## 2 Main Theorem
Noting that \(s=1+O(e)\) and \(\frac{f(x_{n})}{f^{\prime}(x_{n})}=O(e)\), we need a Taylor expansion of \(T_{f}(s)\) about \(s=1\) up to fifth-order terms:  
\[T_{f}(s)=T_{0}+T_{1}(s-1)+T_{2}(s-1)^{2}+T_{3}(s-1)^{3}+T_{4}(s-1)^{4}+T_{5}(s -1)^{5}+O(e^{6}). \tag{8}\]  
Thus, we find  
\[z_{n}=x_{n}-T_{f}(s)\cdot\frac{f(x_{n})}{f^{\prime}(x_{n})}=\alpha+(1-T_{0})e +(T_{0}+2T_{1}\gamma)\theta_{2}e^{2}+\]  
\[[-2T_{0}(\theta_{2}^{2}-\theta_{3})-\gamma(8T_{1}\theta_{2}^{2}+4T_{2}\gamma \theta_{2}^{2}+3T_{1}(-2+\gamma)\theta_{3})]e^{3}+\Sigma_{i=4}^{6}W_{i}e^{i}+O (e^{7}), \tag{9}\]  
where \(W_{i}=W_{i}(\gamma,\theta_{2},\theta_{3},\cdots,\theta_{6},T_{0},\cdots,T_{5})\) for \(4\leq i\leq 6\). In view of the fact that \(f(z_{n})=f(x_{n})|_{e_{n}\rightarrow(z_{n}-\alpha)}\), we get:  
\[f(z_{n})=f^{\prime}(\alpha)[(1-T_{0})e+[(1-T_{0})^{2}\theta_{2}+(T_{0}+2T_{1} \gamma)\theta_{2}]e^{2}+\Sigma_{i=3}^{6}F_{i}e^{i}+O(e^{7})], \tag{10}\]  
where \(F_{i}=F_{i}(\gamma,\theta_{2},\theta_{3},\cdots,\theta_{6},T_{0},\cdots,T_{5})\) for \(3\leq i\leq 6\). Noting that \(s=O(1)\) and \(\frac{f(z_{n})}{f^{\prime}(x_{n})}=O(e)\), we need a Taylor expansion of \(L_{f}(s)\) about \(s=1\) up to fifth-order terms:  
\[L_{f}(s)=L_{0}+L_{1}(s-1)+L_{2}(s-1)^{2}+L_{3}(s-1)^{3}+L_{4}(s-1)^{4}+L_{5}(s -1)^{5}+O(e^{6}). \tag{11}\]  
Hence, we have:  
\[f^{\prime}(x_{n})=f^{\prime}(\alpha)[1+2\theta_{2}e_{n}+3\theta_{3}e_{n}^{2}+4 \theta_{4}e_{n}^{3}+5\theta_{5}e_{n}^{4}+6\theta_{6}e_{n}^{5}+O(\epsilon_{n}^{ 6})]. \tag{4}\]0  
\[f^{\prime}(x_{n})=f^{\prime}(\alpha)[1+2\theta_{2}e_{n}+3\theta_{3}e_{n}^{2}+4 \theta_{4}e_{n}^{3}+5\theta_{5}e_{n}^{4}+6\theta_{6}e_{n}^{5}+O(\epsilon_{n}^{ 6})]. \tag{4}\]1  
where \(G_{i}=G_{i}(\gamma,\theta_{2},\theta_{3},\cdots,\theta_{6},T_{0},\cdots,T_{5},L_{0},\cdots,L_{5})\) for \(3\leq i\leq 6\). By taking  
\[f^{\prime}(x_{n})=f^{\prime}(\alpha)[1+2\theta_{2}e_{n}+3\theta_{3}e_{n}^{2}+4 \theta_{4}e_{n}^{3}+5\theta_{5}e_{n}^{4}+6\theta_{6}e_{n}^{5}+O(\epsilon_{n}^{ 6})]. \tag{4}\]2  
we further obtain  
\[f^{\prime}(x_{n})=f^{\prime}(\alpha)[1+2\theta_{2}e_{n}+3\theta_{3}e_{n}^{2}+4 \theta_{4}e_{n}^{3}+5\theta_{5}e_{n}^{4}+6\theta_{6}e_{n}^{5}+O(\epsilon_{n}^{ 6})]. \tag{4}\]3  
\[f^{\prime}(x_{n})=f^{\prime}(\alpha)[1+2\theta_{2}e_{n}+3\theta_{3}e_{n}^{2}+4 \theta_{4}e_{n}^{3}+5\theta_{5}e_{n}^{4}+6\theta_{6}e_{n}^{5}+O(\epsilon_{n}^{ 6})]. \tag{4}\]4  
From \((\gamma-1)(3\gamma-2)=0\), we find two sets of relations:  
\[f^{\prime}(x_{n})=f^{\prime}(\alpha)[1+2\theta_{2}e_{n}+3\theta_{3}e_{n}^{2}+4 \theta_{4}e_{n}^{3}+5\theta_{5}e_{n}^{4}+6\theta_{6}e_{n}^{5}+O(\epsilon_{n}^{ 6})]. \tag{4}\]5  
for vanishing the fifth-order term in (14). Hence, we eventually have two sets of relations:  
\[f^{\prime}(x_{n})=f^{\prime}(\alpha)[1+2\theta_{2}e_{n}+3\theta_{3}e_{n}^{2}+4 \theta_{4}e_{n}^{3}+5\theta_{5}e_{n}^{4}+6\theta_{6}e_{n}^{5}+O(\epsilon_{n}^{ 6})]. \tag{4}\]6  
\[f^{\prime}(x_{n})=f^{\prime}(\alpha)[1+2\theta_{2}e_{n}+3\theta_{3}e_{n}^{2}+4 \theta_{4}e_{n}^{3}+5\theta_{5}e_{n}^{4}+6\theta_{6}e_{n}^{5}+O(\epsilon_{n}^{ 6})]. \tag{4}\]7which are substituted into \(G_{6}\) in (14) and lead us to the desired relation (2) with \(\{|T_{3}|<\infty,|L_{2}|<\infty\}\) when \(\gamma=\frac{2}{3}\), or with \(\{|L_{3}|<\infty,|T_{2}|<\infty\}\) when \(\gamma=1\). **Remark 1**. : _The fifth-order expansion of the weight functions is considered due to the fact that \(s=1+O(e)\) and \(\frac{f(z_{n})}{f^{\prime}(x_{n})}=O(e)\) or \(\frac{f(x_{n})}{f^{\prime}(x_{n})}=O(e)\). However, the result of Theorem 1 shows that \(z=O(e^{k})\), \(\frac{f(z_{n})}{f^{\prime}(x_{n})}=O(e^{k})\), for \(k\in\{3,4\}\) with \(\gamma\in\{1,\frac{2}{3}\}\). Hence, we require only at most the third-order expansion for both weight functions to achieve the desired sixth-order convergence. This favorable fact is used below to establish the corresponding theorem for a family of sixth-order systems of nonlinear equations._