# An Optimal Order Method for Multiple Roots in Case of Unknown Multiplicity
## 2 Scheme and Analysis of the Local Convergence Rate
By Equation (19), we attain:  
\[= 0,\qquad if\;\psi(x)=0\]5  
Using the expression Equations (19) and (20) in Equation (18) and then using symbolic computation software, Mathematica [28], we get:  
\[= 0,\qquad if\;\psi(x)=0\]6 \[= 0,\qquad if\;\psi(x)=0\]7  
where:  
\[= 0,\qquad if\;\psi(x)=0\]8\[= 0,\qquad if\;\psi(x)=0\]9  
\[y_{n} = x_{n}-\frac{f(x_{n})}{f^{\prime}(x_{n})}\]0 \[y_{n} = x_{n}-\frac{f(x_{n})}{f^{\prime}(x_{n})}\]1  
\[y_{n} = x_{n}-\frac{f(x_{n})}{f^{\prime}(x_{n})}\]2 \[y_{n} = x_{n}-\frac{f(x_{n})}{f^{\prime}(x_{n})}\]3  
\[y_{n} = x_{n}-\frac{f(x_{n})}{f^{\prime}(x_{n})}\]4 \[y_{n} = x_{n}-\frac{f(x_{n})}{f^{\prime}(x_{n})}\]5 \[y_{n} = x_{n}-\frac{f(x_{n})}{f^{\prime}(x_{n})}\]6 \[y_{n} = x_{n}-\frac{f(x_{n})}{f^{\prime}(x_{n})}\]7  
\[y_{n} = x_{n}-\frac{f(x_{n})}{f^{\prime}(x_{n})}\]8 \[y_{n} = x_{n}-\frac{f(x_{n})}{f^{\prime}(x_{n})}\]9 \[u_{n} = y_{n}-\frac{f(y_{n})}{f^{\prime}(y_{n})} \tag{11}\]0 \[u_{n} = y_{n}-\frac{f(y_{n})}{f^{\prime}(y_{n})} \tag{11}\]1 \[u_{n} = y_{n}-\frac{f(y_{n})}{f^{\prime}(y_{n})} \tag{11}\]2 \[u_{n} = y_{n}-\frac{f(y_{n})}{f^{\prime}(y_{n})} \tag{11}\]3  
By virtue of the relation Equation (21), we can obtain the [PERSON]'s series expansion for \(f(z_{n})\), and then, using its expressions together with \(f(x_{n})\), in the first sub-step of scheme Equation (15) and after simplifying, we have:  
\[u_{n} = y_{n}-\frac{f(y_{n})}{f^{\prime}(y_{n})} \tag{11}\]4  
where:  
\[u_{n} = y_{n}-\frac{f(y_{n})}{f^{\prime}(y_{n})} \tag{11}\]5  
\[u_{n} = y_{n}-\frac{f(y_{n})}{f^{\prime}(y_{n})} \tag{11}\]6  
\[u_{n} = y_{n}-\frac{f(y_{n})}{f^{\prime}(y_{n})} \tag{11}\]7 \[u_{n} = y_{n}-\frac{f(y_{n})}{f^{\prime}(y_{n})} \tag{11}\]8 \[u_{n} = y_{n}-\frac{f(y_{n})}{f^{\prime}(y_{n})} \tag{11}\]9\[x_{n+1} = u_{n}-\frac{f(u_{n})}{f^{\prime}(u_{n})}\]0 \[x_{n+1} = u_{n}-\frac{f(u_{n})}{f^{\prime}(u_{n})}\]1 \[x_{n+1} = u_{n}-\frac{f(u_{n})}{f^{\prime}(u_{n})}\]2 \[x_{n+1} = u_{n}-\frac{f(u_{n})}{f^{\prime}(u_{n})}\]3 \[x_{n+1} = u_{n}-\frac{f(u_{n})}{f^{\prime}(u_{n})}\]4 \[x_{n+1} = u_{n}-\frac{f(u_{n})}{f^{\prime}(u_{n})}\]5 \[x_{n+1} = u_{n}-\frac{f(u_{n})}{f^{\prime}(u_{n})}\]6  
