \bigr\Vert \\ &\leq\gamma+ N\Vert x-x_{0}\Vert \leq\gamma+ K\Vert x-x_{0}\Vert \\ &= g'' \bigl(\Vert x-x_{0}\Vert \bigr). \end{aligned}$$
(II) When \(t\in[0,r_{1})\), we know \(g'(t)<0\). Hence when \(x \in \overline{B(x_{0}, r_{1})}\),
$$\begin{aligned} &\bigl\Vert F'(x_{0})^{-1}F'(x)-I \bigr\Vert \\ &\quad= \bigl\Vert F'(x_{0})^{-1} \bigl[F'(x)-F'(x_{0})-F''(x_{0}) (x-x_{0})+F''(x_{0}) (x-x_{0}) \bigr] \bigr\Vert \\ &\quad\le \biggl\Vert { \int_{0}^{1} {F'(x_{0} )^{ - 1} \bigl[ {F'' \bigl( {x_{0} + t(x - x_{0} )} \bigr) - F''(x_{0} )} \bigr]\,dt(x - x_{0} )} } \biggr\Vert + \gamma \Vert {x - x_{0} } \Vert \\ &\quad\le \biggl\Vert { \int_{0}^{1} {Nt\,dt(x - x_{0} )^{2} } } \biggr\Vert + \gamma \Vert {x - x_{0} } \Vert \le\frac{1}{2}K\Vert {x - x_{0} } \Vert ^{2} + \gamma \Vert {x - x_{0} } \Vert \\ &\quad=1+g' \bigl(\Vert x-x_{0}\Vert \bigr)< 1. \end{aligned}$$
By the Banach lemma, we know \((F'(x_{0})^{-1}F'(x) )^{-1}=F'(x)^{-1}F'(x_{0})\) exists and
$$\bigl\Vert {F'(x)^{ - 1} F'(x_{0} )} \bigr\Vert \le\frac{1}{{1 - \Vert {I - F'(x_{0} )^{ - 1} F'(x)} \Vert }} \le - g' \bigl( {\Vert {x - x_{0} } \Vert } \bigr)^{ - 1}. $$
That completes the proof of the lemma. □

Lemma 6

[21]

Assume that the nonlinear operator \(F:\Omega\subset X \rightarrow Y\) is continuously third-order Fréchet differentiable where Ω is an open set and X and Y are Banach spaces. The sequences \(\{x_{n}\}\), \(\{y_{n}\}\) are generated by (3). Then we have
$$\begin{aligned} F(x_{n+1})={}& \frac{1}{2}F^{\prime\prime}(u_{n}) (x_{n+1}-y_{n})^{2}+\frac {1}{6}F'''(x_{n}) (x_{n+1}-y_{n}) (x_{n+1}-x_{n})^{2} \\ &{}- \frac{1}{6} \int^{1}_{0} \biggl[F''' \biggl(x_{n}+\frac{1}{3}t(y_{n}-x_{n}) \biggr)-F'''(x_{n}) \biggr]\,dt(y_{n}-x_{n}) (x_{n+1}-x_{n})^{2} \\ &{}+\frac{1}{2} \int^{1}_{0} \bigl[F''' \bigl(x_{n}+t(x_{n+1}-x_{n}) \bigr)-F'''(x_{n}) \bigr](1-t)^{2}\,dt(x_{n+1}-x_{n})^{3} , \end{aligned}$$
where \(y_{n}=x_{n}-\Gamma_{n}F(x_{n})\) and \(u_{n}=x_{n}+\frac{1}{3}(y_{n}-x_{n})\).

3 Newton-Kantorovich convergence theorem

Now we give a theorem to establish the semilocal convergence of the method (3) in weaker conditions, 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 of at least four of the iterations (3).

