## 5 Numerical Testing
We list them as  
\[\mathbf{x}_{k+1} =\mathbf{x}_{k}-\left(\mathbf{F}^{\prime}(\mathbf{x}_{k})\, \left(\operatorname{diag}(\mathbf{F}(\mathbf{x}_{k}))\,\mathbf{G}^{\prime}( \mathbf{x}_{k})+\operatorname{diag}(\mathbf{G}(\mathbf{x}_{k}))\,\mathbf{F}^ {\prime}(\mathbf{x}_{k})\right)-\mathbf{F}^{\prime\prime}(\mathbf{x}_{k})\, \left(\mathbf{F}(\mathbf{x}_{k})\odot\mathbf{G}(\mathbf{x}_{k})\right)\right)^ {-1} \tag{37}\] \[\mathbf{x}_{k+1} =\mathbf{x}_{k}-\left(\mathbf{F}^{\prime}(\mathbf{x}_{k})+ \operatorname{diag}(\mathbf{F}(\mathbf{x}_{k}))\operatorname{diag}(\mathbf{G}( \mathbf{x}_{k}))^{-1}\,\mathbf{G}^{\prime}(\mathbf{x}_{k})\right)^{-1} \operatorname{diag}(\mathbf{m})\,\mathbf{F}(\mathbf{x}) \tag{38}\]  
When we take \(\mathbf{G}(\mathbf{x})=\mathbf{1}\), the iterative methods Equations (37) and (38) reduce to the following forms\[\mathbf{x}_{k+1} =\mathbf{x}_{k}-\left(\mathbf{F}^{\prime}(\mathbf{x}_{k})^{2}- \mathbf{F}^{\prime\prime}(\mathbf{x}_{k})\,\mathbf{F}(\mathbf{x}_{k})\right)^{- 1}\mathbf{F}^{\prime}(\mathbf{x}_{k})\,\mathbf{F}(\mathbf{x}_{k}) \tag{39}\] \[\mathbf{x}_{k+1} =\mathbf{x}_{k}-\mathbf{F}^{\prime}(\mathbf{x}_{k})^{-1}\,\mathrm{ diag}(\mathbf{m})\,\mathbf{F}(\mathbf{x}) \tag{40}\]  
respectively. \[\text{Problem }2= \begin{cases}\mathrm{F}_{1}(\mathbf{x})=\mathbf{x}_{1}\, \mathbf{x}_{2}=0\\ \mathrm{F}_{2}(\mathbf{x})=\mathbf{x}_{2}\,\mathbf{x}_{3}=0\\ \mathrm{F}_{3}(\mathbf{x})=\mathbf{x}_{3}\,\mathbf{x}_{4}=0\\ \mathrm{F}_{4}(\mathbf{x})=\mathbf{x}_{4}\,\mathbf{x}_{1}=0\end{cases}  \tag{41}\]  
\[\text{Problem }3= \begin{cases}\mathrm{F}_{1}(\mathbf{x})=\sqrt{\mathbf{x}_{1}- \mathbf{1}}\,\mathbf{x}_{2}\,\mathbf{x}_{3}=0\\ \mathrm{F}_{2}(\mathbf{x})=\sqrt{\mathbf{x}_{2}-\mathbf{1}}\,\mathbf{x}_{1}\, \mathbf{x}_{3}=0\\ \mathrm{F}_{3}(\mathbf{x})=\sqrt{\mathbf{x}_{3}-\mathbf{1}}\,\mathbf{x}_{1}\, \mathbf{x}_{2}=0\end{cases}  \tag{42}\]  
In most of the cases, the iterative methods Equations (39) and (40) are badly conditioned. In Tables 1-3, we have shown that the exponential function is not the only and best choice to use as auxiliary function for rapid convergence. In the tables, one can see that a particular choice of auxiliary function for method Equation (37) gives the order of convergence greater than two. The multiplicities of roots in problem 3 are less than one and the iterative method in Equation (37) provides better accuracy in the solution of this problem. In the majority of cases, the performance of iterative method in Equation (37) with unknown multiplicity is better than that of the procedure in Equation (38). \begin{table}
\begin{tabular}{c c c c c c} \hline \hline  & \(\mathbf{G}(\mathbf{x})\) & **Iter** & \(\mid\mid\mathbf{x}-\mathbf{a}\mid\,\mid_{\infty}\) & **Num. Stability** & **COC** \\ \hline Iterative method Equaiton (37) & **1** & 6 & \(\mathrm{O}\left(10^{-43}\right)\) & Badly-conditioned & 2.0 \\  & \(6+\cos(\mathbf{x})/10\) & 6 & \(\mathrm{O}\left(10^{-51}\right)\) & Well-conditioned & 2.05 \\  & \(1+\mathbf{x}^{3}/1000\) & 6 & \(\mathrm{O}\left(10^{-42}\right)\) & Well-conditioned & 2.0 \\  & \(\exp(-\mathbf{x}/100)\) & 6 & \(\mathrm{O}\left(10^{-46}\right)\) & Well-conditioned & 2.0 \\ \hline Iterative method Equaiton (38) & **1** & 6 & \(\mathrm{O}\left(10^{-30}\right)\) & Badly-conditioned & 2.0 \\  & \(6+\cos(\mathbf{x})/10\) & 6 & \(\mathrm{O}\left(10^{-30}\right)\) & Well-conditioned & 2.0 \\  & \(1+\mathbf{x}^{3}/1000\) & 6 & \(\mathrm{O}\left(10^{-30}\right)\) & Well-conditioned & 2.0 \\  & \(\exp(\mathbf{x}/100)\) & 6 & \(\mathrm{O}\left(10^{-30}\right)\) & Well-conditioned & 2.0 \\ \hline \hline \end{tabular}
\end{table} 
Table 1: Problem 1: initial guess \(=[2,1,-2]\), \(\mathrm{m}=[4,5,6]\).