# Local Convergence of an Efficient Multipoint Iterative Method in Banach Space
## 3 Numerical Results  
It is noted that in all examples \(r_{i}\) are found by solving scalar equations \(\bar{p}_{i}(x)=0\), \(i=1,2,3\). Then, \(r\) is obtained using Equation (6). The parameters \(r_{i}\) have been shown to exist above Lemma 1. The divided difference in all examples is chosen as  
\[[x,y,F]=\int_{0}^{1}F^{\prime}(y+\theta(x-y))\mathrm{d}\theta. \tag{28}\]  
All computations are performed in Mathematica software using multi-precision arithmetic.  
**Example 1**.: _Let \(B=C[0,1]\), \(\Omega=\bar{U}(u_{*},1)\). Consider the Hammerstein-type problem as_  
\[u(s)=\int_{0}^{1}K(s,x)\frac{u(x)^{2}}{2}dx,\]_where_  
\[K(s,x)=\left\{\begin{array}{ll}(1-s)x,\ x\leq s,\\ s(1-x),\ s\leq x.\end{array}\right.\]  
_Let \(F:\mathcal{C}[0,1]\) be defined as_  
\[F(u)(s)=u(s)-\int_{0}^{1}K(s,x)\frac{u(x)^{2}}{2}dx.\]  
_But, we get_  
\[\Big{\|}\int_{0}^{1}K(s,x)dx\Big{\|}\leq\frac{1}{8},\]  
_leading to_  
\[F^{\prime}(u)y(s)=y(s)-\int_{0}^{1}K(s,x)u(x)dx,\]  
_since \(F^{\prime}(u_{*}(s))=I\),_  
\[\|F(u_{*})^{-1}(F(u)-F^{\prime}(y))\|<\frac{1}{8}\|u-y\|.\]  
_By Equation (28), we select \(w_{0}(s,x)=w(s,x)=\frac{s+x}{16}\), \(v(x)=w_{1}(x)=8\). The parameters are_  
\[r_{1}=0.88889,r_{2}=0.118175,r_{3}=1.42264\times 10^{-2}\ \text{and}\ \ r=1.42264\times 10^{-2}.\]  
**Example 2**.: _Consider the three-dimensional system_  
\[f_{1}^{\prime}(u_{1})-f_{1}(u_{1})-1=0,\] \[f_{2}^{\prime}(u_{2})-(e-1)u_{2}-1=0,\] \[f_{3}^{\prime}(u_{3})-1=0,\]  
_with \(u_{1}\), \(u_{2}\), \(u_{3}\in\Omega\) for \(f_{1}(0)=f_{2}(0)=f_{3}(0)=0\). Then, the system for \(u=(u_{1},u_{2},u_{3})^{T}\) with \(F:=(f_{1},f_{2},f_{3}):\Omega\to\mathbb{R}^{3}\) is_  
\[F(u)=\Big{(}e^{\mu_{1}}-1,\frac{e-1}{2}{u_{2}}^{2}+u_{2},u_{3}\Big{)}^{T}.\]  
_Hence, we obtain_  
\[F^{\prime}(u)=\quad\begin{bmatrix}e^{\mu_{1}}&0&0\\ 0&(e-1)u_{2}+1&0\\ 0&0&1\end{bmatrix}.\]  
_But \(u_{*}=(0,0,0)^{T}\) and \(F^{\prime}(u_{*})=I\), by the definition in Equation (28), we select \(w_{0}(s,x)=\frac{L_{0}}{2}(s+x)\), \(w(s,x)=\frac{L}{2}(s+x)\), and \(w_{1}(x)=v(x)=\frac{1}{2}(1+e^{\frac{1}{L_{0}}})\), where \(L_{0}=e-1\), \(L=e\). Then, we have_  
\[r_{1}=1.88242\times 10^{-1},r_{2}=3.70077\times 10^{-2},r_{3}=7.34314\times 10^{- 3}\ \text{and}\ \ r=7.34314\times 10^{-3}.\]  
**Example 3**.: _Consider \(F:=(f_{1},f_{2},f_{3}):\Omega\to\mathbb{R}^{3}\) be defined by_  
\[F(u)=\ (10u_{1}+\sin(u_{1}+u_{2})-1,\ 8u_{2}-\cos^{2}(u_{3}-u_{2})-1,\ 12u_{3}+\sin(u_{3})-1)^{T},\]  
_where \(u=(u_{1},u_{2},u_{3})^{T}\)._  
_Then, we obtain_  
\[F^{\prime}(u)=\quad\begin{bmatrix}10+\cos(u_{1}+u_{2})&\cos(u_{1}+u_{2})&0 \\ 0&8+\sin 2(u_{2}-u_{3})&-\sin 2(u_{2}-u_{3})\\ 0&0&12+\cos(u_{3})\end{bmatrix}.\]_Hence, by Equation (28) \(w_{1}(x)=v(x)=0.269812\,x\), and \(w(s,x)=w_{0}(s,x)=\frac{1.08139}{2}(s+x)\). Then we obtain the parameters as_  
\[r_{1}=0.766299,r_{2}=0.658762,r_{3}=0.637403,r=0.637403.\]  
**Example 4**.: _Define function \(F\) on \(\Omega=\ \bar{U}(0,1)\), given as_  
\[F(\phi)(s)=\ \phi(s)-10\int_{0}^{1}s\,x\phi(x)dx.\]  
_Then, we get_  
\[F^{\prime}(\phi(\zeta))(s)=\ \xi(s)-30\int_{0}^{1}s\,x\phi(x)^{2}\xi(x)dx,\ \ \text{for all}\ \ \xi\in\Omega.\]  
_By Equation (28), we can choose \(w_{1}(x)=v(x)=30\) and \(w(s,x)=w_{0}(s,x)=\frac{15}{2}(s+x)\). The parameter values are given as_  
\[r_{1}=2.15054\times 10^{-3},r_{2}=2.24959\times 10^{-9},r_{3}=9.81657\times 10^{ -16},r=9.81657\times 10^{-16}.\]  
**Example 5**.: _The Van der Waals equation of state for a vapor is (see [16])_  
\[\Big{(}P+\frac{a}{V^{2}}\Big{)}(V-b)=R\,T,\]  
_This equation leads to,_  
\[PV^{3}-(Pb+RT)V^{2}+aV-ab=0\]  
_in \(V\), where all constants have a physical meaning whose values can be found in [16]. Choose \(P\) = 10,000 kPa and \(T\) = 800 K. Then, \(u_{*}=36.9167\ldots\). By Equation (28), we can set \(w_{1}(x)=v(x)=10\) and \(w(s,x)=w_{0}(s,x)=0.386121(s+x)\). The parameter values are given as_  
\[r_{1}=1.17721\times 10^{-1},r_{2}=8.79936\times 10^{-4},r_{3}=2.96866\times 10^{ -6},r=2.96866\times 10^{-6}.\]