# Local Convergence Study for an Iterative Scheme with a High Order of Convergence
## 2 Local Convergence Analysis
### Local Convergence Analysis Using Condition (\(C_{4}\))  
In the preceding section, we refrained from using condition (\(C_{4}\)), which sometimes is a downside when dealing with practical problems. In this section, condition (\(C_{4}\)) is used, enabling us to make a comparison between the radii of local convergence obtained when using \(M\) with those obtained excluding this bound. The local convergence analysis is hereby completed. Proceeding with the notation already established in (\(C_{1}\))-(\(C_{4}\)), we now define new auxiliary functions \(\overline{g_{i}}\) that depend on \(M\). For the study of local convergence of (1), the solution \(x^{*}\) is assumed to exist and conditions are established on operators \((\mathcal{T},\,\mathcal{T}^{\prime})\) in that solution. Therefore, we assume conditions (\(C_{1}\))-(\(C_{4}\)) for all \(x,\,y\in D\) and for real numbers \(L_{0}>0,\,L>0\). Let us define function the following function on the interval \(\left[0,\frac{1}{L_{0}}\right[\):  
\[\overline{g_{1}}(t)=\frac{1}{1-L_{0}t}\left(\frac{L}{2}t+\frac{1}{3}M\right) \tag{12}\]  
Now, consider the function \(\overline{h_{1}}(t)=\overline{g_{1}}(t)-1\), then \(\overline{h_{1}}(0)=\frac{M}{3}-1=\frac{M-3}{3}<0\) for \(M<3\) and \(\overline{h_{1}}(1/L_{0})=+\infty\). Consequently, \(\overline{h_{1}}(t)\) has at least one zero in \(]0,\frac{1}{L_{0}}[\). Let \(\overline{r_{1}}\) the smallest zero in \(]0,L_{0}[\), which implies that:  
\[0<\overline{r_{1}}<1/L_{0}\quad y\quad 0\leq\overline{g_{1}}(t)<1,\,\,\,\forall t \in[0,\overline{r_{1}}[. \tag{13}\]  
and  
\[\overline{r_{1}}=\frac{2(3-M)}{3(L+2L_{0})}\]  
Moreover, define function \(g_{2}\) in the interval \([0,\overline{r_{1}}[\) by:  
\[\overline{g_{2}}(t) =\frac{Lt}{2(1-L_{0}t)}+\left|\frac{3}{8}+\alpha\right|\frac{M}{1 -L_{0}\,t}+\left|\alpha\right|\frac{M}{1-L_{0}\,t}\,\overline{g_{1}(t)}+\left| \frac{\alpha}{3}\right|\frac{M^{2}}{1-L_{0}\,t}\] \[+\left|\frac{3}{8}-\frac{\alpha}{3}\right|\left(\frac{M}{1-L_{0} \,t}\,\overline{g_{1}(t)}\right)^{2}.\]  
Consider the function \(\overline{h_{2}}(t)=\overline{g_{2}}(t)-1\). Following that:  
\[\overline{g_{1}}(0) =\frac{M}{3},\quad\overline{g_{2}}(0)=\left|\frac{3}{8}+\alpha \right|M+\left|\alpha\right|M+\left|\frac{\alpha}{3}\right|M^{2}+\left|\frac{ 3}{8}-\frac{\alpha}{3}\right|M^{2},\,\,\,\text{then}\] \[\overline{h_{2}}(0) =\left|\frac{3}{8}+\alpha\right|M+\left|\alpha\right|M+\left| \frac{\alpha}{3}\right|M^{2}+\left|\frac{3}{8}-\frac{\alpha}{3}\right|M^{2}-1 \,\,\,\text{and}\,\,\,\,\overline{h_{2}}(0)<0\,\,\,\text{if}\] \[\left|\frac{3}{8}+\alpha\right|M+\left|\alpha\right|M+\left| \frac{\alpha}{3}\right|M^{2}+\left|\frac{3}{8}-\frac{\alpha}{3}\right|M^{2}<1,\,\,\,\text{also}\] \[\overline{h_{2}}(\overline{r_{1}}) =\frac{L\overline{r_{1}}}{2(1-L_{0}\overline{r_{1}})}+\left| \frac{3}{8}+\alpha\right|\frac{M}{1-L_{0}\,\overline{r_{1}}}+\left|\alpha \right|\frac{M}{1-L_{0}\,\overline{r_{1}}\,\overline{g_{1}}(\overline{r_{1}}) }+\left|\frac{\alpha}{3}\right|\frac{M^{2}}{1-L_{0}\,\overline{r_{1}}}\] \[+\left|\frac{3}{8}-\frac{\alpha}{3}\right|\left(\frac{M}{1-L_{0} \,\overline{r_{1}}\,\overline{g_{1}}(\overline{r_{1}})}\right)^{2}-1.\]  
From (13), we have that \(1-L_{0}\,\overline{r_{1}}>0\), \(\overline{g_{1}}(\overline{r_{1}})=1\) and \(1-L_{0}\overline{r_{1}}\,\overline{g_{1}}(\overline{r_{1}})>0\), and conclude that \(\overline{h_{2}}(\overline{r_{1}})>0\). Consequently, \(\overline{h_{2}}(t)\) has at least one zero in \(]0,\overline{r_{1}}[\). Let \(\overline{r}\) be the smallest zero in \(]0,\overline{r_{1}}[\), we have:  
\[0<\overline{r_{1}}<1/L_{0}\quad y\quad 0\leq\overline{g_{1}}(t)<1,\,\,\,\forall t \in[0,\overline{r_{1}}[. \tag{13}\]0We will now consider the following lemma. **Lemma 2**. : _Provided that operator \(\mathcal{T}\) satisfies (C\({}_{1}\)) and (C\({}_{4}\)), then the inequalities listed below are verified, for all \(x\in D\) and \(t\in[0,1]\):_  
\[