# Local Convergence Study for an Iterative Scheme with a High Order of Convergence
## 2 Local Convergence Analysis
### Local Convergence Analysis Using Conditions \((C_{1})\), \((C_{2})\) and \((C_{3})\)
By using mathematical induction, we replace \(x_{0},y_{0},x_{1}\) with \(x_{k},y_{k},x_{k+1}\) and have:  
\[0<r_{1}<1/L_{0}\quad\text{and}\quad 0\leq g_{1}(t)<1,\,\,\,\,\forall t\in[0,r_{1}[. \tag{3}\]9  
\[g_{2}(t) =\frac{Lt}{2(1-L_{0}\,t)}+\left|\frac{3}{8}+\alpha\right|\frac{1 +L_{0}\,t}{1-L_{0}\,t}+\left|\alpha\right|\frac{1+L_{0}\,t}{1-L_{0}\,t\,g_{1} (t)}+\left|\frac{\alpha}{3}\right|\frac{1+L_{0}\,t\,g_{1}(t)}{1-L_{0}\,t}(1+ L_{0}\,t)\]0  
where the functions \(g_{i}\) have been previously obtained. Using \(\|x_{k+1}-x^{*}\|\leq\|x_{k}-x^{*}\|<r\), it is hereby concluded that \(x_{k+1}\in B(x^{*},r)\). Function \(g_{2}\) is increasing over its domain, and therefore:  
\[g_{2}(t) =\frac{Lt}{2(1-L_{0}\,t)}+\left|\frac{3}{8}+\alpha\right|\frac{1 +L_{0}\,t}{1-L_{0}\,t}+\left|\alpha\right|\frac{1+L_{0}\,t}{1-L_{0}\,t\,g_{1} (t)}+\left|\frac{\alpha}{3}\right|\frac{1+L_{0}\,t\,g_{1}(t)}{1-L_{0}\,t}(1+ L_{0}\,t)\]1 \[g_{2}(t) =\frac{Lt}{2(1-L_{0}\,t)}+\left|\frac{3}{8}+\alpha\right|\frac{1 +L_{0}\,t}{1-L_{0}\,t}+\left|\alpha\right|\frac{1+L_{0}\,t}{1-L_{0}\,t\,g_{1} (t)}+\left|\frac{\alpha}{3}\right|\frac{1+L_{0}\,t\,g_{1}(t)}{1-L_{0}\,t}(1+ L_{0}\,t)\]2 \[g_{2}(t) =\frac{Lt}{2(1-L_{0}\,t)}+\left|\frac{3}{8}+\alpha\right|\frac{1 +L_{0}\,t}{1-L_{0}\,t}+\left|\alpha\right|\frac{1+L_{0}\,t}{1-L_{0}\,t\,g_{1} (t)}+\left|\frac{\alpha}{3}\right|\frac{1+L_{0}\,t\,g_{1}(t)}{1-L_{0}\,t}(1+ L_{0}\,t)\]3 \[g_{2}(t) =\frac{Lt}{2(1-L_{0}\,t)}+\left|\frac{3}{8}+\alpha\right|\frac{1 +L_{0}\,t}{1-L_{0}\,t}+\left|\alpha\right|\frac{1+L_{0}\,t}{1-L_{0}\,t\,g_{1} (t)}+\left|\frac{\alpha}{3}\right|\frac{1+L_{0}\,t\,g_{1}(t)}{1-L_{0}\,t}(1+ L_{0}\,t)\]4  
By applying limits in (11), we get:  
\[g_{2}(t) =\frac{Lt}{2(1-L_{0}\,t)}+\left|\frac{3}{8}+\alpha\right|\frac{1 +L_{0}\,t}{1-L_{0}\,t}+\left|\alpha\right|\frac{1+L_{0}\,t}{1-L_{0}\,t\,g_{1} (t)}+\left|\frac{\alpha}{3}\right|\frac{1+L_{0}\,t\,g_{1}(t)}{1-L_{0}\,t}(1+ L_{0}\,t)\]5  
where \(\lim_{k\to\infty}[g_{2}(\|x_{0}-x^{*}\|)]^{k+1}=0\), and therefore, \(\lim_{k\to\infty}x_{k}=x^{*}\). Subsequently, the method converges to the solution. In order to demonstrate uniqueness, if we have \(y^{*}\in B(x^{*},R)\) with \(\mathcal{T}(y^{*})=0\). Let the operator \(P=\int_{0}^{1}\mathcal{T}^{\prime}(y^{*}+t(x^{*}-y^{*}))dt\). Therefore, using (C2):  
\[g_{2}(t) =\frac{Lt}{2(1-L_{0}\,t)}+\left|\frac{3}{8}+\alpha\right|\frac{1 +L_{0}\,t}{1-L_{0}\,t}+\left|\alpha\right|\frac{1+L_{0}\,t}{1-L_{0}\,t\,g_{1} (t)}+\left|\frac{\alpha}{3}\right|\frac{1+L_{0}\,t\,g_{1}(t)}{1-L_{0}\,t}(1+ L_{0}\,t)\]6 \[g_{2}(t) =\frac{Lt}{2(1-L_{0}\,t)}+\left|\frac{3}{8}+\alpha\right|\frac{1 +L_{0}\,t}{1-L_{0}\,t}+\left|\alpha\right|\frac{1+L_{0}\,t}{1-L_{0}\,t\,g_{1} (t)}+\left|\frac{\alpha}{3}\right|\frac{1+L_{0}\,t\,g_{1}(t)}{1-L_{0}\,t}(1+ L_{0}\,t)\]7 \[g_{2}(t) =\frac{Lt}{2(1-L_{0}\,t)}+\left|\frac{3}{8}+\alpha\right|\frac{1 +L_{0}\,t}{1-L_{0}\,t}+\left|\alpha\right|\frac{1+L_{0}\,t}{1-L_{0}\,t\,g_{1} (t)}+\left|\frac{\alpha}{3}\right|\frac{1+L_{0}\,t\,g_{1}(t)}{1-L_{0}\,t}(1+ L_{0}\,t)\]8  
Thus:  
\[g_{2}(t) =\frac{Lt}{2(1-L_{0}\,t)}+\left|\frac{3}{8}+\alpha\right|\frac{1 +L_{0}\,t}{1-L_{0}\,t}+\left|\alpha\right|\frac{1+L_{0}\,t}{1-L_{0}\,t\,g_{1} (t)}+\left|\frac{\alpha}{3}\right|\frac{1+L_{0}\,t\,g_{1}(t)}{1-L_{0}\,t}(1+ L_{0}\,t)\]9  
For the aforementioned reasons and Banach Lemma, \(P^{-1}\) exists and:  
\[+\left|\frac{3}{8}-\frac{\alpha}{3}\right|\left(\frac{1+L_{0}\,t }{1-L_{0}\,t\,g_{1}(t)}\right)^{2}.\]0  
Using the inverse operator \(P^{-1}\):  
\[+\left|\frac{3}{8}-\frac{\alpha}{3}\right|\left(\frac{1+L_{0}\,t }{1-L_{0}\,t\,g_{1}(t)}\right)^{2}.\]1  
hence, \(x^{*}-y^{*}=0\), and therefore, \(x^{*}=y^{*}\). It is concluded that the root \(x^{*}\) is unique.