\[x_{n+1} = u_{n}-\frac{f(u_{n})}{f^{\prime}(u_{n})}\]7 \[x_{n+1} = u_{n}-\frac{f(u_{n})}{f^{\prime}(u_{n})}\]8 \[x_{n+1} = u_{n}-\frac{f(u_{n})}{f^{\prime}(u_{n})}\]9 \[f^{\prime}(x_{n})\approx\frac{f(z_{n})-f(x_{n})}{f(x_{n})}=g_{1}(x_{n},z_{n}) \tag{12}\]0 \[f^{\prime}(x_{n})\approx\frac{f(z_{n})-f(x_{n})}{f(x_{n})}=g_{1}(x_{n},z_{n}) \tag{12}\]1 \[f^{\prime}(x_{n})\approx\frac{f(z_{n})-f(x_{n})}{f(x_{n})}=g_{1}(x_{n},z_{n}) \tag{12}\]2 \[f^{\prime}(x_{n})\approx\frac{f(z_{n})-f(x_{n})}{f(x_{n})}=g_{1}(x_{n},z_{n}) \tag{12}\]3 \[f^{\prime}(x_{n})\approx\frac{f(z_{n})-f(x_{n})}{f(x_{n})}=g_{1}(x_{n},z_{n}) \tag{12}\]4 \[f^{\prime}(x_{n})\approx\frac{f(z_{n})-f(x_{n})}{f(x_{n})}=g_{1}(x_{n},z_{n}) \tag{12}\]5 \[f^{\prime}(x_{n})\approx\frac{f(z_{n})-f(x_{n})}{f(x_{n})}=g_{1}(x_{n},z_{n}) \tag{12}\]6 \[f^{\prime}(x_{n})\approx\frac{f(z_{n})-f(x_{n})}{f(x_{n})}=g_{1}(x_{n},z_{n}) \tag{12}\]7  
\[f^{\prime}(x_{n})\approx\frac{f(z_{n})-f(x_{n})}{f(x_{n})}=g_{1}(x_{n},z_{n}) \tag{12}\]8 \[f^{\prime}(x_{n})\approx\frac{f(z_{n})-f(x_{n})}{f(x_{n})}=g_{1}(x_{n},z_{n}) \tag{12}\]9 \[f^{\prime}(y_{n})\approx\frac{f[x_{n},y_{n}]f[y_{n},z_{n}]}{f[x_{n},z_{n}]}=g _{2}(x_{n},y_{n},z_{n}) \tag{13}\]0 \[f^{\prime}(y_{n})\approx\frac{f[x_{n},y_{n}]f[y_{n},z_{n}]}{f[x_{n},z_{n}]}=g _{2}(x_{n},y_{n},z_{n}) \tag{13}\]1 \[f^{\prime}(y_{n})\approx\frac{f[x_{n},y_{n}]f[y_{n},z_{n}]}{f[x_{n},z_{n}]}=g _{2}(x_{n},y_{n},z_{n}) \tag{13}\]2 \[f^{\prime}(y_{n})\approx\frac{f[x_{n},y_{n}]f[y_{n},z_{n}]}{f[x_{n},z_{n}]}=g _{2}(x_{n},y_{n},z_{n}) \tag{13}\]3 \[f^{\prime}(y_{n})\approx\frac{f[x_{n},y_{n}]f[y_{n},z_{n}]}{f[x_{n},z_{n}]}=g _{2}(x_{n},y_{n},z_{n}) \tag{13}\]4 \[f^{\prime}(y_{n})\approx\frac{f[x_{n},y_{n}]f[y_{n},z_{n}]}{f[x_{n},z_{n}]}=g _{2}(x_{n},y_{n},z_{n}) \tag{13}\]5 \[f^{\prime}(y_{n})\approx\frac{f[x_{n},y_{n}]f[y_{n},z_{n}]}{f[x_{n},z_{n}]}=g _{2}(x_{n},y_{n},z_{n}) \tag{13}\]6 \[f^{\prime}(y_{n})\approx\frac{f[x_{n},y_{n}]f[y_{n},z_{n}]}{f[x_{n},z_{n}]}=g _{2}(x_{n},y_{n},z_{n}) \tag{13}\]7 \[f^{\prime}(y_{n})\approx\frac{f[x_{n},y_{n}]f[y_{n},z_{n}]}{f[x_{n},z_{n}]}=g _{2}(x_{n},y_{n},z_{n}) \tag{13}\]8 \[f^{\prime}(y_{n})\approx\frac{f[x_{n},y_{n}]f[y_{n},z_{n}]}{f[x_{n},z_{n}]}=g _{2}(x_{n},y_{n},z_{n}) \tag{13}\]9 \[f^{\prime}(u_{n})\approx b_{2}-b_{1}b_{4}=g_{3}(x_{n},y_{n},z_{n},u_{n}) \tag{14}\]0 \[f^{\prime}(u_{n})\approx b_{2}-b_{1}b_{4}=g_{3}(x_{n},y_{n},z_{n},u_{n}) \tag{14}\]1 \[f^{\prime}(u_{n})\approx b_{2}-b_{1}b_{4}=g_{3}(x_{n},y_{n},z_{n},u_{n}) \tag{14}\]2  
In a similar way to the first sub-step, the [PERSON]'s series for \(f(y_{n})\) can be derived using Equation (22), and then, using the involved expressions from the above relations, we can write:  
