## 2 Iterative Method with Fifth-Order Convergence for Solving Multiple Roots
By Equation (18), we obtain  
\[h^{\prime}(x_{n})=h(x^{*})[b_{1}+2b_{2}e_{n}+3b_{3}e_{n}^{2}+4b_{4}e_{n}^{3}+5 b_{5}e_{n}^{4}+6b_{6}e_{n}^{5}+o(e_{n}^{6})]. \tag{19}\]  
Substituting Equations (18) and (19) into Equation (17), we get  
\[F(x_{n})=\dfrac{e_{n}h(x_{n})}{mh(x_{n})+e_{n}h^{\prime}(x_{n})}=c_{1}e_{n}+c_ {2}e_{n}^{2}+c_{3}e_{n}^{3}+c_{4}e_{n}^{4}+o(e_{n}^{5}) \tag{20}\]where  
\[c_{1}=\frac{1}{m},c_{2}=-\frac{(a-b)b_{1}+n}{m^{2}(a-b)} \tag{21}\] \[c_{3}= \frac{1}{(a-b)^{2}m^{3}}((a-b)^{2}b_{1}^{2}m+a^{2}b_{1}^{2}-2a^{2}b_ {2}m-2 abb_{1}^{2}+4 abb_{2}m\] (22) \[+b^{2}b_{1}^{2}-2b^{2}b_{2}m+2 ab_{1}n-2 bb_{1}n+mn+n^{2})\] \[c_{4}= -\frac{1}{m^{4}(a-b)^{3}}(a^{3}b_{1}^{3}m^{2}-3a^{2}bb_{1}^{3}m^{2 }+3 ab^{2}b_{1}^{3}m^{2}-b^{3}b_{1}^{3}m^{2}+2a^{3}b_{1}^{3}m\] (23) \[f^{\prime}(x)=m(x-x^{*})^{m-1}(x-x_{1})^{n}h(x)+n(x-x^{*})^{m}(x-x_{1})^{n-1}h (x)+(x-x^{*})^{m}(x-x_{1})^{n}h^{\prime}(x). \tag{16}\]0 \[f^{\prime}(x)=m(x-x^{*})^{m-1}(x-x_{1})^{n}h(x)+n(x-x^{*})^{m}(x-x_{1})^{n-1}h (x)+(x-x^{*})^{m}(x-x_{1})^{n}h^{\prime}(x). \tag{16}\]1 \[f^{\prime}(x)=m(x-x^{*})^{m-1}(x-x_{1})^{n}h(x)+n(x-x^{*})^{m}(x-x_{1})^{n-1}h (x)+(x-x^{*})^{m}(x-x_{1})^{n}h^{\prime}(x). \tag{16}\]2 \[f^{\prime}(x)=m(x-x^{*})^{m-1}(x-x_{1})^{n}h(x)+n(x-x^{*})^{m}(x-x_{1})^{n-1}h (x)+(x-x^{*})^{m}(x-x_{1})^{n}h^{\prime}(x). \tag{16}\]3 \[f^{\prime}(x)=m(x-x^{*})^{m-1}(x-x_{1})^{n}h(x)+n(x-x^{*})^{m}(x-x_{1})^{n-1}h (x)+(x-x^{*})^{m}(x-x_{1})^{n}h^{\prime}(x). \tag{16}\]4 \[f^{\prime}(x)=m(x-x^{*})^{m-1}(x-x_{1})^{n}h(x)+n(x-x^{*})^{m}(x-x_{1})^{n-1}h (x)+(x-x^{*})^{m}(x-x_{1})^{n}h^{\prime}(x). \tag{16}\]5  
Substituting Equation (20) into Equation (11), we obtain  
\[g_{1}(x_{n})= c_{1}+c_{2}(2+c_{1})e_{n}+(3c_{3}+3c_{1}c_{3}+c_{1}^{2}c_{3}+c_{2 }^{2})e_{n}^{2}+(4c_{2}c_{3}+2c_{1}c_{2}c_{3}+4c  
From Equations (25)-(27), it follows that  
