# Hybrid Newton-like Inverse Free Algorithms for Solving Nonlinear Equations
## 6 Numerical Examples
: _The approximated computational order of convergence of a sequence \(\{x_{n}\}_{n\geq 0}\) is defined by_  
\[\widehat{\rho}_{n}=\frac{\ln|\widehat{e}_{n+1}/\widehat{e}_{n}|}{\ln|\widehat {e}_{n}/\widehat{e}_{n-1}|},\]  
_where \(\widehat{e}_{n}=x_{n}-x_{n-1}\). \(x_{n},x_{n-1},x_{n-2}\) are three consecutive iterates [6]._  
Table 5 demonstrates that the convergence of the proposed methods closely corresponds with the convergence of [PERSON]'s method, particularly for values of \(k\) ranging from \(4\) to \(5\) with the convergence order closely approximating \(2\). \begin{table}
\begin{tabular}{c c c c} \hline
**Algorithm** & **Iterations** & **Algorithm** & **Iterations** \\ \hline (2) [PERSON] & 7 & (2) [PERSON] & 7 \\ (37), \(k=1\) & 8 & (42), \(k=1\) & 12 \\ (38), \(k=2\) & 7 & (42), \(k=2\) & 8 \\ (39), \(k=3\) & 7 & (42), \(k=3\) & 8 \\ (40), \(k=4\) & 7 & (42), \(k=4\) & 7 \\ (41), \(k=5\) & 7 & (42), \(k=5\) & 7 \\ \hline \end{tabular}
\end{table} 
Table 4: Number of iterations to achieve tolerance \(\varepsilon=10^{-12}\), where \(x_{0}=(-15,-15)\) and \(\|I-G^{\prime}(x_{0})\|=0.257<1\). \begin{table}
\begin{tabular}{c c c c} \hline
**Algorithm** & **Iterations** & **Algorithm** & **Iterations** \\ \hline (2) [PERSON] & 5 & (2) [PERSON] & 5 \\ (37), \(k=1\) & 7 & (42), \(k=1\) & 9 \\ (38), \(k=2\) & 5 & (42), \(k=2\) & 7 \\ (39), \(k=3\) & 5 & (42), \(k=3\) & 6 \\ (40), \(k=4\) & 5 & (42), \(k=4\) & 6 \\ (41), \(k=5\) & 5 & (42), \(k=5\) & 5 \\ \hline \end{tabular}
\end{table} 
Table 3: Number of iterations to achieve tolerance \(\varepsilon=10^{-9}\), where \(x_{0}=(-15,-15)\) and \(\|I-G^{\prime}(x_{0})\|=0.257<1\). \begin{table}
\begin{tabular}{c c c c} \hline
**Algorithm** & **Iterations** & **Algorithm** & **Iterations** \\ \hline (2) Newton & 7 & (2) Newton & 7 \\ (37), \(k=1\) & 8 & (42), \(k=1\) & 12 \\ (38), \(k=2\) & 7 & (42), \(k=2\) & 8 \\ (39), \(k=3\) & 7 & (42), \(k=3\) & 8 \\ (40), \(k=4\) & 7 & (42), \(k=4\) & 7 \\ (41), \(k=5\) & 7 & (42), \(k=5\) & 7 \\ \hline \end{tabular}
\end{table} 
Table 4: Number of iterations to achieve tolerance \(\varepsilon=10^{-12}\), where \(x_{0}=(-15,-15)\) and \(\|I-G^{\prime}(x_{0})\|=0.257<1\). \begin{table}
\begin{tabular}{c c c c} \hline
**Algorithm** & **Iterations** & **Algorithm** & **Iterations** \\ \hline (2) Newton & 1.8624 & 1.9697 \\ (37), \(k=1\) & 0.863 & 1 \\ (38), \(k=2\) & 0.2695 & 1.0438 \\ (39), \(k=3\) & 1.9714 & 2.3569 \\ (40), \(k=4\) & 1.8354 & 1.9453 \\ (41), \(k=5\) & 1.8642 & 1.9661 \\ \hline \end{tabular}
\end{table} 
Table 5: Computational Order of Convergence and the Approximated Computational Order of Convergence, where \(x_{0}=(-15,-15)\), \(\varepsilon=10^{-12}\). **Example 2**. : _Let \(T_{1}=T_{2}=\mathbb{R}^{3}\) and \(\Omega=U[x^{*},1]\). The mapping \(G\) is defined on \(\Omega\) for \(\Theta=(\Theta_{1},\Theta_{2},\Theta_{3})^{tr}\in\mathbb{R}^{3}\) as_  
\[G(\Theta)=(\Theta_{1},e^{\Theta_{2}}-1,\frac{e-1}{2}\Theta_{3}^{2}+\Theta_{3})^ {tr}.\]  
_Then, the definition of the derivative according to [PERSON] [2; 3; 8; 13; 30] gives for the mapping \(G\) that_  
\[G^{\prime}(\Theta)=\left[ \begin{array}{ccc}1&0&0\\ 0&e^{\Theta_{2}}&0\\ 0&0&(e-1)\Theta_{3}+1\end{array} \right].\]  
_The point \(x^{*}=(0,0,0)^{tr}\) solves the equation \(G(a)=0\). Moreover, \(G^{\prime}(x^{*})=I\). The conditions of Theorem 3 hold, provided that \(\gamma=\gamma(t)=(e-1)t,\psi(t)=(e-1)t\), and \(k=1\). Then, we can take \(\rho_{4}\in(0,0.2909883534)\)._  
**Example 3**. : _Let \(U[0,1]\) stand the space of continuous functions mapping the interval \([0,1]\) into the real number system. Let \(T_{1}=T_{2}=K[0,1]\) and \(\Omega=U[x^{*},1]\) with \(x^{*}(\kappa)=0\). The operator \(G\) is defined on \(U[0,1]\) as_  
\[G(z)(\kappa)=z(\kappa)-4\int_{0}^{1}\kappa z(\tau)^{3}d\tau.\]  
_Then, the definition of the derivative according to [PERSON] [2; 3; 8; 13; 30] gives for the operator \(G\)_  
\[G^{\prime}(z(w))(\kappa)=w(\kappa)-12\int_{0}^{1}\kappa z\tau z(\tau)^{2}w( \tau)d\tau\]  
_for each \(w\in U[0,1]\).