Theorem 2

Let X and Y be two Banach spaces, and \(F:\Omega\subseteq X \rightarrow Y\) be a third-order Fréchet differentiable on a non-empty open convex subset Ω. Assume that all conditions (C6)-(C9) hold true and \(x_{0} \in\Omega\). If \(\eta< \beta\), \(\overline{B(x_{0}, r_{1})}\subset\Omega\), then the sequence \(\{ x_{n}\}\) generated by (3) is well defined, \(\{x_{n}\} \in\overline{B(x_{0}, r_{1})} \) and converges to the unique solution \(x^{*} \in B(x_{0}, \alpha)\), and \(\Vert x_{n}-x^{*}\Vert \leq r_{1}-t_{n}\). Further, we have
$$ \bigl\Vert x_{n}-x^{*} \bigr\Vert \leq \frac{{(\sqrt[3]{\lambda_{1}} \theta)^{4^{n} } }}{{\sqrt[3]{\lambda_{1}} - (\sqrt[3]{\lambda_{1}} \theta)^{4^{n} } }}(r_{2} - r_{1} ),\quad n = 0,1, \ldots, $$
(30)
where \(\theta=\frac{{r_{1} }}{{r_{2} }}, \lambda_{1} = q(r_{1} )\), \(\alpha=\frac{2}{\gamma+\sqrt{\gamma^{2}+2K}}\).

Proof

We will prove the following formula by induction:

\((I_{n})\) \(x_{n} \in\overline{B(x_{0}, t_{n})}\),

\((II_{n})\) \(\Vert F'(x_{n})^{-1}F'(x_{0})\Vert \leq-g'(t_{n})^{-1}\),

\((III_{n})\) \(\Vert F'(x_{0})^{-1}F''(x_{n}) \Vert \leq g''(\Vert x_{n}-x_{0}\Vert )\leq g''(t_{n})\),

\((IV_{n})\) \(\Vert y_{n}-x_{n}\Vert \leq s_{n}-t_{n}\),

\((V_{n})\) \(y_{n}\in\overline{B(x_{0}, s_{n})}\),

\((VI_{n})\) \(\Vert x_{n+1}-y_{n}\Vert \leq t_{n+1}-s_{n}\).

Estimate that (\(I_{n}\))-(\(VI_{n}\)) are true for \(n=0\) by the initial conditions. Now, assume that \((I_{n})\)-\((VI_{n})\) are true for all integers \(k \leq n\).

