# Two Efficient Derivative-Free Iterative Methods for Solving Nonlinear Systems
## 1 Introduction
: The first order divided difference operator of \(F\) as a mapping \([\cdot,\cdot;F]:D\times D\subset R^{m}\times R^{m}\to L(R^{m})\) (see [5; 10; 11]) is given by  
\[[x+h,x;F]=\int_{0}^{1}F^{\prime}(x+th)dt,\forall(x,h)\in R^{m}\times R^{m} \tag{9}\]  
Expanding \(F^{\prime}(x+th)\) in Taylor series at the point \(x\) and integrating, we obtain  
\[\int_{0}^{1}F^{\prime}(x+th)dt= F^{\prime}(x)+\frac{1}{2}F^{\prime\prime}(x)h+\frac{1}{6}F^{ \prime\prime\prime}(x)h^{2}+O(h^{3}) \tag{10}\]  
Developing \(F(x^{(k)})\) in a neighborhood of \(\alpha\) and assuming that \(\Gamma=[F^{\prime}(\alpha)]^{-1}\) exists, we have  
\[F(x^{(k)})=F^{\prime}(\alpha)[e+A_{2}e^{2}+A_{3}e^{3}+A_{4}e^{4}+A_{5}e^{5}+O( \varepsilon^{6})] \tag{11}\]  
where \(A_{i}=\frac{1}{i! }\Gamma F^{(i)}(\alpha)\in L_{i}(R^{m},R^{m})\). The derivatives of \(F(x^{(k)})\) can be given by  
\[F^{\prime}(x^{(k)})=F^{\prime}(\alpha)[I+2A_{2}e+3A_{3}e^{2}+4A_{4}e^{3}+5A_{ 5}e^{4}+O(\varepsilon^{5})] \tag{12}\]  
\[F^{\prime\prime}(x^{(k)})=F^{\prime}(\alpha)[2A_{2}+6A_{3}e+12A_{4}e^{2}+20A_ {5}e^{3}+O(\varepsilon^{4})] \tag{13}\]  
\[F^{\prime\prime\prime}(x^{(k)})=F^{\prime}(\alpha)[6A_{3}+24A_{4}e+60A_{5}e^{2 }+O(\varepsilon^{3})] \tag{14}\]  
Setting \(y=x+h\) and \(E=y-\alpha\), we have \(h=E-e\). Replacing the previous expressions Equations (12)-(14) into Equation (10) we get  
\[[x^{(k)},y^{(k)};F]=F^{\prime}(\alpha)(I+A_{2}(E+e)+A_{3}(E^{2}+Ee+e^{2})+O( \varepsilon^{5})) \tag{15}\]  
Noting that \(w^{(k)}-\alpha=e+F(x^{k})\) and \(s^{(k)}-\alpha=e-F(x^{k})\), we replace in Equation (15) \(E\) by \(e+F(x^{k})\), \(e\) by \(e-F(x^{k})\), we obtain  
\[[w^{(k)},s^{(k)};F]=F^{\prime}(\alpha)(I+2A_{2}e+(3A_{3}+A_{3}F^{\prime}( \alpha)^{2})e^{2}+O(\varepsilon^{3}))=F^{\prime}(\alpha)D(e)+O(\varepsilon^{3}  
Solving the system Equation (19), we obtain  
\[X_{2}=-2A_{2} \tag{20}\]  
\[e_{n+1}=2A_{2}E\varepsilon-(A_{3}+A_{3}F^{\prime}(\alpha)^{2}-4A_{2}^{2})e^{2} \varepsilon+O(\varepsilon^{7}) \tag{8}\]0  
then,  
\[e_{n+1}=2A_{2}E\varepsilon-(A_{3}+A_{3}F^{\prime}(\alpha)^{2}-4A_{2}^{2})e^{2} \varepsilon+O(\varepsilon^{7}) \tag{8}\]1  
\[e_{n+1}=2A_{2}E\varepsilon-(A_{3}+A_{3}F^{\prime}(\alpha)^{2}-4A_{2}^{2})e^{2} \varepsilon+O(\varepsilon^{7}) \tag{8}\]2  
Similar to Equation (11), we have  
\[e_{n+1}=2A_{2}E\varepsilon-(A_{3}+A_{3}F^{\prime}(\alpha)^{2}-4A_{2}^{2})e^{2} \varepsilon+O(\varepsilon^{7}) \tag{8}\]3  
From Equations (15) and (22)-(24), we get  
\[ \begin{array}{l}\mu_{1}=(3I-2[w^{(k)},s^{(k)};F]^{-1}[y^{(k)},x^{(k)};F])[w^ {(k)},s^{(k)};F]^{-1}\\ =(I-2A_{2}E+(A_{3}+A_{3}F^{\prime}(\alpha)^{2}-4A_{2}^{2})e^{2})\Gamma\end{array}  \tag{25}\]  
Taking into account Equations (4), (24) and (25), we obtain  
\[ \begin{array}{l}\varepsilon=\psi_{4}(x^{(k)},w^{(k)},s^{(k)},y^{(k)})- \alpha=E-\mu_{1}F(y^{(k)})\\ =E-(I-2A_{2}E+(A_{3}+A_{3}F^{\prime}(\alpha)^{2}-4A_{2}^{2})e^{2})(E+A_{2}E^{2 }+O(E^{3}))\\ =(4A_{2}^{2}-A_{3}-A_{3}F^{\prime}(\alpha)^{2})E\epsilon^{2}+A_{2}E^{2}+O(e^{ 5})\end{array}  \tag{26}\]  
This means that the Equation (4) is of fourth-order convergence. Therefore, from Equations (5) and (24)-(26), we obtain the error equation:  
\[ \begin{array}{l}e_{n+1}=x^{(k+1)}-\alpha=\varepsilon-\mu_{1}F(z^{(k)})\\ =\varepsilon-(I-2A_{2}E+(A_{3}+A_{3}F^{\prime}(\alpha)^{2}-4A_{2}^{2})e^{2})( \varepsilon+O(\varepsilon^{2}))\\ =2A_{2}E\varepsilon-(A_{3}+A_{3}F^{\prime}(\alpha)^{2}-4A_{2}^{2})e^{2} \varepsilon+O(e^{\prime})\end{array}  \tag{27}\]  
This means that the Equation (5) is of sixth-order convergence.