# An Efficient Sixth-Order Newton-Type Method for Solving Nonlinear Systems
## 2 The New Method and Analysis of Convergence  
Based on the method (3), we construct the following iterative scheme:  
\[\left\{ \begin{array}{l}y^{(k)}=x^{(k)}-F^{\prime}(x^{(k)})^{-1}F(x^{(k)}), \\ z^{(k)}=y^{(k)}-\big{(}2I-F^{\prime}(x^{(k)})^{-1}F^{\prime}(y^{(k)})\big{)} F^{\prime}(x^{(k)})^{-1}F(y^{(k)}),\\ x^{(k+1)}=z^{(k)}-\big{(}2I-F^{\prime}(x^{(k)})^{-1}F^{\prime}(y^{(k)})\big{)} F^{\prime}(x^{(k)})^{-1}F(z^{(k)}),\end{array} \right. \tag{6}\]where \(I\) is the identity matrix. We note that the first two steps of method (6) are the same as those of method (3), and the third step of method (6) increases one function evaluation \(F(z^{(k)})\). Method (6) will be denoted by M6. For method (6), we have the following convergence analysis. **Theorem 1. **_Let \(\alpha\in R^{m}\) be a solution of the system \(F(x)=0\) and \(F:D\subset R^{m}\to R^{m}\) be sufficiently differentiable in an open neighborhood \(D\) of \(\alpha\). Suppose that \(F^{\prime}(x)\) is nonsingular in \(D\). Then, for an initial approximation sufficiently close to \(\alpha\), the iterations converge with order 6._  
**Proof. ** By using the notation introduced in [6], we have the following [PERSON]'s expansion of \(F(x^{(k)})\) around \(\alpha\):  
\[F(x^{(k)})=F^{\prime}(\alpha)[e+A_{2}e^{2}+A_{3}e^{3}+O(e^{4})], \tag{7}\]  
where \(A_{i}=\frac{1}{\pi}F^{\prime}(\alpha)^{-1}F^{(i)}(\alpha)\in L_{i}(R^{m},R^{m})\), \(e=x^{(k)}-\alpha\ A_{i}e^{i}\in R^{m}\), \(F^{(i)}(\alpha)\in L(R^{m}\times\cdots\times R^{m},R^{m})\), \(F^{-1}(\alpha)\in L(R^{m})\) and \(e^{i}\) denotes \((e,\overset{i}{\cdot},e)\). From (7) the derivatives of \(F(x^{(k)})\) can be written as:  
\[F^{\prime}(x^{(k)})=F^{\prime}(\alpha)[I+2A_{2}e+3A_{3}e^{2}+O(e^{3})]=F^{ \prime}(\alpha)D(e)+O(e^{3}), \tag{8}\]  
where \(D(e)=I+2A_{2}e+3A_{3}e^{2}\). The inverse of (8) can be written as:  
\[F^{\prime}(x^{(k)})^{-1}=D(e)^{-1}F^{\prime}(\alpha)^{-1}+O(e^{3}). \tag{9}\]  
Then, we compel the inverse of \(D(e)\) to be (see [6]):  
\[D(e)^{-1}=I+X_{2}e+X_{3}e^{2}+O(e^{3}), \tag{10}\]  
such that \(X_{2}\) verifies:  
\[D(e)D(e)^{-1}=D(e)^{-1}D(e)=I. \tag{11}\]  
Solving system (11), we get:  
\[X_{2}=-2A_{2}, \tag{12}\]  
\[X_{3}=4A_{2}^{2}-3A_{3}, \tag{13}\]  
then,  
\[F^{\prime}(x^{(k)})^{-1}=[I-2A_{2}e+(4A_{2}^{2}-3A_{3})e^{2}]F^{\prime}(\alpha )^{-1}+O(e^{3}). \tag{14}\]  
Let us denote \(E=y^{(k)}-\alpha\). From (6), (7) and (14), we arrive at:  
\[E=e-F^{\prime}(x^{(k)})^{-1}F(x^{(k)})=A_{2}e^{2}+O(e^{3}). \tag{15}\]  
By a similar argument to that of (7), we obtain:  
\[F(y^{(k)})=F^{\prime}(\alpha)[E+A_{2}E^{2}+O(E^{3})]=F^{\prime}(\alpha)[A_{2} e^{2}+A_{2}^{3}e^{4}+O(e^{5})], \tag{16}\]  
\[F^{\prime}(x^{(k)})=F^{\prime}(\alpha)[I+2A_{2}e+3A_{3}e^{2}+O(e^{3})]=F^{ \prime}(\alpha)D(e)+O(e^{3}), \tag{8}\]0  
From (14) and (17), we have:  
\[F^{\prime}(x^{(k)})=F^{\prime}(\alpha)[I+2A_{2}e+3A_{3}e^{2}+O(e^{3})]=F^{ \prime}(\alpha)D(e)+O(e^{3}), \tag{8}\]1  
Let us denote \(\varepsilon=z^{(k)}-\alpha\). From (14), (15) and (18), we get:  
\[ \begin{array}{ll}\varepsilon=z^{(k)}-\alpha&=E-[I-2A_{2}E-4A_{2}^{2}e^{2}][E +A_{2}E^{2}+O(E^{3})]\\ &=A_{2}E^{2}+4A_{2}^{2}e^{2}E+O(e^{5})\\ &=5A_{3}^{3}e^{4}+O(e^{5}),\end{array}  \tag{19}\]\[F(z^{(k)})=F^{\prime}(\alpha)[\varepsilon+A_{2}\varepsilon^{2}+O(\varepsilon^{3})] =F^{\prime}(\alpha)[5A_{2}^{3}e^{4}+O(\varepsilon^{5})]. \tag{20}\]  
Therefore, from (15) and (18)--(20), we obtain the error equation:  
\[ \begin{array}{ll}e_{n+1}=x^{(k+1)}-\alpha&=\varepsilon-[I-2A_{2}E-4A_{2}^{2 }e^{2}][\varepsilon+O(\varepsilon^{2})]\\ &=2A_{2}E\varepsilon+4A_{2}^{2}e^{2}\varepsilon+O(\varepsilon^{7})\\ &=30A_{2}^{5}e^{6}+O(\varepsilon^{7}).\end{array}  \tag{21}\]  
This implies that method (6) is of sixth-order convergence. This completes the proof. In order to simplify the calculation, the new method (6) can be written as:  
\[\left\{