# An Efficient Sixth-Order Newton-Type Method for Solving Nonlinear Systems
## 4 Numerical Examples  
In this section, we compare the related methods by numerical experiments. The numerical experiments are performed using the MAPLE computer algebra system with 2048 digits. The method M6 is compared with NM, CM4, SNAAM, CHM, CTVM by solving some nonlinear systems. The stopping criterion used is \(\left|\left|x^{(k+1)}-x^{(k)}\right|\right|<10^{-200}\) or \(\left|\left|F(x^{(k)})\right|\right|<10^{-200}\). Following nonlinear systems are used: \(F_{1}(x_{1},x_{2})=(2-e^{x_{1}}+\arctan(x_{2}),\ \arctan(x_{1}^{2}+x_{2}^{2}-5))\) and the initial value is \(x^{(0)}=(1.35,\ 2)^{T}\). The solution is \(\alpha\approx(\ 1.1290650391602,1.9300808629035)^{T}\). \(F_{2}(x_{1},x_{2},x_{3})=(x_{2}+x_{3}-e^{-x_{1}},\ \ x_{1}+x_{3}-e^{-x_{3}},x_{1}+x_{2}-e^{-x_{ 3}})\) and the initial value is \(x^{(0)}=(0.2,1.5,1.5)^{T}\). The solution is \(\alpha\approx(\ 0.351733711249,\ 0.351733711249,\ 0.351733711249)^{T}\). \begin{table}
\begin{tabular}{c c c c c c c} \hline \(m\) & \(CEI_{1}\) & \(CEI_{2}\) & \(CEI_{3}\) & \(CEI_{4}\) & \(CEI_{5}\) & \(CEI_{6}\) \\ \hline
5 & 1.0055606 & 1.0052450 & 1.0049895 & 1.0052838 & 1.0055283 & 1.0050600 \\
7 & 1.0025422 & 1.0025753 & 1.0023729 & 1.0025126 & 1.0026423 & 1.0025375 \\
9 & 1.0013845 & 1.0014870 & 1.0013340 & 1.0014096 & 1.0014831 & 1.0014905 \\
11 & 1.0008405 & 1.0009480 & 1.0008314 & 1.0008761 & 1.0009207 & 1.0009643 \\
20 & 1.0001777 & 1.0002326 & 1.0001898 & 1.0001978 & 1.0002060 & 1.0002482 \\
50 & 1.0000141 & 1.0000224 & 1.0000165 & 1.0000169 & 1.0000173 & 1.0000258 \\
100 & 1.0000019 & 1.0000034 & 1.0000023 & 1.0000024 & 1.0000024 & 1.0000040 \\
200 & 1.0000002 & 1.0000005 & 1.0000003 & 1.0000003 & 1.0000003 & 1.0000006 \\ \hline \end{tabular}
\end{table} 
Table 2: Computational efficiency indices of the methods (\(\mu=2\)). Figure 1: The boundary functions \(\mathrm{H}_{6,1}\) and \(\mathrm{H}_{6,2}\) in \((\mu,m)\)—plain. \begin{table}
\begin{tabular}{c c c c c c c} \hline \(m\) & \(CEI_{1}\) & \(CEI_{2}\) & \(CEI_{3}\) & \(CEI_{4}\) & \(CEI_{5}\) & \(CEI_{6}\) \\ \hline
5 & 1.0028332 & 1.0027489 & 1.0028941 & 1.0029907 & 1.0030675 & 1.0029177 \\
7 & 1.0013956 & 1.0014055 & 1.0014554 & 1.0015068 & 1.0015525 & 1.0015157 \\
9 & 1.0008054 & 1.0008390 & 1.0008536 & 1.0008839 & 1.0009123 & 1.0009150 \\
11 & 1.0005124 & 1.0005505 & 1.0005504 & 1.0005697 & 1.0005882 & 1.0006057 \\
20 & 1.0001242 & 1.0001488 & 1.0001391 & 1.0001434 & 1.0001476 & 1.0001681 \\
50 & 1.0000117 & 1.0000168 & 1.0000139 & 1.000141 & 1.000144 & 1.0000199 \\
100 & 1.0000017 & 1.0000028 & 1.0000021 & 1.0000021 & 1.0000022 & 1.0000034 \\
200 & 1.0000002 & 1.0000004 & 1.0000003 & 1.0000003 & 1.0000003 & 1.0000005 \\ \hline \end{tabular}
\end{table} 
Table 3: Computational efficiency indices of the methods (\(\mu=6\)). \(F_{3}(x)=(f_{1}(x),f_{2}(x),\ldots,f_{n}(x))\), where \(x=\left(x_{1},x_{2},\cdots,x_{n}\right)^{T}\) and \(f_{i}:R^{n}\to R\), \(i=1,2,\cdots,n\), such that  
\[f_{i}(x)=x_{i}x_{i+1}-1,\ i=1,2,\cdots,n-1,\]  
\[f_{n}(x)=x_{n}x_{1}-1.\]  
When \(n\) is odd, the exact zeros of \(F_{4}(x)\) are \(\alpha_{1}=(1,1,\cdots,1)\) and \(\alpha_{2}=(-1,-1,\cdots,-1)\). The initial value is \(x^{(0)}=(2.5,\ 0.5,\ 1.5,\ 2.5,\ 2.5,\ 1.5,\ 2.5,\ 0.5,\ 2.5,\ 1.5,\ 8.5)^{T}\). Table 4 presents the results showing the following information: the number of iterations \(k\) needed to converge to the solution, the value of the stopping factors at the last step and the computational order of convergence \(\rho\). The computational order of convergence \(\rho\) is defined by [18]:  
\[\rho\approx\frac{\ln(\left|\left|x^{(k+1)}-x^{(k)}\right|\right|\left|/\left| x^{(k)}-x^{(k-1)}\right|\right|)}{\ln(\left|\left|x^{(k)}-x^{(k-1)}\right| \left|/\left|x^{(k-1)}-x^{(k-2)}\right|\right|)}. \tag{37}\]  
The numerical results shown in Table 4 are in concordance with the theory developed in this paper. The order of convergence of our method (M6) is 6, which is higher than the methods NM and CM4. The iterative method SNAM is not convergent (nc) for \(F_{3}\) with the corresponding initial value.