\((I_{n+1})\) From the above assumptions, we have
$$\begin{aligned} \Vert x_{n+1}-x_{0} \Vert & \leq \Vert x_{n+1}-y_{n}\Vert +\Vert y_{n}-x_{n}\Vert +\Vert x_{n}-x_{0} \Vert \\ & \leq(t_{n+1}-s_{n})+(s_{n}-t_{n})+(t_{n}-t_{0})=t_{n+1}. \end{aligned}$$
(31)
\((II_{n+1})\) From (II) of Lemma 5, we can obtain
$$ \bigl\Vert {F'(x_{n + 1} )^{ - 1} F'(x_{0} )} \bigr\Vert \le - g' \bigl( { \Vert {x_{n + 1} - x_{0} } \Vert } \bigr)^{ - 1} \le - g'(t_{n + 1} )^{ - 1}. $$
(32)
\((III_{n+1})\) From (I) of Lemma 5, we can obtain
$$ \bigl\Vert {F'(x_{0} )^{ - 1} F''(x_{n + 1} )} \bigr\Vert \le g'' \bigl( {\Vert {x_{n + 1} - x_{0} } \Vert } \bigr) \le g''(t_{n + 1} ). $$
(33)
\((IV_{n+1})\) From Lemma 5, we have
$$\begin{aligned} \bigl\Vert {F'(x_{0} )^{ - 1} F(x_{n + 1} )} \bigr\Vert \le{}& \frac{1}{2}g''(r_{n} ) (t_{n + 1} - s_{n} )^{2} + \frac{1}{6}N(t_{n + 1} - s_{n} ) (t_{n + 1} - t_{n} )^{2} \\ &{} + \frac{L}{{36}}(s_{n} - t_{n} )^{2} (t_{n + 1} - t_{n} )^{2} + \frac{L}{{24}}(t_{n + 1} - t_{n} )^{4} \\ \le{}&\frac{1}{2}g''(r_{n}) (t_{n + 1} - s_{n} )^{2} \\ &{}+ \frac{1}{6} \biggl( N + \frac{L}{{36}}\cdot\frac{(s_{n} - t_{n})^{2}}{t_{n + 1} - s_{n}} +\frac{L}{24} \cdot \frac{(t_{n + 1} - t_{n} )^{2} }{t_{n + 1} - s_{n} } \biggr) ( {t_{n + 1} - s_{n} } ) (t_{n + 1} - t_{n} )^{2} \\ \le{}& \frac{1}{2}g''(r_{n}) (t_{n + 1} - s_{n} )^{2} + \frac{1}{6}K ( {t_{n + 1} - s_{n} } ) (t_{n + 1} - t_{n} )^{2} \\ \le{}& g(t_{n+1}). \end{aligned}$$
(34)
Thus, we have
$$\begin{aligned} \Vert y_{n+1}-x_{n+1} \Vert & \leq \bigl\Vert -F'(x_{n+1})F'(x_{0}) \bigr\Vert \bigl\Vert F'(x_{0})^{-1}F(x_{n+1}) \bigr\Vert \\ &\leq- \frac{g(t_{n+1})}{g'(t_{n+1})}=s_{n+1}-t_{n+1}. \end{aligned}$$
(35)
\((V_{n+1})\) From the above assumptions and (35), we obtain
$$\begin{aligned} \Vert y_{n+1}-x_{0} \Vert & \leq \Vert y_{n+1}-x_{n+1}\Vert +\Vert x_{n+1}-y_{n}\Vert +\Vert y_{n}-x_{n} \Vert +\Vert x_{n}-x_{0}\Vert \\ & \leq (s_{n+1}-t_{n+1})+(t_{n+1}-s_{n})+(s_{n}-t_{n})+(t_{n}-t_{0})=s_{n+1}, \end{aligned}$$
(36)
so \(y_{n+1}\in\overline{B(x_{0}, s_{n+1})}\).
\((VI_{n+1})\) Since
$$\begin{aligned} \bigl\Vert I-K_{F}(x_{n+1}) \bigr\Vert &\geq1- \bigl\Vert K_{F}(x_{n+1}) \bigr\Vert \geq 1- \bigl(-g'(t_{n+1}) \bigr)^{-1}g''(r_{n+1}) (s_{n+1}-t_{n+1}) \\ &=1+g'(t_{n+1})^{-1}g''(r_{n+1}) (s_{n+1}-t_{n+1})=1-h_{n+1}, \end{aligned}$$
we have
