## 5 Extension to a Family of the Sixth-Order Methods for Nonlinear Systems of Equations
\(e_{z}=z-\alpha=x-\alpha-T_{f}(s)f^{\prime}(x)^{-1}f(x)=e-(T_{0}I+T_{1}(s-I)+T_{2} (s-I)^{2}+T_{3}(s-I)^{3})f^{\prime}(x)^{-1}f(x)=e(1-T_{0})+c_{2}e^{2}(T_{0}+2T_{1 }\gamma)\)\(+(c_{3}(2T_{0}-3T_{1}(-2+\gamma)\gamma)-2c_{2}^{2}(T_{0}+2\gamma(2T_{1}+T_{2 }\gamma)))e^{3}+\sum_{j=4}^{6}\mathcal{F}_{j}e^{j}\big{]}+O(||e||^{7})\) with \(\mathcal{E}_{j}=\mathcal{F}_{j}(\gamma,c_{2},c_{3},c_{4},c_{5},c_{6},T_{0},T_{1}, T_{2})\). Equating the coefficients of the first and second-order terms in \(e_{z}\) yields:  
\[T_{0}=1,T_{1}=-1/(2\gamma). \tag{34}\]  
We obtain: \(f(z)=f(x)\big{|}_{e->e_{z}}\) and \(f^{\prime}(x)^{-1}f(z)=(1-T_{0})e+(-2+3T_{0}+2T_{1}\gamma)c_{2}e^{2}+((-2+2T_{0}+ 3T_{0}^{2}-T_{0}^{3}-3T_{1}(-2+\gamma)\gamma)c_{3}-4(-1+2T_{0}+3T_{1}\gamma+T_{2 }\gamma^{2}))c_{2}^{2}e^{3}+\sum_{j=4}^{6}\mathcal{G}_{j}e^{j}+O(||e||^{7})\), with \(\mathcal{G}_{j}=\mathcal{G}_{j}(\gamma,c_{2},c_{3},c_{4},c_{5},c_{6},T_{0},T_{1}, T_{2})\). \(e^{(n+1)}=x^{(n+1)}-\alpha=z^{(n)}-\alpha-L_{f}(s)f^{\prime}(x^{(n)})^{-1}f(z^{(n)})= e_{z}-(L_{0}I+L_{1}(s-I)+L_{2}(s-I)^{2}+L_{3}(s-I)^{3})f^{\prime}(x^{(n)})^{-1}f(z^{(n)})=(L_{0}-1)(T_{0}-1 )e+(T_{0}-2L_{1}T_{0}\gamma+2(L_{1}+T_{1}\gamma)+L_{0}(2-3T_{0}-2T_{1}\gamma)) c_{2}e^{2}+\sum_{j=3}^{6}\mathcal{H}_{j}e^{j}\big{]}+The discussions thus far lead us to the following theorem for nonlinear systems of equations. Theorem 2: Let \(f:\Omega\subset\mathbb{C}^{d}\rightarrow\mathbb{C}^{d}\) with \(d\in\mathbb{N}\) have a simple root \(\alpha\) and be sufficiently Fretchet differentiable in \(\Omega\) containing \(\alpha\). Let \(x^{(0)}\) be an initial guess chosen close to \(\alpha\). Let \(T_{f},L_{f}:\mathbb{C}^{d\times d}\rightarrow\mathbb{C}^{d\times d}\) be matrix functions sufficiently Fretchet differentiable in a neighborhood of \(I\), being defined by:  
\(T_{f}(s)=T_{0}I+T_{1}(s-I)+T_{2}(s-I)^{2}+T_{3}(s-I)^{3}+O(||(s-I)^{4}||)\),  
\(L_{f}(s)=L_{0}I+L_{1}(s-I)+L_{2}(s-I)^{2}+L_{3}(s-I)^{3}+O(||(s-I)^{4}||)\) with \(T_{i},L_{i}\in\mathbb{C}\) for \(i=0,1,2,3\) and \(s=f^{\prime}(x_{n})^{-1}f^{\prime}(x_{n})\). If \(\{\gamma=\frac{2}{3},T_{0}=1,T_{1}=-\frac{3}{4},T_{2}=\frac{9}{8},L_{0}=1,L_ {1}=-\frac{3}{2},|T_{3}|<\infty,|L_{2}|<\infty\}\) or \(\{\gamma=1,T_{0}=1,T_{1}=-\frac{1}{2},L_{0}=1,L_{1}=-1,L_{2}=\frac{3}{2},|L_ {3}|<\infty,|T_{2}|<\infty\}\) are given, then iterative scheme \((1)\) reduces to a family of sixth-order methods satisfying the error equation below. For \(n=0,1,2,\cdots,\)  
\[e^{(n+1)}=\left\{ \begin{array}{l}\big{[}\frac{2(27-8L_{2})(13+64T_{3})}{243} c_{2}^{2}c_{3}^{2}+\frac{2(27-8L_{2})}{81}c_{2}^{2}c_{4}-\frac{(135+64T_{3})}{2 }c_{3}c_{2}^{3}-\frac{c_{3}c_{4}}{9}+\\ c_{3}^{2}c_{2}-\frac{2(27-8L_{2})}{9}c_{2}^{2}c_{3}c_{2}\big{]}e^{(n)}{}^{6}+O( ||e^{(n)^{7}}||),\text{ if }\gamma=\frac{2}{3}.\\ \big{[}-4c_{2}^{2}(-1+2T_{2})(9+4L_{3}+2T_{2})+(-6+12T_{2})c_{3}c_{2}^{2}+(1-2 T_{2})c_{2}c_{3}c_{2}^{2}+\\ \frac{1}{4}(c_{2}c_{3}^{2}+4(9+4L_{3}+2T_{2})c_{2}^{3}c_{3}-6c_{3}c_{2}c_{3}) \big{]}e^{(n)}{}^{6}+O(||e^{(n)^{7}}||),\text{ if }\gamma=1.\end{array} \right. \tag{37}\]  
where \(c_{j}=\frac{1}{j! }f^{\prime}(\alpha)^{-1}f^{(j)}(\alpha)\) for \(j=2,3,\cdots\). Equation (37) clearly reduces to (2) for a scalar function by identifying \(c_{i}\) with \(\theta_{i}\). In what follows, we employ several test examples for the zeros of vector-valued functions to verify the convergence behavior claimed here. In terms of Euclidean norm \(||\bullet||\), we display the error sizes for \(e_{k}=||x^{(k+1)}-x^{(k)}||\), residual error \(||f(x^{(k+1)})||\) and ACOC using the error criterion of \(||x^{(k+1)}-x^{(k)}||<10^{-140}\) within 20 iterations. Test Example 1  
We consider a nonlinear algebraic vector equation \(f:\mathbb{R}^{3}\rightarrow\mathbb{R}^{3}\) defined by \(f(x)=0\) with \(x=(x_{1},x_{2},x_{3})^{T}\) as follows:  
\[\left\{ \begin{array}{l}\pi(x_{1}^{2}+x_{2}^{2}/2)-3x_{3}=0,\\ x_{1}^{2}+x_{2}/2+2\cos x_{3}=0,\\ x_{1}x_{2}-\cos x_{2}\cdot\sin(2x_{3})-2=0.\end{array} \right. \tag{38}\]  
The exact solution is given by \(x=(1,2,\pi)^{T}\).