## 5 Numerical Results  
Numerical performance of the presented methods is compared with [PERSON]'s method and some known methods given in the beginning of this paper. All the numerical calculations are done using MATLAB package for the test examples given below. We have taken 500 digits accuracy while calculating the approximate numerical solutions. The following error residual is adopted to stop the iteration:  
\[err_{min}=\|x^{(r+1)}-x^{(r)}\|_{2}<10^{-100}.\]  
The following formula is used to find approximated computational order of convergence \(p_{c}\):  
\[p_{c}\approx\frac{\log\left(\|x^{(r+1)}-x^{(r)}\|_{2}/\|x^{(r)}-x^{(r-1)}\|_{ 2}\right)}{\log\left(\|x^{(r)}-x^{(r-1)}\|_{2}/\|x^{(r-1)}-x^{(r-2)}\|_{2} \right)}.\]  
Here N denotes the number of iterations needed for obtaining the minimum residual \((err_{min})\), \(n_{inv}\) represents the total number of Frechet derivative's inverse and \(n_{total}\) represents the total number of function evaluations \((\Phi\)_and_\(\Phi^{\prime})\) to reach the minimum residual as in [7]. **Test Example 1** (Te1):  
\[\Phi(y_{1},y_{2})=(y_{1}+exp(y_{2})-cos(y_{2}),\quad 3y_{1}-y_{2}-sin(y_{2})).\]  
The Jacobian matrix is given by \(\Phi^{\prime}(y)=\left( \begin{array}{cc}1&exp(y_{2})+sin(y_{2})\\ 3&-1-cos(y_{2})\end{array} \right).\) Initial approximation is taken as \(y^{(0)}=(1.5,2)^{T}\) and the analytic solution is given by \(\alpha=(0,0)^{T}\). **Test Example 2** (TE2):  
\[\left\{ \begin{array}{ll}y_{2}y_{3}+y_{4}(y_{2}+y_{3})=0,\\ y_{1}y_{3}+y_{4}(y_{1}+y_{3})=0,\\ y_{1}y_{2}+y_{4}(y_{1}+y_{2})=0,\\ y_{1}y_{2}+y_{1}y_{3}+y_{2}y_{3}=1.\end{array} \right.\]  
The above system is solved by taking the starting approximation \(y^{(0)}=(0.5,0.5,0.5,-0.2)^{T}\). The solution is given by \(\alpha\approx(0.577350,0.577350,0.577350,-0.288675)^{T}\). The Jacobian matrix is given by  
\[\Phi^{\prime}(y)=\left( \begin{array}{cccc}0&y_{3}+y_{4}&y_{2}+y_{4}&y_{2}+y_ {3}\\ y_{3}+y_{4}&0&y_{1}+y_{4}&y_{1}+y_{3}\\ y_{2}+y_{4}&y_{1}+y_{4}&0&y_{1}+y_{2}\\ y_{2}+y_{3}&y_{1}+y_{3}&y_{1}+y_{2}&0\end{array} \right).\]  
**Test Example 3** (TE3):  
\[\left\{ \begin{array}{ll}\cos y_{2}-\sin y_{1}=0,\\ y_{3}^{y_{1}}-\dfrac{1}{y_{2}}=0,\\ \exp y_{1}-y_{3}^{2}=0.\end{array} \right.\]  
The solution for the above system is \(\alpha\approx(0.909569,0.661227,1.575834)^{T}\). The initial vector for the iteration is taken as \(x^{(0)}=(1,0.5,1.5)^{T}\). The Jacobian matrix produced thus is given by  
\[\Phi^{\prime}(y)=\left( \begin{array}{cccc}-\cos y_{1}&-\sin y_{2}&0\\ y_{3}^{y_{1}}\ln y_{3}&1/y_{2}^{2}&y_{3}^{y_{1}}y_{1}/y_{3}\\ \exp y_{1}&0&-2y_{3}\end{array} \right).\]  
**Test Example 4** (TE4): The following boundary value problem is considered  
\[y^{\prime\prime}+y^{3}=0,\;y(0)=0,\;y(1)=1,\]  
where equal mesh is used for dividing the interval \([0,1]\) which is given below  
\[u_{0}=0<u_{1}<u_{2}<...<u_{m-1}<u_{m}=1,\;u_{j+1}=u_{j}+h,\;h=1/m.\]  
Denote \(y_{0}=y(u_{0})=0,y_{1}=y(u_{1}),...,y_{m-1}=y(u_{m-1}),y_{m}=y(u_{m})=1\). Discretizing the second derivative by the following difference formula  
\[y^{\prime\prime}\approx\dfrac{y_{r-1}-2y_{r}+y_{r+1}}{h^{2}},\;r=1,2,3,...,m-1,\]  
we obtain \(m-1\) nonlinear equations in \(m-1\) variables as given below  
\[y_{r-1}-2y_{r}+y_{r+1}+h^{2}y_{r}^{3}=0,r=1,2,3,...,m-1.\]The above equations are solved by taking \(m=16\) (i.e., \(n=15\)) and \(y^{(0)}=(1,1,...,1)^{T}\) as the initial approximation, where we get the Jacobian matrix with 43 non-zero elements as below. \[\left( \begin{array}{cccccccc}3h^{2}y_{1}^{2}-2&1&0&0&...&0&0&0\\ 1&3h^{2}y_{2}^{2}-2&1&0&...&0&0&0\\ 0&1&3h^{2}y_{3}^{2}-2&1&...&0&0&0\\.&&&&&&&&\\.&&&&&&&&\\.&&&&&&&&\\ 0&0&0&0&...&1&3h^{2}y_{14}^{2}-2&1\\ 0&0&0&0&...&0&1&3h^{2}y_{15}^{2}-2\end{array} \right).\]  
The following solution is obtained for the differential equation at the given mesh points. \[