+\section{Aberth method and difficulties}\r
+A cubically convergent iteration method for finding zeros of\r
+polynomials was proposed by O.Aberth[?].The Aberth method is a\r
+purely algebraic derivation.To illustrate the derivation, we let\r
+$w_{i}(z)$ be the product of linear factor $ w_{i}(z)=\prod_{j=1,j\r
+\neq i}^{n} (z-x_{j})$\r
+\r
+and rational function $R_{i}(z)$ be the correction term of\r
+Weistrass method (?)\r
+$$R_{i}(z)=\dfrac{p(z)}{w_{i}(Z)} , i=1,2,...,n. $$\r
+\r
+Differentiating the rational function $R_{i}(z)$ and applying the\r
+Newton method, we have\r
+$$\dfrac{R_{i}(z)}{R_{i}^{'}(z)}= \dfrac{p(z)}{p^{'}(z)-p(z)\dfrac{w_{i}(z)}{w_{i}^{'}(z)}}= \dfrac{p(z)}{p^{'}(z)-p(z) \sum _{j=1,j \neq i}^{n}\dfrac{1}{z-x_{i}}}, i=1,2,...,n $$\r
+Substituting $x_{j}$ for z we obtain the Aberth iteration method\r
+\r
+Let present the means stages of Aberth's method.\r
+\r
+\subsection{Polynomials Initialization}\r
+ The initialization of polynomial P(z) with complex coefficients\r
+ are given by:\r
+ $$ p(z)=\sum{a_{i}z^{n-i}}. where a_{n} \neq 0,a_{0}=1, a_{i}\subset C $$\r
+\r
+\r
+\subsection{Vector $Z^{0)}$ Initialization}\r
+\r
+The choice of the initial points $z^{(0)}_{i} , i = 1, . . . , n,$\r
+from which starting the iteration (2) or (3), is rather delicate\r
+since the number of steps needed by the iterative method to reach\r
+a given approximation strongly depends on it. In [1] the Aberth\r
+iteration is started by selecting n equispaced points on a circle\r
+of center 0 and radius r, where r is an upper bound to the moduli\r
+of the zeros. After[18] performs this choice by selecting complex\r
+numbers along different circles and relies on the result of [19].\r
+\r
+$$\sigma_{0}=\frac{u+v}{2};u=\frac{\sum_{i=1}^{n}u_{i}}{n.max_{i=1}^{n}u_{i}}; v=\frac{\sum_{i=0}^{n-1}v_{i}}{n.min_{i=0}^{n-1}v_{i}};u_{i}=2.|a_{i}|^{\frac{1}{i}}; v_{i}=\frac{|\frac{a_{n}}{a_{i}}|^{\frac{1}{n-i}}}{2} $$\r
+\r
+\subsection{Iterative Function Hi}\r
+The operator used with Aberth's method is corresponding to the\r
+following equation which will enable the convergence towards\r
+polynomial solutions, provided all the roots are distinct.\r
+\r
+$$ H_{i}(z)=z_{i}-\frac{1}{\frac{P^{'}(z_{i})}{P(z_{i})}-\sum_{j\neq i}{\frac{1}{z_{i}-z_{j}}}} $$\r
+\r
+\subsection{Convergence condition}\r
+determines the success of the termination. It consists in stopping\r
+the iterative function $H_{i}(z)$ when the are stable,the method\r
+converge sufficiently:\r
+$$ \forall i \in [1,n]; \frac{z_{i}^{(k)}-z_{i}^{(k-1)}}{z_{i}^{(k)}}< \xi$$\r
+\r
+\section{Difficulties and amelioration}\r
+the Aberth method implementation suffer of overflow problems. This\r
+situation occurs, for instance, in the case where a polynomial\r
+having positive coefficients and large degree is computed at a\r
+point $\xi$ where $|\xi| > 1$.Indeed the limited number in the\r
+mantissa of floating takings the computation of P(z) wrong when z\r
+is large. for example $(10^{50}) +1+ (- 10_{50})$ will give result\r
+0 instead of 1 in reality.consequently we can't compute the roots\r
+for large polynomial's degree. This problem was discuss in [17]\r
+for the Durand-Kerner method, the authors propose to use the\r
+logratihm and the exponential of a complex:\r
+\r
+$$ \forall(x,y)\in R^{*2}; \ln (x+i.y)=\ln(x^{2}+y^{2}) 2+i.\arcsin(y\sqrt{x^{2}+y^{2}})_{\left] -\pi, \pi\right[ } $$\r
+$$ \forall(x,y)\in R^{*2}; \exp(x+i.y)= \exp(x).\exp(i.y)$$\r
+$$ =\exp(x).\cos(y)+i.\exp(x).\sin(y)$$\r
+\r
+\r
+The application of logarithm can replace any multiplications and\r
+divisions with additions and subtractions; consequently it\r
+manipulates lower absolute values and can be compute the roots for\r
+large polynomial's degree exceed 1000[17].\r
+\r
+Applying this solution for the Aberth method we obtain the\r
+iteration function with logarithm:\r
+%%$$ \exp \bigl( \ln(p(z)_{k})-ln(\ln(p(z)_{k}^{'}))- \ln(1- \exp(\ln(p(z)_{k})-ln(\ln(p(z)_{k}^{'})+\ln\sum_{i\neq j}^{n}\frac{1}{z_{k}-z_{j}})$$\r
+\r
+$$ H_{i}(z)=z_{i}^{k}-\exp \left(\ln \left( p(z_{k})\right)-\ln\left(p(z_{k}^{'})\right)- \ln\left(1- \exp\left( \ln (p(z_{k}))-\ln(p(z_{k}^{'}))+\ln \left( \sum_{k\neq j}^{n}\frac{1}{z_{k}-z_{j}}\right)\right) \right) \right)$$\r
+\r
+\r
+\r
+this solution is applying when it is necessary\r
+\r
+\section{The implementation of simultaneous methods in a parallel computer}\r
+\r
