Just MAJ experiment part

diff --git a/paper.tex b/paper.tex
index e06aa394efb228903ee1c4b2720de4b00228bb8d..7907494f1d577ecf3be95f67c24ca4a0c32b5034 100644 (file)
--- a/paper.tex
+++ b/paper.tex
@@ -321,6 +321,7 @@ Applying this solution for the Aberth method we obtain the
 iteration function with 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}
 iteration function with 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}

 H_{i}(z)=z_{i}^{k}-\exp \left(\ln \left(
 p(z_{k})\right)-\ln\left(p(z_{k}^{'})\right)- \ln
 \left(1-Q(z_{k})\right)\right),
 H_{i}(z)=z_{i}^{k}-\exp \left(\ln \left(
 p(z_{k})\right)-\ln\left(p(z_{k}^{'})\right)- \ln
 \left(1-Q(z_{k})\right)\right),


@@ -329,12 +330,17 @@ p(z_{k})\right)-\ln\left(p(z_{k}^{'})\right)- \ln
 where:
 
 \begin{equation}
 where:
 
 \begin{equation}

+\label{Log_H1}

 Q(z_{k})=\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).
 \end{equation}
 
 This solution is applied when the root except the circle unit, represented by the radius $R$ evaluated as:
 Q(z_{k})=\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).
 \end{equation}
 
 This solution is applied when the root except the circle unit, represented by the radius $R$ evaluated as:

-$$R = \exp( \log(DBL\_MAX) / (2*n) )$$ where $DBL\_MAX$ stands for the maximum representable double value.
+\begin{equation}
+\label{R}
+R = \exp( \log(DBL\_MAX) / (2*n) )
+\end{equation}
+ where $DBL\_MAX$ stands for the maximum representable double value.

 
 \section{The implementation of simultaneous methods in a parallel computer}
 \label{secStateofArt}   
 
 \section{The implementation of simultaneous methods in a parallel computer}
 \label{secStateofArt}   


@@ -663,9 +669,9 @@ In this experiment we report the performance of log.exp solution describe in ~\r
 \label{fig:01}
 \end{figure}
 
 \label{fig:01}
 \end{figure}
 

-The figure 3, show a comparison between the execution time of the Ehrlisch-Aberth algorithm applying log-exp solution and the execution time of the Ehrlisch-Aberth algorithm without applying log-exp solution, with full polynomials degrees. We can see that the execution time for the both algorithms are the same while the polynomials degrees are less than 4500.After,we show clearly that the Ehrlisch-Aberth algorithm without applying log.exp stop to converge consequently,can not solving polynomial exceed 4500, in counterpart, applying log.exp solution the Ehrlisch-Aberth algorithm can solving very high and large full polynomial exceed 500,000 degrees.
+The figure 3, show a comparison between the execution time of the Ehrlisch-Aberth algorithm applying log-exp solution and the execution time of the Ehrlisch-Aberth algorithm without applying log-exp solution, with full polynomials degrees. We can see that the execution time for the both algorithms are the same while the polynomials degrees are less than 4500. After,we show clearly that the classical version of Ehrlisch-Aberth algorithm (without applying log.exp) stop to converge and can not solving polynomial exceed 4500, in counterpart, the new version of Ehrlisch-Aberth algorithm (applying log.exp solution) can solve very high and large full polynomial exceed 100,000 degrees.

 
 

-in fact, when the modulus of the roots are up than R given in (~\ref{eq:radiusR}),this exceed the the limited number in the mantissa of floating points representations who justify the divergence of the Ehrlisch-Aberth algorithm without log.exp. However, applying log.exp solution given in equation~\ref{alg3-update} took into account the limit of floating using the iterative function given in~\ref{eq:Aberth-H-GS}.
+in fact, when the modulus of the roots are up than R given in (~\ref{R}),this exceed the limited number in the mantissa of floating points representations and can not compute the iterative function given in ~\ref{eq:Aberth-H-GS} to obtain the root solution, who justify the divergence of the classical Ehrlisch-Aberth algorithm. However, applying log.exp solution given in ~\ref{sec2} took into account the limit of floating using the iterative function in~\ref{Log_H1} ~\ref{Log_H2}.