# An Efficient Sixth-Order Newton-Type Method for Solving Nonlinear Systems
## 3 Computational Efficiency
The relation between CHM and M6 is given by: \[R_{6,4}=\frac{\log(\rho_{6})C_{4}(\mu,m)}{\log(\rho_{4})C_{6}(\mu,m)}=\frac{m (2m+3)\mu+2(m^{3}-m)/3+5m^{2}+m}{m(2m+3)\mu+(m^{3}-m)/3+7m^{2}+2m}.\] (29) Subtracting the denominator from the numerator of (29), we have: \[\frac{1}{3}m(m^{2}-6m-4).\] (30) \begin{table}
\begin{tabular}{c c c c c} \hline \hline
**Methods** & \(\rho\) & _a(m)_ & _p(m)_ & \(C(\mu,m)\) \\ \hline NM & 2 & \(m(m+1)\) & \((m^{3}-m)/3+m^{2}\) & \(C_{1}=m(m+1)\mu+(m^{3}-m)/3+m^{2}\) \\ CM4 & 4 & \(2m(m+1)\) & \((m^{3}-m)/3+4m^{2}+m\) & \(C_{2}=2m(m+1)\mu+(m^{3}-m)/3+4m^{2}+m\) \\ SNAM & 6 & \(m(2m+3)\) & \(2(m^{3}-m)/3+6m^{2}\) & \(C_{3}=m(2m+3)\mu+2(m^{3}-m)/3+6m^{2}\) \\ CHM & 6 & \(m(2m+3)\) & \(2(m^{3}-m)/3+5m^{2}+m\) & \(C_{4}=m(2m+3)\mu+2(m^{3}-m)/3+5m^{2}+m\) \\ CTVM & 6 & \(m(2m+3)\) & \(2(m^{3}-m)/3+4m^{2}+3m\) & \(C_{5}=m(2m+3)\mu+2(m^{3}-m)/3+4m^{2}+3m\) \\ M6 & 6 & \(m(2m+3)\) & \((m^{3}-m)/3+7m^{2}+2m\) & \(C_{6}=m(2m+3)\mu+(m^{3}-m)/3+7m^{2}+2m\) \\ \hline \hline \end{tabular}
\end{table} 
Table 1: Computational cost of the iterative methods. Equation (30) is positive for \(m\geq 6.606\). Thus, we get \(CEI_{6}>CEI_{4}\) for all \(m\geq 7\) and \(\mu>0\). 3. The relation between CHM and M6 can be given by: \[R_{6,4}=\frac{\log(\rho_{6})C_{5}(\mu,m)}{\log(\rho_{5})C_{6}(\mu,m)}=\frac{m(2 m+3)\mu+2(m^{3}-m)/3+4m^{2}+3m}{m(2m+3)\mu+(m^{3}-m)/3+7m^{2}+2m}.\] (31) Subtracting the denominator from the numerator of (31), we have: \[\frac{1}{3}m(m^{2}-9m+2).\] (32) Equation (32) is positive for \(m\geq 8.772\). Thus, we obtain \(CEI_{6}>CEI_{5}\) for all \(m\geq 9\) and \(\mu>0\). This completes the proof. **Theorem 3**. : _For all \(m\geq 2\) we have:_  
1. \(CEI_{6}>CEI_{1}\) _for all_ \(\mu>\frac{m^{2}\log(3)+3(\log(3)-6\log(2))m-(6\log(2)+\log(3))}{3(m\log(2/3)+ \log(4/3))}\)_,_
2. \(CEI_{6}>CEI_{2}\) _for all_ \(\mu>\frac{m^{2}\log(3/2)+6(2\log(3)-5\log(2))m+2(\log(3)-4\log(2))}{6(m\log(2 /3)+\log(4/3))}\)_._  
**Proof. **  
1. From expression (26) and Table 1, we get the following relation between NM and M6: \[R_{6,1}=\frac{\log(\rho_{6})C_{1}(\mu,m)}{\log(\rho_{1})C_{6}(\mu,m)}=\frac{ \log(6)}{\log(2)}\frac{m(m+1)\mu+(m^{3}-m)/3+m^{2}}{m(2m+3)\mu+(m^{3}-m)/3+7m^{ 2}+2m}.\] (33) We consider the boundary \(R_{6,1}=1\). The boundary can be given by the following equation: \[\mu=\mathrm{H}_{6,1}(m)=\frac{m^{2}\log(3)+3(\log(3)-6\log(2))m-(6\log(2)+ \log(3))}{3(m\log(2/3)+\log(4/3))},\] (34) where \(CEI_{6}>CEI_{1}\) over it (see Figure 1). Boundary (34) cuts axes at points \((\mu,m)=(0,\ 8.8948)\) and \((12.2470,\ 2)\). Thus, we get \(CEI_{6}>CEI_{1}\) since \(R_{6,1}>1\) for all \(m\geq 2\) and \(\mu>\mathrm{H}_{6,1}(m)\). 2. The relation between CM4 and M6 is given by: \[R_{6,2}=\frac{\log(\rho_{6})C_{2}(\mu,m)}{\log(\rho_{2})C_{6}(\mu,m)}=\frac{ \log(6)}{\log(4)}\frac{2m(m+1)\mu+(m^{3}-m)/3+4m^{2}+m}{m(2m+3)\mu+(m^{3}-m)/3+ 7m^{2}+2m}.\] (35) We consider the boundary \(R_{6,2}=1\). The boundary can be given by the following equation: \[\mu=\mathrm{H}_{6,2}(m)=\frac{m^{2}\log(3/2)+6(2\log(3)-5\log(2))m+2(\log(3)-4 \log(2))}{6(m\log(2/3)+\log(4/3))},\] (36) where \(CEI_{6}>CEI_{2}\) over it (see Figure 1). Boundary (36) cuts axes at points \((\mu,m)=(0,\ 19.2012)\) and \((5.3984,\ 2)\). Thus, we get \(CEI_{6}>CEI_{2}\) since \(R_{6,2}>1\) for all \(m\geq 2\) and \(\mu>\mathrm{H}_{6,2}(m)\). This completes the proof. In Tables 2 and 3, we show the computational efficiency indices of NM, CM4, SNAM, CHM, CTVM and M6, for different values of the size of the nonlinear systems. The results shown in Tables 2 and 3 are in concordance with the Theorems 2 and 3. We can see that, for small- and medium-sized systems, the computational efficiency index of our method (M6) is similar to the other methods in this paper.