$$\begin{aligned} \Vert x_{n+2}-y_{n+1}\Vert & = \frac{1}{2} \bigl\Vert K_{F}(x_{n+1}) \bigl[I-K_{F}(x_{n+1}) \bigr]^{-1} \bigr\Vert \bigl\Vert F'(x_{n+1})^{-1}F(x_{n+1}) \bigr\Vert \\ & \leq \frac{1}{2}\frac{h_{n+1}}{1-h_{n+1}}\cdot\frac {g(t_{n+1})}{-g'(t_{n+1})}=t_{n+2}-s_{n+1}. \end{aligned}$$
(37)
Further, we have
$$ \Vert x_{n+2}-x_{n+1}\Vert \leq \Vert x_{n+2}-y_{n+1}\Vert +\Vert y_{n+1}-x_{n+1} \Vert \leq t_{n+2}-t_{n+1}, $$
(38)
and when \(m > n\)
$$ \Vert x_{m}-x_{n}\Vert \leq \Vert x_{m}-x_{m-1}\Vert +\cdots+\Vert x_{n+1}-x_{n} \Vert \leq t_{m}-t_{n}. $$
(39)
It then follows that the sequence \(\{x_{n}\}\) is convergent to a limit \(x^{*}\). Take \(n \rightarrow\infty\) in (34), we deduce \(F(x^{*})=0\). From (39), we also get
$$ \bigl\Vert x^{*}-x_{n} \bigr\Vert \leq r_{1}-t_{n}. $$
(40)
Now, we prove the uniqueness. Suppose \(x^{**}\) is also the solution of \(F(x)\) on \(B(x_{0}, \alpha)\). By Taylor expansion, we have
$$ 0=F \bigl(x^{**} \bigr)-F \bigl(x^{*} \bigr)= \int^{1}_{0}F^{\prime} \bigl((1-t)x^{*}+tx^{**} \bigr)\,dt \bigl(x^{**}-x^{*} \bigr). $$
(41)
Since
$$\begin{aligned} & \biggl\Vert F'(x_{0})^{-1} \int^{1}_{0} \bigl[F' \bigl((1-t)x^{*}+tx^{**} \bigr)-F'(x_{0}) \bigr]\,dt \biggr\Vert \\ &\quad \leq \biggl\Vert F'(x_{0})^{-1} \int^{1}_{0}{ \int^{1}_{0}F'' \bigl[x_{0}+t \bigl(x^{*}-x_{0} \bigr)+t \bigl(x^{**}-t^{*} \bigr) \bigr]}\,ds\,dt \bigl[x^{*}-x_{0}+t \bigl(x^{**}-x^{*} \bigr) \bigr] \biggr\Vert \\ &\quad \leq \int^{1}_{0}{ \int^{1}_{0}g'' \bigl[s \bigl\Vert x^{*}-x_{0}+t \bigl(x^{**}-x^{*} \bigr) \bigr\Vert \bigr]}\,ds\,dt \bigl\Vert x^{*}-x_{0}+t \bigl(x^{**}-x^{*} \bigr) \bigr\Vert \\ &\quad = \int^{1}_{0}g' \bigl( \bigl\Vert \bigl(x^{*}-x_{0} \bigr)+t \bigl(x^{**}-x^{*} \bigr) \bigr\Vert \bigr)\,dt-g'(0) \\ &\quad = \int^{1}_{0}g' \bigl( \bigl\Vert (1-t) \bigl(x^{*}-x_{0} \bigr)+t \bigl(x^{**}-x_{0} \bigr) \bigr\Vert \bigr)+1 \\ &\quad < \frac{g'(r_{1})+g'(\alpha)}{2}+1 \leq1, \end{aligned}$$
(42)
we can find that the inverse of \(\int^{1}_{0}F' ((1-t)x^{*}+tx^{**} )\,dt\) exists, so \(x^{**}=x^{*}\).
From Lemma 3, we get
$$ \bigl\Vert x_{n}-x^{*} \bigr\Vert \leq \frac{{(\sqrt[3]{\lambda_{1}} \theta)^{4^{n} } }}{{\sqrt[3]{\lambda_{1}} - (\sqrt[3]{\lambda_{1}} \theta)^{4^{n} } }}(r_{2} - r_{1} ), \quad n = 0,1, \ldots. $$
(43)
This completes the proof of the theorem. □

