## 5 Conclusions  
The inclusion of preconditioners in the existing iterative methods for finding zeros with multiplicities for solving a system of nonlinear equations gives benefits in numerical stability and numerical accuracy. The proposed methodology is equally effective for nonlinear and systems of nonlinear equations. It is assumed in all cases that the preconditioners should be non-zero, because in this way, it does not affect the zeros of nonlinear or systems of nonlinear equations. The different selections of preconditioners provide different families of iterative methods. The claimed order of convergence is also verified by computing the computational order of convergence in all numerical simulations. Study of the dynamics of nonlinear preconditioners for finding zeros with multiplicities of nonlinear equations and systems of nonlinear equations could be an interesting topic for research.  
First author established the idea and all other authors contributed equally in the article.  
**Conflicts of Interest:** The authors declare no conflict of interest.  
\begin{table}
\begin{tabular}{c c c c c c} \hline \hline  & \(\boldsymbol{\Lambda(\mathbf{z})}\) & \(\boldsymbol{\Omega(\mathbf{z})}\) & Iter. & \(||\boldsymbol{\Phi(\mathbf{z})}||_{\infty}\) & **CCO** \\ \hline Iterative method (21) & \(\boldsymbol{1}\) & \(\boldsymbol{1}\) & \(1\) & - & - \\  & \(6+cos(\mathbf{z})/10\) & \(\boldsymbol{1}\) & \(7\) & \(O\left(10^{-2042}\right)\) & \(3.0\) \\  & \(1+\boldsymbol{x}^{3}/1000\) & \(\boldsymbol{1}\) & \(7\) & \(O\left(10^{-8482}\right)\) & \(3.98\) \\  & \(exp(\mathbf{z}/100)\) & \(\boldsymbol{1}\) & \(7\) & \(O\left(10^{-376}\right)\) & \(2.00\) \\ \hline Iterative method (11) & \(\boldsymbol{1}\) & - & \(1\) & - & - \\  & \(6+cos(\mathbf{z})/10\) & - & \(20\) & \(O\left(10^{-23}\right)\) & \(1.0\) \\  & \(1+\boldsymbol{x}^{3}/1000\) & - & \(20\) & Not converging & - \\  & \(exp(\mathbf{z}/100)\) & - & \(7\) & \(O\left(10^{-443}\right)\) & \(2.0\) \\ \hline \hline \end{tabular}
\end{table}
Table 2: Problem 2: initial guess \(=[1,2,4,3]\), \(m=[2,2,2,2]\).  
\begin{table}
\begin{tabular}{c c c c c} \hline \hline  & \(\boldsymbol{\Lambda(\mathbf{z})}\) & \(\boldsymbol{\Omega(\mathbf{z})}\) & Iter. & \(||\boldsymbol{\Phi(\mathbf{z})}||_{\infty}\) & **CCO** \\ \hline Iterative method (21) & \(\boldsymbol{1}\) & \(\boldsymbol{1}\) & \(12\) & \(O\left(10^{-2011}\right)\) & \(2.00\) \\  & \(6+cos(\mathbf{z})/10\) & \(\boldsymbol{1}\) & \(12\) & \(O\left(10^{-1914}\right)\) & \(2.00\) \\  & \(1+\boldsymbol{x}^{3}/1000\) & \(\boldsymbol{1}\) & \(12\) & \(O\left(10^{-1248}\right)\) & \(2.00\) \\  & \(exp(-\mathbf{z}/10)\) & \(\boldsymbol{1}\) & \(12\) & \(O\left(10^{-2767}\right)\) & \(2.00\) \\  & \(exp(-\mathbf{z}/10)\) & \(exp(\mathbf{z}/10000)\) & \(12\) & \(O\left(10^{-2110}\right)\) & \(2.00\) \\  & \(exp(-\mathbf{z}/10)\) & \(exp(-\mathbf{z}/10000)\) & \(12\) & \(O\left(10^{-2771}\right)\) & \(2.00\) \\ \hline Iterative method (11) & \(\boldsymbol{1}\) & - & \(1\) & - & - \\  & \(6+cos(\mathbf{z})/10\) & - & \(12\) & \(O\left(10^{-56}\right)\) & \(2.00\) \\  & \(1+\boldsymbol{x}^{3}/1000\) & - & \(20\) & Not converging & - \\  & \(exp(-\mathbf{z}/10)\) & - & \(7\) & \(O\left(10^{-35}\right)\) & \(2.00\) \\ \hline \hline \end{tabular}
\end{table}
Table 3: Problem 3: initial guess \(=[2,4,3]\), \(m=[1/2,1/2,1/2]\).