## 2 Iterative Method with Fifth-Order Convergence for Solving Multiple Roots  
We consider the simple transformation (see [16,25]):  
\[F(x)= \begin{cases}\frac{f(x)}{f^{\prime}(x)},&\text{if }f(x)\
eq 0,\\ 0,&\text{if }f(x)=0,\end{cases}  \tag{9}\]  
and use a Newton-like iterative method:  
\[\left\{ \begin{array}{ll}y_{n}=x_{n}-\frac{F(x_{n})}{F^{\prime}(x_{n})},\\ z_{n}=y_{n}-\frac{F(y_{n})}{F^{\prime}(y_{n})},\\ x_{n+1}=z_{n}-\frac{F(z_{n})}{F^{\prime}(z_{n})}.\end{array} \right. \tag{10}\]  
In order to avoid computing the first derivatives of function \(F(x_{n})\), \(F(y_{n})\) and \(F(z_{n})\), we approximate them as follows:  
\[F^{\prime}(x_{n})\approx\frac{F(x_{n}+F(x_{n}))-F(x_{n})}{F(x_{n})}=g_{1}(x_{ n}), \tag{11}\]  
\[F^{\prime}(y_{n})\approx\frac{2(F(y_{n})-F(x_{n}))}{y_{n}-x_{n}}-g_{1}(x_{n})=g_{2}(x_{n}) \tag{12}\]  
\[F^{\prime}(z_{n})\approx F[z_{n},y_{n}]+F[z_{n},x_{n},x_{n}](z_{n}-y_{n}) \tag{13}\] \[= \frac{F(z_{n})-F(y_{n})}{z_{n}-y_{n}}+\frac{\frac{F(z_{n})-F(x_{n })}{z_{n}-x_{n}}-g_{1}(x_{n})}{z_{n}-x_{n}}(z_{n}-y_{n})\] \[= g_{3}(x_{n})\](See [25, 26, 27, 28] for the detail discussions of Equations (11)-(13) respectively.) Substituting the approximations of \(F^{\prime}(x_{n})\), \(F^{\prime}(y_{n})\) and \(F^{\prime}(z_{n})\) given by Equations (11)-(13) in Equation (10), we establish the following new iterative method:  
\[\left\{ \begin{array}{l}y_{n}=x_{n}-\dfrac{F(x_{n})}{g_{1}(x_{n})},\\ z_{n}=y_{n}-\dfrac{F(y_{n})}{g_{2}(x_{n})},\\ x_{n+1}=z_{n}-\dfrac{F(z_{n})}{g_{3}(x_{n})}.\end{array} \right. \tag{14}\]  
We give the following convergence theorem for the proposed method Equation (14) as follows. **Theorem 1**. : _Suppose that \(f\in C^{1}(D)(D\subseteq R\to R)\) has multiple roots \(x^{*}\in D\), whose multiplicity \(m\) is the highest multiplicity of nonlinear equation, where \(D\) is an open interval. If the initial point \(x_{0}\) is sufficiently close to \(x^{*}\), the iterative method defined by (14) has fifth-order convergence._  
Proof. : Without loss of generality, we assume that \(f(x)\) has two multiple roots  
\[f(x)=(x-x^{*})^{m}(x-x_{1})^{n}h(x) \tag{15}\]  
where \(x^{*}\) is a multiple root of Equation (15) with multiplicity \(m\) and \(x_{1}\) is a multiple root of Equation (15) with multiplicity \(n\)\((m>n)\), \(h(x)\) is a continuous function with \(h(x^{*})\
eq 0\) and \(h(x_{1})\
eq 0\). According to Equation (15), we have  
\[f^{\prime}(x)=m(x-x^{*})^{m-1}(x-x_{1})^{n}h(x)+n(x-x^{*})^{m}(x-x_{1})^{n-1}h (x)+(x-x^{*})^{m}(x-x_{1})^{n}h^{\prime}(x). \tag{16}\]  
Dividing Equation (15) by Equation (16), we get  
\[F(x)=\dfrac{f(x)}{f^{\prime}(x)}=\dfrac{(x-x^{*})(x-x_{1})h(x)}{m(x-x_{1})h(x) +n(x-x^{*})h(x)+(x-x^{*})(x-x_{1})h^{\prime}(x)} \tag{17}\]  
From Equation (17), we can see that the problem of computing multiple roots of \(f(x)=0\) can be reduced to the equivalent problem of computing simple root \(x^{*}\) of \(F(x)=0\). Using [PERSON]'s expansion, we have  
\[h(x_{n})=h(x^{*})[1+b_{1}e_{n}+b_{2}e_{n}^{2}+b_{3}e_{n}^{3}+b_{4}e_{n}^{4}+b _{5}e_{n}^{5}+b_{6}e_{n}^{6}+o(e_{n}^{7})] \tag{18}\]  
where \(b_{k}=\dfrac{h^{(k)}(x^{*})}{k!h(x^{*})}\), \(k=1,2,...\), and \(e_{n}=x_{n}-x^{*}\).