4 Numerical examples

In this section, we illustrate the previous study with an application to the following nonlinear equations.

Example 1

Let \(X=Y=R\), and
$$ F(x)=\frac{1}{6}x^{3}+\frac{1}{6}x^{2}- \frac{5}{6}x+\frac{1}{3}=0. $$
(44)
We consider the initial point \(x_{0}=0\), \(\Omega= [-1, 1]\), we can get
$$\eta=\gamma=\frac{2}{5},\qquad N=\frac{6}{5},\qquad L=0. $$
Hence, from (8), we have \(K=N=\frac{6}{5}\) and
$$\beta= \frac{{2 ( {\gamma+ 2\sqrt{\gamma^{2} + 2K} } )}}{{3 ( {\gamma + \sqrt{\gamma^{2} + 2K} } )^{2} }} = \frac{3}{5},\quad \eta< \beta. $$
This means that the hypotheses of Theorem 2 are satisfied, we can get the sequence \(\{x_{n}\}_{(n \geq 0)}\) generated by the method (3) is well defined and converges.

Example 2

Consider an interesting case as follows:
$$ x(s)=1+\frac{1}{4}x(s) \int^{1}_{0}\frac{s}{s+t}x(t)\,dt, $$
(45)
where we have the space \(X=C[0,1]\) with norm
$$\Vert x \Vert = \mathop{\max } _{0 \le s \le1} \bigl\vert {x(s)} \bigr\vert . $$
This equation arises in the theory of the radiative transfer, neutron transport and the kinetic theory of gases.
Let us define the operator F on X by
$$ F(x)=\frac{1}{4}x(s) \int^{1}_{0}{\frac{s}{s+t}x(t)\,dt}-x(s)+1. $$
(46)
Then for \(x_{0}=1\) we can obtain
$$\begin{aligned} &N=0, \qquad L=0,\qquad K=0,\qquad \bigl\Vert F'(x_{0})^{-1} \bigr\Vert =1.5304,\qquad \eta=0.2652, \\ &\gamma = \bigl\Vert {F'(x_{0} )^{ - 1} F''(x_{0} )} \bigr\Vert = 1.5304 \times 2 \cdot\frac{1}{4}\mathop{\max} _{0 \le s \le1} \biggl\vert { \int _{0}^{1} {\frac{s}{{s + t}}\,dt} } \biggr\vert = 1.5304 \times\frac{{\ln2}}{2} = 0.5303, \\ &\frac{{2 ( {\gamma + 2\sqrt{\gamma^{2} + 2K} } )}}{{3 ( {\gamma + \sqrt{\gamma^{2} + 2K} } )^{2} }} = 0.9429 > \eta. \end{aligned}$$
That means that the hypotheses of Theorem 2 are satisfied.

Example 3

Consider the problem of finding the minimizer of the chained Rosenbrock function [22]:
$$ g(\mathbf{x})=\sum_{i=1}^{m} \bigl[4 \bigl(x_{i}-x^{2}_{i+1} \bigr)^{2}+ \bigl(1-x^{2}_{i+1} \bigr) \bigr],\quad \mathbf{x} \in \mathbf{R}^{m} . $$
(47)

For finding the minimum of g one needs to solve the nonlinear system \(F(\mathbf{x})=0\), where \(F(\mathbf{x})=\nabla g(\mathbf{x})\). Here, we apply the method (3), and compare it with Chebyshev method (CM), the Halley method (HM), and the super-Halley method (SHM).

In a numerical tests, the stopping criterion of each method is \(\Vert \mathbf{x}_{k}-\mathbf{x}^{*}\Vert _{2} \leq1e-15 \), where \(\mathbf{x}^{*}=(1,1,\ldots,1)^{T}\) is the exact solution. We choose \(m = 30\) and \(x_{0} = 1.2\mathbf{ x}^{*}\). Listed in Table 1 are the iterative errors (\(\Vert \mathbf{x}_{k}-\mathbf{x}^{*}\Vert _{2}\)) of various methods. From Table 1, we know that, as tested here, the performance of the method (3) is better.
Table 1

The iterative errors ( \(\pmb{\Vert \mathbf{x}_{n}-\mathbf{x}^{*}\Vert _{2}}\) ) of various methods

n

CM

HM

SHM

Method ( 3 )

1

3.086657e−001

2.169261e−001

5.962251e−002

3.119924e−002

2

1.198959e−003

4.506048e−003

1.913033e−005

1.989913e−006

3

1.099899e−007

2.048709e−008

1.435454e−014

7.771561e−016

4

3.140185e−016

5.438960e−016

8.599751e−016

 

5 Conclusions

In this paper, a new Newton-Kantorovich convergence theorem of a fourth-order super-Halley method is established. As compared with the method in [21], the differentiability conditions of the method in the paper are mild. Finally, some examples are provided to show the application of the convergence theorem.

Declarations

Acknowledgements

This work is supported by the National Natural Science Foundation of China (11371243, 11301001, 61300048), and the Natural Science Foundation of Universities of Anhui Province (KJ2014A003).

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
School of Statistics and Applied Mathematics, Anhui University of Finance and Economics
(2)
Department of Mathematics, Shanghai University

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