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[kahina_paper1.git] / paper.tex

diff --git a/paper.tex b/paper.tex
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+++ b/paper.tex
@@ -269,7 +269,7 @@ The initialization of a polynomial p(z) is done by setting each of the $n$ compl
 
 \subsection{Vector $z^{(0)}$ Initialization}
 
 
 \subsection{Vector $z^{(0)}$ Initialization}
 

-Like for any iterative method, we need to choose $n$ initial guess points $z^{(0)}_{i}, i = 1, . . . , n.$
+As for any iterative method, we need to choose $n$ initial guess points $z^{(0)}_{i}, i = 1, . . . , n.$

 The initial guess is very important since the number of steps needed by the iterative method to reach
 a given approximation strongly depends on it.
 In~\cite{Aberth73} the Ehrlich-Aberth iteration is started by selecting $n$
 The initial guess is very important since the number of steps needed by the iterative method to reach
 a given approximation strongly depends on it.
 In~\cite{Aberth73} the Ehrlich-Aberth iteration is started by selecting $n$


@@ -303,9 +303,12 @@ Here we give a second form of the iterative function used by Ehrlich-Aberth meth
 EA2: z^{k+1}=z_{i}^{k}-\frac{\frac{p(z_{i}^{k})}{p'(z_{i}^{k})}}
 {1-\frac{p(z_{i}^{k})}{p'(z_{i}^{k})}\sum_{j=1,j\neq i}^{j=n}{\frac{1}{(z_{i}^{k}-z_{j}^{k})}}}, i=0,. . . .,n
 \end{equation}
 EA2: z^{k+1}=z_{i}^{k}-\frac{\frac{p(z_{i}^{k})}{p'(z_{i}^{k})}}
 {1-\frac{p(z_{i}^{k})}{p'(z_{i}^{k})}\sum_{j=1,j\neq i}^{j=n}{\frac{1}{(z_{i}^{k}-z_{j}^{k})}}}, i=0,. . . .,n
 \end{equation}

-we notice that the function iterative in Eq.~\ref{Eq:Hi} it the same those presented in Eq.~\ref{Eq:EA}, but we prefer used the last one seen the advantage of its use to improve the Ehrlich-Aberth method and resolve very high degrees polynomials. More detail in the section ~\ref{sec2}.    
+It can be noticed that this equation is equivalent to Eq.~\ref{Eq:EA},
+but we prefer the latter one because we can use it to improve the
+Ehrlich-Aberth method and find the roots of very high degrees polynomials. More
+details are given in Section ~\ref{sec2}.

 \subsection{Convergence Condition}
 \subsection{Convergence Condition}

-The convergence condition determines the termination of the algorithm. It consists in stopping from running the iterative function  when the roots are sufficiently stable. We consider that the method converges sufficiently when:
+The convergence condition determines the termination of the algorithm. It consists in stopping the iterative function  when the roots are sufficiently stable. We consider that the method converges sufficiently when:

 
 \begin{equation}
 \label{eq:Aberth-Conv-Cond}
 
 \begin{equation}
 \label{eq:Aberth-Conv-Cond}


@@ -316,7 +319,8 @@ The convergence condition determines the termination of the algorithm. It consis
 
 \section{Improving the Ehrlich-Aberth Method for high degree polynomials with exp.log formulation}
 \label{sec2}
 
 \section{Improving the Ehrlich-Aberth Method for high degree polynomials with exp.log formulation}
 \label{sec2}

-The Ehrlich-Aberth method implementation suffers of overflow problems. This
+With high degree polynomial, the Ehrlich-Aberth method implementation,
+as well as the Durand-Kerner implement, suffers from overflow problems. This

 situation occurs, for instance, in the case where a polynomial
 having positive coefficients and a large degree is computed at a
 point $\xi$ where $|\xi| > 1$, where $|x|$ stands for the modolus of a complex $x$. Indeed, the limited number in the
 situation occurs, for instance, in the case where a polynomial
 having positive coefficients and a large degree is computed at a
 point $\xi$ where $|\xi| > 1$, where $|x|$ stands for the modolus of a complex $x$. Indeed, the limited number in the


@@ -344,7 +348,7 @@ Using the logarithm (eq.~\ref{deflncomplex}) and the exponential (eq.~\ref{defex
 manipulate lower absolute values and the roots for large polynomial's degrees can be looked for successfully~\cite{Karimall98}.
 
 Applying this solution for the Ehrlich-Aberth method we obtain the
 manipulate lower absolute values and the roots for large polynomial's degrees can be looked for successfully~\cite{Karimall98}.
 
 Applying this solution for the Ehrlich-Aberth method we obtain the

-iteration function with logarithm:
+iteration function with exponential and logarithm:

 %%$$ \exp \bigl(  \ln(p(z)_{k})-ln(\ln(p(z)_{k}^{'}))- \ln(1- \exp(\ln(p(z)_{k})-ln(\ln(p(z)_{k}^{'})+\ln\sum_{i\neq j}^{n}\frac{1}{z_{k}-z_{j}})$$
 \begin{equation}
 \label{Log_H2}
 %%$$ \exp \bigl(  \ln(p(z)_{k})-ln(\ln(p(z)_{k}^{'}))- \ln(1- \exp(\ln(p(z)_{k})-ln(\ln(p(z)_{k}^{'})+\ln\sum_{i\neq j}^{n}\frac{1}{z_{k}-z_{j}})$$
 \begin{equation}
 \label{Log_H2}