relecture section III B

[kahina_paper2.git] / paper.tex

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--- a/paper.tex
+++ b/paper.tex
@@ -663,29 +663,26 @@ 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,\ldots,n
 \end{equation}
 
-This method contains 4 steps. The first step consists of the initial
-approximations of all the roots of the polynomial. The second step
-initializes the solution vector $Z$ using the Guggenheimer
-method~\cite{Gugg86} to ensure the distinction of the initial vector
-roots. In step 3, the iterative function based on the Newton's
-method~\cite{newt70} and Weiestrass operator~\cite{Weierstrass03} is
-applied. With this step the computation of roots will converge,
-provided that all roots are different.
+This method contains 4 steps. The first step consists of the initial approximations of all the roots of the polynomial.\LZK{Pas compris??}
+The second step initializes the solution vector $Z$ using the Guggenheimer method~\cite{Gugg86} to ensure the distinction of the initial vector roots.\LZK{Quelle est la différence entre la 1st step et la 2nd step? Que veut dire " to ensure the distinction of the initial vector roots"?} 
+In step 3, the iterative function based on the Newton's method~\cite{newt70} and Weiestrass operator~\cite{Weierstrass03} is applied. With this step the computation of roots will converge, provided that all roots are different.\LZK{On ne peut pas expliquer un peu plus comment? Donner des formules comment elle se base sur la méthode de Newton et de l'opérateur de  Weiestrass?}
+\LZK{Elle est où la 4th step??}
+\LZK{Conclusion: Méthode mal présentée et j'ai presque rien compris!}
 
 
 In order to stop the iterative function, a stop condition is
 applied. This condition checks that all the root modules are lower
-than a fixed value $\xi$.
+than a fixed value $\epsilon$.
 
 \begin{equation}
 \label{eq:Aberth-Conv-Cond}
-\forall i \in [1,n];\vert\frac{z_{i}^{k}-z_{i}^{k-1}}{z_{i}^{k}}\vert<\xi
+\forall i\in[1,n],~\vert\frac{z_i^k-z_i^{k-1}}{z_i^k}\vert<\epsilon
 \end{equation}
+
+\LZK{On ne dit pas plutôt "the relative errors" à la place de "root modules"? Raph nous confirmera quelle critère d'arrêt a utilisé.}
+
 \subsection{Improving Ehrlich-Aberth method}
-With high degree polynomials, the Ehrlich-Aberth method suffers from
-floating point overflows due to the mantissa of floating points
-representations. This induces errors in the computation of $p(z)$ when
-$z$ is large.
+With high degree polynomials, the Ehrlich-Aberth method suffers from floating point overflows due to the mantissa of floating points representations. This induces errors in the computation of $p(z)$ when $z$ is large.
  
 %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: 
 
@@ -702,28 +699,23 @@ 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:
+100,000 and to reach a degree up to more than 1,000,000. The reformulation of the iteration~(\ref{Eq:EA1}) of the Ehrlich-Aberth method with exponential and logarithm is defined as follows, for $i=1,\dots,n$:
 
 \begin{equation}
 \label{Log_H2}
-z^{k+1}_{i}=z_{i}^{k}-\exp \left(\ln \left(
-p(z_{i}^{k})\right)-\ln\left(p'(z^{k}_{i})\right)- \ln\left(1-Q(z^{k}_{i})\right)\right),
+z^{k+1}_i = z_i^k - \exp(\ln(p(z_i^k)) - \ln(p'(z^k_i)) - \ln(1-Q(z^k_i))),
 \end{equation}
 
 where:
 
-\begin{eqnarray}
+\begin{equation}
 \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) \nonumber  \\
-i=1,...,n  
-\end{eqnarray}
+Q(z^k_i) = \exp(\ln(p(z^k_i)) - \ln(p'(z^k_i)) + \ln(\sum_{i\neq j}^n\frac{1}{z^k_i-z^k_j})).
+\end{equation}
 
 
 %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~\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}. \LZK{Je n'ai pas compris cette dernière phrase?}
 
 %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}.