\[f^{\prime}(u_{n})\approx b_{2}-b_{1}b_{4}=g_{3}(x_{n},y_{n},z_{n},u_{n}) \tag{14}\]3where:  
\[f^{\prime}(u_{n})\approx b_{2}-b_{1}b_{4}=g_{3}(x_{n},y_{n},z_{n},u_{n}) \tag{14}\]4  
\[f^{\prime}(u_{n})\approx b_{2}-b_{1}b_{4}=g_{3}(x_{n},y_{n},z_{n},u_{n}) \tag{14}\]5 \[f^{\prime}(u_{n})\approx b_{2}-b_{1}b_{4}=g_{3}(x_{n},y_{n},z_{n},u_{n}) \tag{14}\]6  
\[f^{\prime}(u_{n})\approx b_{2}-b_{1}b_{4}=g_{3}(x_{n},y_{n},z_{n},u_{n}) \tag{14}\]7 \[f^{\prime}(u_{n})\approx b_{2}-b_{1}b_{4}=g_{3}(x_{n},y_{n},z_{n},u_{n}) \tag{14}\]8 \[f^{\prime}(u_{n})\approx b_{2}-b_{1}b_{4}=g_{3}(x_{n},y_{n},z_{n},u_{n}) \tag{14}\]9 \[y_{n} = x_{n}-\frac{f(x_{n})}{g_{1}(x_{n},z_{n})}\]0 \[y_{n} = x_{n}-\frac{f(x_{n})}{g_{1}(x_{n},z_{n})}\]1  
\[y_{n} = x_{n}-\frac{f(x_{n})}{g_{1}(x_{n},z_{n})}\]2 \[y_{n} = x_{n}-\frac{f(x_{n})}{g_{1}(x_{n},z_{n})}\]3 \[y_{n} = x_{n}-\frac{f(x_{n})}{g_{1}(x_{n},z_{n})}\]4 \[y_{n} = x_{n}-\frac{f(x_{n})}{g_{1}(x_{n},z_{n})}\]5 \[y_{n} = x_{n}-\frac{f(x_{n})}{g_{1}(x_{n},z_{n})}\]6 \[y_{n} = x_{n}-\frac{f(x_{n})}{g_{1}(x_{n},z_{n})}\]7 \[y_{n} = x_{n}-\frac{f(x_{n})}{g_{1}(x_{n},z_{n})}\]8 \[y_{n} = x_{n}-\frac{f(x_{n})}{g_{1}(x_{n},z_{n})}\]9 \[u_{n} = y_{n}-\frac{f(y_{n})}{g_{2}(x_{n},z_{n},y_{n})} \tag{15}\]0  
\[u_{n} = y_{n}-\frac{f(y_{n})}{g_{2}(x_{n},z_{n},y_{n})} \tag{15}\]1 \[u_{n} = y_{n}-\frac{f(y_{n})}{g_{2}(x_{n},z_{n},y_{n})} \tag{15}\]2 \[u_{n} = y_{n}-\frac{f(y_{n})}{g_{2}(x_{n},z_{n},y_{n})} \tag{15}\]3 \[u_{n} = y_{n}-\frac{f(y_{n})}{g_{2}(x_{n},z_{n},y_{n})} \tag{15}\]4 \[u_{n} = y_{n}-\frac{f(y_{n})}{g_{2}(x_{n},z_{n},y_{n})} \tag{15}\]5 \[u_{n} = y_{n}-\frac{f(y_{n})}{g_{2}(x_{n},z_{n},y_{n})} \tag{15}\]6 \[u_{n} = y_{n}-\frac{f(y_{n})}{g_{2}(x_{n},z_{n},y_{n})} \tag{15}\]7 \[u_{n} = y_{n}-\frac{f(y_{n})}{g_{2}(x_{n},z_{n},y_{n})} \tag{15}\]8 \[u_{n} = y_{n}-\frac{f(y_{n})}{g_{2}(x_{n},z_{n},y_{n})} \tag{15}\]9 \[= 0,\qquad if\;\psi(x)=0\]00 \[= 0,\qquad if\;\psi(x)=0\]01 \[= 0,\qquad if\;\psi(x)=0\]02 \[= 0,\qquad if\;\psi(x)=0\]03 \[= 0,\qquad if\;\psi(x)=0\]04Finally, we obtain the Taylor's series expansion of \(f(u_{n})\), with the help of just the previous equation, and using the required relations obtained above, in the last sub-step of scheme Equation (15), we get the final error expression as follows:  
\[= 0,\qquad if\;\psi(x)=0\]05 \[= 0,\qquad if\;\psi(x)=0\]06  
Thus, the proof is completed.