## 2 Main Theorem  
The main theorem for a nonlinear scalar equation will be pursued and extended later in Section 5 to a system of nonlinear vector equations:  
**Theorem 1**. : _Assume that \(f:\mathbf{D}\subset\mathbb{C}\rightarrow\mathbb{C}\) has a simple root \(\alpha\) and is analytic in a region \(\mathbf{D}\) containing \(\alpha\). Let \(\theta_{j}=\frac{f^{(j)}(\alpha)}{j!\,f^{(\alpha)}}\) for \(j=2,3,\cdots\). Let \(x_{0}\) be an initial guess chosen in a sufficiently small neighborhood of \(\alpha\). Let \(T_{f},L_{f}:\mathbb{C}\rightarrow\mathbb{C}\) be analytic in a neighborhood of \(1\). Let \(T_{i}=\frac{1}{i! }\frac{d^{i}}{ds^{i}}T_{f}(s)\big{|}_{s=1}\) and \(L_{i}=\frac{1}{i! }\frac{d^{i}}{ds^{i}}L_{f}(s)\big{|}_{s=1}\) for \(0\leq i\leq 5\). If \(T_{0}=1,T_{1}=-\frac{1}{2\gamma},L_{0}=1,L_{1}=-\frac{1}{\gamma}\), then iterative scheme \((1)\) defines a family of fifth-order methods for \(\gamma\in\mathbb{R}-\{0\}\). If we add further constraints either with \(\{\gamma=\frac{2}{3},T_{2}=\frac{9}{8},|T_{3}|<\infty,|L_{2}|<\infty\}\) or with \(\{\gamma=1,L_{2}=\frac{3}{2},|L_{3}|<\infty,|T_{2}|<\infty\}\), then iterative scheme \((1)\) reduces to a family of sixth-order methods satisfying the error equation below. For \(n=0,1,2,\cdots\),_  
\[e_{n+1}=\left\{ \begin{array}{l}-\frac{1}{243}\left[2(-27+8L_{2})\theta_{2}^{ 2}+9\theta_{3}\right]\cdot\left[(135+64T_{3})\theta_{2}^{3}-27\theta_{2}\theta _{3}+3\theta_{4}\right]\epsilon_{n}^{6}+O(\epsilon_{n}^{7}),\text{ if }\gamma=\frac{2}{3},\\ \left[4(1-2T_{2})(9+4L_{3}+2T_{2})\theta_{2}^{5}+4(1+L_{3}+3T_{2})\theta_{2}^{ 3}\theta_{3}-\frac{5}{4}\theta_{2}\theta_{3}^{2}\right]\epsilon_{n}^{6}+O( \epsilon_{n}^{7}),\text{ if }\gamma=1.\end{array} \right. \tag{2}\]  
Proof. : Taylor series expansion of \(f(x_{n})\) about \(\alpha\) up to sixth-order terms with \(f(\alpha)=0\) leads us to the following:  
\[f(x_{n})=f^{\prime}(\alpha)[e_{n}+\theta_{2}e_{n}^{2}+\theta_{3}e_{n}^{3}+ \theta_{4}e_{n}^{4}+\theta_{5}e_{n}^{5}+\theta_{6}e_{n}^{6}+O(\epsilon_{n}^{7 })]. \tag{3}\]  
It follows that  
\[f^{\prime}(x_{n})=f^{\prime}(\alpha)[1+2\theta_{2}e_{n}+3\theta_{3}e_{n}^{2}+4 \theta_{4}e_{n}^{3}+5\theta_{5}e_{n}^{4}+6\theta_{6}e_{n}^{5}+O(\epsilon_{n}^{ 6})]. \tag{4}\]  
For brevity of notation, we denote \(e_{n}\) by \(e\), unless otherwise specified from now on. Symbolic computation of Mathematica [20] yields:  
\[y_{n}=x_{n}-\gamma\frac{f(x_{n})}{f^{\prime}(x_{n})}=\alpha+(1-\gamma)e+\gamma \theta_{2}e^{2}-2\gamma(\theta_{2}^{2}-\theta_{3})e^{3}+Y_{4}e^{4}+Y_{5}e^{5}+Y_ {6}e^{6}+O(\epsilon^{7}), \tag{5}\]  
where \(Y_{4}=\gamma(4\theta_{2}^{3}-7\theta_{2}\theta_{3}+3\theta_{4})\), \(Y_{5}=-2\gamma(4\theta_{2}^{4}-10\theta_{2}^{2}\theta_{3}+3\theta_{3}^{2}+5 \theta_{2}\theta_{4}-2\theta_{5})\), \(Y_{6}=\gamma(16\theta_{2}^{5}-52\theta_{2}^{3}\theta_{3}+33\theta_{2}\theta_{3 }^{2}+28\theta_{2}^{2}\theta_{4}-17\theta_{3}\theta_{4}-13\theta_{2}\theta_{5}+ 5\theta_{6})\). In view of the fact that \(f^{\prime}(y_{n})=f^{\prime}(x_{n})|_{\alpha_{n}\rightarrow(y_{n}-\alpha)}\), we get:  
\[f^{\prime}(y_{n})=f^{\prime}(\alpha)[1-2(\gamma-1)\theta_{2}e+[2\gamma\theta_{2 }^{2}+3(\gamma-1)^{2}\theta_{3}]e^{2}+\Sigma_{i=3}^{5}D_{i}e^{i}+O(\epsilon^{ 6})], \tag{6}\]where \(D_{i}=D_{i}(\gamma,\theta_{2},\theta_{3},\cdots,\theta_{6})\) for \(3\leq i\leq 5\). Hence, we have:  
\[s=\frac{f^{\prime}(y_{n})}{f^{\prime}(x_{n})}=1-2\gamma\theta_{2}e+3\gamma[2 \theta_{2}^{2}+(-2+\gamma)\theta_{3}])e^{2}+\Sigma_{i=3}^{5}E_{i}e^{i}+O(e^{6}), \tag{7}\]  
where \(E_{i}=E_{i}(\gamma,\theta_{2},\theta_{3},\cdots,\theta_{6})\) for \(3\leq i\leq 5\).