representations. This induces errors in the computation of $p(z)$ when
$z$ is large.
-Experimentally, it is very difficult to solve polynomials with Ehrlich-Aberth method and have roots which except the circle of unit, represented by the radius $r$ evaluated as:
-
-\begin{equation}
-\label{R.EL}
-R = exp(log(DBL\_MAX)/(2*n) );
-\end{equation}
-
+%Experimentally, it is very difficult to solve polynomials with the Ehrlich-Aberth method and have roots which except the circle of unit, represented by the radius $r$ evaluated as:
%\begin{equation}
-
-%R = \exp( \log(DBL\_MAX) / (2*n) )
+%\label{R.EL}
+%R = exp(log(DBL\_MAX)/(2*n) );
%\end{equation}
- where \verb=DBL_MAX= stands for the maximum representable \verb=double= value.
-
-In order to hold into account the limit of size of floats, we propose to modifying the iterative function and compute the logarithm of:
-\begin{equation}
-EA: z^{k+1}_{i}=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=1,. . . .,n
-\end{equation}
-This method allows, indeed, to exceed the computation of the polynomials of degree 100,000 and to reach a degree upper to 1,000,000. For that purpose, it is necessary to use the logarithm and the exponential of a complex. The iterative function of Ehrlich-Aberth method with exponential and logarithm is given as following:
+
+% where \verb=DBL_MAX= stands for the maximum representable \verb=double= value.
+
+In order to solve this problem, we propose to modify the iterative
+function by using the logarithm and the exponential of a complex and
+we propose a new version of the Ehrlich-Aberth method. This method
+allows us to exceed the computation of the polynomials of degree
+100,000 and to reach a degree up to more than 1,000,000. This new
+version of the Ehrlich-Aberth method with exponential and logarithm is
+defined as follows:
\begin{equation}
\label{Log_H2}
where:
-\begin{equation}
+\begin{eqnarray}
\label{Log_H1}
Q(z^{k}_{i})=\exp\left( \ln (p(z^{k}_{i}))-\ln(p'(z^{k}_{i}))+\ln \left(
-\sum_{i\neq j}^{n}\frac{1}{z^{k}_{i}-z^{k}_{j}}\right)\right)i=1,...,n,
-\end{equation}
+\sum_{i\neq j}^{n}\frac{1}{z^{k}_{i}-z^{k}_{j}}\right)\right)\\
+i=1,...,n, \nonumber
+\end{eqnarray}
%We propose to use the logarithm and the exponential of a complex in order to compute the power at a high exponent.
-Using the logarithm and the exponential operators, we can replace any multiplications and divisions with additions and subtractions. Consequently, computations manipulate lower absolute values and the roots for large polynomial degrees can be looked for successfully~\cite{Karimall98}.
+Using the logarithm and the exponential operators, we can replace any multiplications and divisions with additions and subtractions. Consequently, computations manipulate lower absolute values~\cite{Karimall98}.
%This problem was discussed earlier in~\cite{Karimall98} for the Durand-Kerner method. The authors
%propose to use the logarithm and the exponential of a complex in order to compute the power at a high exponent. Using the logarithm and the exponential operators, we can replace any multiplications and divisions with additions and subtractions. Consequently, computations manipulate lower absolute values and the roots for large polynomial degrees can be looked for successfully~\cite{Karimall98}.
