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[kahina_paper1.git] / paper.tex

diff --git a/paper.tex b/paper.tex
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--- a/paper.tex
+++ b/paper.tex
@@ -268,7 +268,7 @@ The initialization of a polynomial $p(z)$ is done by setting each of the $n$ com
 
 \subsection{Vector $Z^{(0)}$ Initialization}
 \label{sec:vec_initialization}
 
 \subsection{Vector $Z^{(0)}$ Initialization}
 \label{sec:vec_initialization}

-As 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 $n$


@@ -299,7 +299,7 @@ Here we give a second form of the iterative function used by Ehrlich-Aberth meth
 
 \begin{equation}
 \label{Eq:Hi}
 
 \begin{equation}
 \label{Eq:Hi}

-EA2: z^{k+1}=z_{i}^{k}-\frac{\frac{p(z_{i}^{k})}{p'(z_{i}^{k})}}
+EA2: 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=0,. . . .,n
 \end{equation}
 It can be noticed that this equation is equivalent to Eq.~\ref{Eq:EA},
 {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}
 It can be noticed that this equation is equivalent to Eq.~\ref{Eq:EA},


@@ -321,8 +321,8 @@ 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
 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
-mantissa of floating points representations makes the computation of p(z) wrong when z
+point $\xi$ where $|\xi| > 1$, where $|z|$ stands for the modolus of a complex $z$. Indeed, the limited number in the
+mantissa of floating points representations makes the computation of $p(z)$ wrong when z

 is large. For example $(10^{50}) +1+ (- 10^{50})$ will give the wrong result
 of $0$ instead of $1$. Consequently, we can not compute the roots
 for large degrees. This problem was early discussed in
 is large. For example $(10^{50}) +1+ (- 10^{50})$ will give the wrong result
 of $0$ instead of $1$. Consequently, we can not compute the roots
 for large degrees. This problem was early discussed in


@@ -350,7 +350,7 @@ 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}

-EA.EL: z^{k+1}=z_{i}^{k}-\exp \left(\ln \left(
+EA.EL: 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),
 \end{equation}
 p(z_{i}^{k})\right)-\ln\left(p'(z^{k}_{i})\right)- \ln
 \left(1-Q(z^{k}_{i})\right)\right),
 \end{equation}


@@ -360,7 +360,7 @@ where:
 \begin{equation}
 \label{Log_H1}
 Q(z^{k}_{i})=\exp\left( \ln (p(z^{k}_{i}))-\ln(p'(z^{k}_{i}))+\ln \left(
 \begin{equation}
 \label{Log_H1}
 Q(z^{k}_{i})=\exp\left( \ln (p(z^{k}_{i}))-\ln(p'(z^{k}_{i}))+\ln \left(

-\sum_{k\neq j}^{n}\frac{1}{z^{k}_{i}-z^{k}_{j}}\right)\right).
+\sum_{i\neq j}^{n}\frac{1}{z^{k}_{i}-z^{k}_{j}}\right)\right)i=1,...,n,

 \end{equation}
 
 This solution is applied when the root except the circle unit, represented by the radius $R$ evaluated in C language as:
 \end{equation}
 
 This solution is applied when the root except the circle unit, represented by the radius $R$ evaluated in C language as: