## 3 Numerical Results  
We employ the proposed modification of [PERSON]'s method with three-step Equation (14) (MNM) to solve some nonlinear equations. All the computations were done by using Visual C++ 6.0 and were satisfied the condition such that \(|f(x_{n})|<1.E-17\), \(|x_{n}-x^{*}|<1.E-17\). In order to show the effectiveness of our iterative method, we provide at least three different initial iterative values. From different initial iterative values, they can be convergent to the same iterative solution whose multiplicity is the highest multiplicity of nonlinear equation. We used the following test functions and obtained the approximate zeros \(x^{*}\) round up to the 17-th decimal place:  
\[g_{1}(x)=\frac{(x-\sqrt{5})^{7}(x-\sqrt{3})^{4}}{(x-1)^{2}+1},x_{0}=-12.0,23.0, 1.9,m=7,n=4,x^{*}=2.236067977499790\]  
\[g_{2}(x)=(\sin(x)^{2}-2x+1)^{5}(x-2)^{3},x_{0}=-23,23,1.5,1.7,m=5,n=3,x^{*}=0. 71483582544138924\]  
\[g_{3}(x)=(8x\exp(-x^{2})-2x-3)^{8}(x-2)^{5},x_{0}=-20,20,0.0,1.5,m=8,n=5,x^{*} =-1.7903531791589544\]  
\[g_{4}(x)=\frac{(2x\cos(x)+x^{2}-3)^{10}(x-3)^{8}}{(x^{2}+1)},x_{0}=-23,23,3. 1,2.99,m=10,n=8,x^{*}=2.9806452794385368\]  
\[g_{5}(x)=(\exp(-x^{2}+x+3)-x+2)^{9}(x-\frac{13}{5})^{6},x_{0}=-18,18,2.5,2.55,m=9,n=6,x^{*}=2.4905398276083051\]  
\[g_{6}(x)=(\exp(-x)+2\sin(x))^{4}(x-2)^{3},x_{0}=4.0,2.5,m=4,n=3,x^{*}=3.16274 88709263654\]  
\[g_{7}(x)=(\ln(x^{2}+3x+5)-2x+7)^{15}(x-\sqrt{37})^{10},x_{0}=34,5.5,8,m=15,n= 10,x^{*}=5.4690123359101421\]  
\[g_{8}(x)=(\sqrt{x^{2}+2x+5}-2\sin(x)-x^{2}+3)^{20}(x-\sqrt{7})^{11},x_{0}=9,2,3,m=20,n=11,x^{*}=2.3319676558839640\]  
\[g_{9}(x)=\frac{(x-2)^{7}(x-\sqrt{5})^{8}}{(x-1)^{2}+1},x_{0}=-12,12,2.1,2.5,m =8,n=7,x^{*}=2.2360679774997897\]  
\[g_{10}(x)=(x-\frac{5}{2})^{\frac{14}{4}}\exp(x)(x-2)^{2},x_{0}=-13,2.4,2.1,m= \frac{15}{4},n=2,x^{*}=2.50000000000\]  
\[g_{2}(x)=(\sin(x)^{2}-2x+1)^{5}(x-2)^{3},x_{0}=-23,23,1.5,1.7,m=5,n=3,x^{*}=0. 71483582544138924\]0  
\[g_{2}(x)=(\sin(x)^{2}-2x+1)^{5}(x-2)^{3},x_{0}=-23,23,1.5,1.7,m=5,n=3,x^{*}=0. 71483582544138924\]1  
\[g_{2}(x)=(\sin(x)^{2}-2x+1)^{5}(x-2)^{3},x_{0}=-23,23,1.5,1.7,m=5,n=3,x^{*}=0. 71483582544138924\]2  
\[g_{2}(x)=(\sin(x)^{2}-2x+1)^{5}(x-2)^{3},x_{0}=-23,23,1.5,1.7,m=5,n=3,x^{*}=0. 71483582544138924\]3  
\[g_{2}(x)=(\sin(x)^{2}-2x+1)^{5}(x-2)^{3},x_{0}=-23,23,1.5,1.7,m=5,n=3,x^{*}=0. 71483582544138924\]4  
\[g_{2}(x)=(\sin(x)^{2}-2x+1)^{5}(x-2)^{3},x_{0}=-23,23,1.5,1.7,m=5,n=3,x^{*}=0. 71483582544138924\]5  
\[g_{2}(x)=(\sin(x)^{2}-2x+1)^{5}(x-2)^{3},x_{0}=-23,23,1.5,1.7,m=5,n=3,x^{*}=0. 71483582544138924\]6  
\[g_{2}(x)=(\sin(x)^{2}-2x+1)^{5}(x-2)^{3},x_{0}=-23,23,1.5,1.7,m=5,n=3,x^{*}=0. 71483582544138924\]7  
The computational results indicate that our proposed iterative method can converge to multiple roots whose multiplicity is the highest multiplicity of nonlinear equation. In the next section, for the special case which the multiplicity of the roots of nonlinear equation is a single multiplicity, we present the analysis result for comparison with previous methods. **Remark 1**. : _The method of the Formula (14) can also solve the problem of Euler equation for higher-order linear ordinary differential equation with variable coefficients.