# Expanding the Applicability of Some High Order Householder-Like Methods
## 2 Semilocal Convergence
Indeed, we need the following definition. **Definition 3**. : _Let \(F:D\to B_{2}\) be a five-times Frechet-differentiable, which satisfies the center \(\gamma_{0}\)-condition. We say that operator \(F\) satisfies the \(\delta\)-\(\gamma\)-condition at \(x_{0}\), if for each \(x\in D_{0}\):_  
\[\|\Gamma_{0}F(x_{0})\| \leq \beta,\] \[\|\Gamma_{0}F^{\prime\prime}(x_{0})\| \leq \frac{2\delta}{(1-\delta\|x-x_{0}\|)^{2}},\] \[\|\Gamma_{0}F^{\prime\prime}(x)\| \leq \frac{6\delta^{2}}{(1-\delta\|x-x_{0}\|)^{4}}=f_{1}^{\prime \prime\prime}(\|x-x_{0}\|),\] \[t_{n+1} = s_{n}-\frac{A_{0}(f;t_{n})}{f^{\prime}(t_{n})}-\left(\sum_{k=2}^ {\infty}\frac{A_{k}(f;t_{n})}{f^{\prime}(t_{n})}\left[\frac{f(t_{n})}{f^{ \prime}(t_{n})}\right]^{k-1}\right)\frac{f(t_{n})}{f^{\prime}(t_{n})}.\]0 \[t_{n+1} = s_{n}-\frac{A_{0}(f;t_{n})}{f^{\prime}(t_{n})}-\left(\sum_{k=2}^ {\infty}\frac{A_{k}(f;t_{n})}{f^{\prime}(t_{n})}\left[\frac{f(t_{n})}{f^{ \prime}(t_{n})}\right]^{k-1}\right)\frac{f(t_{n})}{f^{\prime}(t_{n})}.\]1  
_Clearly, we have that \(\delta\leq\gamma\)._  
_Now, we define the scalar sequences \(\{s_{n}\}\), \(\{\bar{t}_{n}\}\) by:_  
\[ \begin{array}{lll}\bar{t}_{0}&=&0,\quad\bar{s}_{0}=\beta,\\ \bar{s}_{n}&=&\bar{t}_{n}-\frac{f_{1}(\bar{t}_{n})}{f_{1}^{\prime}(\bar{t}_{n} )}\quad\quad n\geq 0,\\ \bar{t}_{n+1}&=&\bar{s}_{n}-\frac{A_{0}(f_{1};\bar{t}_{n})}{f_{0}^{\prime}(\bar {t}_{n})}-\left(\sum_{k=2}^{\infty}\frac{A_{k}(f_{1};\bar{t}_{n})}{f_{1}^{ \prime}(\bar{t}_{n})}\left[\frac{f_{1}(\bar{t}_{n})}{f_{1}^{\prime}(\bar{t}_{ n})}\right]^{k-1}\right)\frac{f_{1}(\bar{t}_{n})}{f_{1}^{\prime}(\bar{t}_{n})}. \end{array}  \tag{14}\]  
_where function \(f_{1}\) is defined on the interval \([0,1/\delta)\) by:_  
\[f_{1}(t)=\beta-t+\frac{\delta t^{2}}{1-\delta t}, \tag{15}\]  
_for some \(t\in(0,\frac{1}{\delta})\)._  
Moreover, define constants \(\alpha_{0}\), \(\bar{t}^{*}\), \(\bar{t}^{**}\) and \(R_{0}\) by \(\alpha_{0}=\delta\beta\), \(\bar{t}^{*}=\frac{1+\alpha_{0}-\sqrt{(1+\alpha_{0})^{2}-8\alpha_{0}}}{4\delta}\), \(\bar{t}^{**}=\frac{1+\alpha_{0}+\sqrt{(1+\alpha_{0})^{2}-8\alpha_{0}}}{4\delta}\) and \(R_{0}=(1-\frac{1}{\sqrt{2}})\frac{1}{\gamma_{0}}\). If condition:  
\[\alpha_{0}\leq 3-2\sqrt{2} \tag{16}\]  
holds, then function \(f_{1}\) has two real roots \(\bar{t}^{*}\) and \(\bar{t}^{**}\) such that:  
\[\beta\leq\bar{t}^{*}\leq\left(1+\frac{1}{\sqrt{2}}\right)\beta\leq\left(1- \frac{1}{\sqrt{2}}\right)\frac{1}{\delta}\leq\bar{t}^{**}\leq\frac{1}{2\delta}.\]  
Moreover, we have for each \(t\in[0,\bar{t}^{*})\): \(f_{1}(t)>0\), \(f_{1}^{\prime}(t)=\frac{1-2(1-\delta t)^{2}}{(1-\delta t)^{2}}<0\) and:  
\[\bar{f}_{1}^{(i)}(t)=\frac{i!\delta^{i-1}}{(1-\delta t)(2+i)},\quad i=2,3,\dots.\]  
Notice that:  
\[\alpha\leq 3-2\sqrt{2}\Rightarrow\alpha_{0}\leq 3-2\sqrt{2} \tag{17}\]  
but not necessarily vice versa, unless, if \(\gamma=\delta\). We need a series of auxiliary results. **Lemma 1**. : _Suppose that \(\alpha_{0}\leq 3-2\sqrt{2}\), \(A_{0}(f_{1};t)\geq 0\) and \(\sum_{k=2}^{\infty}\frac{A_{k}(f_{1};t)}{f_{1}^{\prime}(t)}\left[\frac{f_{1}(t)} {f_{1}^{\prime}(t)}\right]^{k-1}\geq 0.\) Then, sequences \(\{\overline{s}_{n}\}\), \(\{\overline{t}_{n}\}\) generated by (14) are well defined for each \(n=0,1,2,\ldots\) and converge monotonically to \(\bar{t}^{*}\) so that:_  
\[0\leq\overline{t}_{n}\leq\overline{s}_{n}\leq\overline{t}_{n+1}<\bar{t}^{*}. \tag{18}\]  
Proof. : Simply replace \(f,\alpha,t^{*},s_{n}\), \(t_{n}\) in the proof of Lemma 2 [4] by \(f_{1},\alpha_{0},\bar{t}^{*},\bar{s}_{n},\bar{t}_{n}\), respectively. **Lemma 2**. :
1. _Let_ \(f_{1}\) _be the real function given in (_15_).