# An Efficient Sixth-Order Newton-Type Method for Solving Nonlinear Systems
## 3 Computational Efficiency  
Here, we want to compare the computational efficiency index of our sixth-order method (M6, (6)) with [PERSON]'s method (NM, (1)), [PERSON]'s fourth-order method (CM4, (3)), [PERSON]'s sixth-order method (SNAM, (5)) and [PERSON]'s sixth-order methods (CHM, (4)) and (CTVM, (18)). The method CTVM [7] is as follows:  
\[\left\{ \begin{array}{ll}y^{(k)}=x^{(k)}-1/2F^{\prime}(x^{(k)})^{-1}F(x^{(k) }),\\ z^{(k)}=x^{(k)}+[F^{\prime}(x^{(k)})-2F^{\prime}(y^{(k)})]^{-1}[3F(x^{(k)})-4F (y^{(k)})],\\ x^{(k+1)}=z^{(k)}+[F^{\prime}(x^{(k)})-2F^{\prime}(y^{(k)})]^{-1}F(z^{(k)}). \end{array} \right. \tag{23}\]  
We define, respectively, the computational efficiency index (\(CEI\)) of the methods NM, CM4, SNAM, CHM, CTVM and M6 as:  
\[CEI_{i}(\mu,\;m)=\rho_{i}\frac{1}{i^{(i,m-)}},\;i=1,2,3,4,5,6, \tag{24}\]  
where \(\rho_{i}\) is the convergence order of the method and \(C_{i}(\mu,m)\) is the computational cost of method. The \(C_{i}(u,m)\) is given by [8]:  
\[C_{i}(\mu,\;m)=a_{i}(m)\mu+p_{i}(m), \tag{25}\]  
where \(a_{i}(m)\) represents the number of evaluations of the scalar functions used in the evaluations of \(F\), \(F^{\prime}\) and \([y,x;F]\). The \(p_{i}(m)\) represents the computational cost per iteration. \(\mu>0\) is the ratio between multiplications (and divisions) and evaluations of functions that are required to express \(C_{i}(\mu,m)\) in terms of multiplications (and divisions). The divided difference \([y,x;F]\) is defined by [9]:  
\[[y,x;F]_{ij}=(F_{i}(y_{1}\cdots,y_{j-1},y_{j},x_{j+1},\cdots,x_{m})-F_{i}(y_{1} \cdots,y_{j-1},x_{j},x_{j+1},\cdots,x_{m}))/(y_{j}-x_{j}),\;1\leq i,j\leq m,\]  
where \(F(x)\) and \(F(y)\) are computed separately. When we compute a divided difference, we need \(m^{2}\) quotients and \(m(m-1)\) scalar function evaluations. We must add \(m\) multiplications for the scalar product, \(m^{2}\) multiplications for the matrix-vector multiplication and \(m^{2}\) evaluations of scalar functions for any new derivative \(F^{\prime}\). In order to factorize a matrix, we require \((m^{3}-m)/3\) multiplicationsand divisions in the LU decomposition. We require \(m^{2}\) multiplications and divisions for solving two triangular linear systems. Taking into account the previous considerations, we give the computational cost of each iterative method in Table 1.  
From Table 1, we can see that these sixth-order iterative methods have the same number of function evaluations. Our method (M6) needs less LU decompositions than other methods with the same order. Therefore, the computational cost of our method is lower. We use the following expressions [10] to compare the computational efficiency index of each iterative method:  
\[R_{i,j}=\frac{\log CEI_{i}}{\log CEI_{j}}=\frac{\log(\rho_{i})C_{j}(\mu,m)}{ \log(\rho_{j})C_{i}(\mu,m)},i,j=1,2,3,4,5,6. \tag{26}\]  
For \(R_{i,j}>1\) the iterative method \(M_{i}\) is more efficient than \(M_{j}\). We have the following theorem:  
**Theorem 2**. : _For all \(\mu>0\) we have:_  
1. \(CEI_{6}>CEI_{3}\) _for all_ \(m\geq 5\)_,_
2. \(CEI_{6}>CEI_{4}\) _for all_ \(m\geq 7\)_,_
3. \(CEI_{6}>CEI_{5}\) _for all_ \(m\geq 9\)_._  
**Proof**. We note that the methods SNAM, CHM, CTVM and method M6 have the same order \(\rho_{3}=\rho_{4}=\rho_{5}=\rho_{6}=6\) and the same number of function evaluations \(a_{3}(m)=a_{4}(m)=a_{5}(m)=a_{6}(m)=m(2m+3)\). 1. Based on the expression (26), the relation between SNAM and M6 can be given by: \[R_{6,3}=\frac{\log(\rho_{6})C_{3}(\mu,m)}{\log(\rho_{3})C_{6}(\mu,m)}=\frac{m( 2m+3)\mu+2(m^{3}-m)/3+6m^{2}}{m(2m+3)\mu+(m^{3}-m)/3+7m^{2}+2m}.\] (27) Subtracting the denominator from the numerator of (27), we have: \[\frac{1}{3}m(m^{2}-3m-7).\] (28) Equation (28) is positive for \(m\geq 4.541\). Thus, we get \(CEI_{6}>CEI_{3}\) for all \(m\geq 5\) and \(\mu>0\). 2.