\[ \begin{split} z_{n}=& x^{*}-\frac{c_{2}^{2}(1+c_{1})}{c_ {1}}e_{n}^{3}-\frac{c_{2}}{c_{1}^{3}}(-3c_{2}^{2}c_{1}-2c_{2}^{2}c_{1}^{2}-c_{2 }^{2}+6c_{3}c_{1}^{2}+c_{3}c_{1}+7c_{3}c_{1}^{3}+2c_{1}^{4}c_{3})e_{n}^{4}\\ &-\frac{1}{c_{1}^{4}}(9c_{3}^{2}c_{1}^{3}+2c_{3}^{2}c_{1}^{2}+4c_ {2}^{2}+16c_{1}^{4}c_{2}c_{4}+2c_{1}^{6}c_{2}c_{4}+9c_{2}c_{4}c_{1}^{5}-21c_{1} ^{2}c_{2}^{2}c_{3}-10c_{1}^{3}c_{2}^{2}c_{3}-8c_{1}c_{2}^{2}c_{3}\\ &+7c_{1}c_{2}^{4}+3c_{1}^{2}c_{2}^{4}+2c_{1}^{3}c_{2}^{4}+11c_{2}c_ {4}c_{1}^{3}+2c_{2}c_{4}c_{1}^{2}+12c_{1}^{4}c_{3}^{2}+6c_{1}^{5}c_{3}^{2}+c_{1 }^{6}c_{3}^{2}+c_{1}^{4}c_{2}^{4})e_{n}^{5}+o(e_{n}^{6})\end{split}  \tag{28}\]  
It is similar to Equation (26), we have  
\[ \begin{split} F(z_{n})=&-c_{2}^{2}(1+c_{1})e_{n}^{3 }-\frac{c_{2}}{c_{1}^{2}}(-3c_{2}^{2}c_{1}-c_{2}^{2}c_{1}^{2}-c_{2}^{2}+6c_{1} ^{2}c_{3}+c_{3}c_{1}+7c_{3}c_{1}^{3}+2c_{3}c_{1}^{4})e_{n}^{4}\\ &+o(e_{n}^{5})\end{split}  \tag{29}\]  
Moreover, substituting Equations (20), (24)-(26), (28) and (29) into Equation (13), we get  
\[g_{3}(x_{n})=c_{1}-c_{2}^{2}(1+c_{1})e_{n}^{2}-\frac{c_{2}}{c_{1}}(2c_{3}c_{1 }^{3}+7c_{3}c_{1}^{2}+7c_{3}c_{1}+2c_{2}^{2}c_{1}+c_{2}^{2}+2c_{3})e_{n}^{e}+o (e_{n}^{4}) \tag{30}\]  
Substituting Equations (28)-(30) into the third formula of Equation (14), we get  
\[x_{n+1}=z_{n}-\frac{F(z_{n})}{g_{3}(x_{n})}=x^{*}+\frac{(1+2c_{1}+c_{1}^{2})c _{2}^{4}}{c_{1}^{2}}e_{n}^{5}+o(e_{n}^{6}) \tag{31}\]  
which just means that the iterative method defined by Equation (14) has fifth-order convergence. The proof is completed. We further consider how to find the highest multiplicity of the root \(x^{*}\) in the iterative method. If \(x_{n}\) is the \(n\)-th iteration computed by an iterative method applied to \(f\), then from Equation (9), we have  
\[ \begin{split} f_{n}&\approx\frac{(x_{n}-x^{*})(x_{n }-x_{1})h(x_{n})}{m(x_{n}-x_{1})h(x_{n})+n(x_{n}-x^{*})h(x_{n})+(x_{n}-x^{*})(x _{n}-x_{1})h^{\prime}(x_{n})}\\ &=\frac{\varepsilon_{n}(x_{n}-x_{1})h(x_{n})}{m(x_{n}-x_{1})h(x_ {n})+n\varepsilon_{n}h(x_{n})+\varepsilon_{n}(x_{n}-x_{1})h^{\prime}(x_{n})},\end{split}  \tag{32}\]  
where \(f_{n}=f(x_{n})\). Because \(\varepsilon_{n}\) is small, we get \(f_{n}\approx\frac{\varepsilon_{n}}{m}\).