From: Christophe Guyeux <guyeux@gmail.com>
Date: Fri, 10 Oct 2014 13:49:20 +0000 (+0200)
Subject: avancées dans la preuve
X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/GMRES2stage.git/commitdiff_plain/093b40c0c2a74ebfa6e3366e75a49bac1d4f59f2?hp=5b9bad14b26c5a92c18a688312debc08072dc241

avancées dans la preuve
---

diff --git a/paper.tex b/paper.tex
index 6add44a..012e7f1 100644
--- a/paper.tex
+++ b/paper.tex
@@ -768,9 +768,9 @@ The base case is obvious, as for $k=1$, the TSIRM algorithm simply consists in a
 Suppose now that the claim holds for all $m=1, 2, \hdots, k-1$, that is, $\forall m \in \{1,2,\hdots, k-1\}$, $||r_m|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||$.
 We will show that the statement holds too for $r_k$. Two situations can occur:
 \begin{itemize}
-\item If $k \mod m \neq 0$, then the TSIRM algorithm consists in executing GMRES once. In that case, we obtain $||r_k|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_{k-1}||\leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_0||$.
-
-\item Else, let $\operatorname{span}(S) = \left \{ {\sum_{i=1}^k \lambda_i v_i \Big| k \in \mathbb{N}, v_i \in S, \lambda _i \in \mathbb{R}} \right \}$ be the linear span of a set of real vectors $S$. So,\\
+\item If $k \mod m \neq 0$, then the TSIRM algorithm consists in executing GMRES once. In that case, we obtain $||r_k|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_{k-1}||\leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||$ by the inductive hypothesis.
+\item Else, the TSIRM algorithm consists in two stages: a first GMRES($m$) execution leads to a temporary $x_k$ whose residue satisfies $||r_k|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_{k-1}||\leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||$, and a least squares resolution.
+Let $\operatorname{span}(S) = \left \{ {\sum_{i=1}^k \lambda_i v_i \Big| k \in \mathbb{N}, v_i \in S, \lambda _i \in \mathbb{R}} \right \}$ be the linear span of a set of real vectors $S$. So,\\
 $\min_{\alpha \in \mathbb{R}^s} ||b-R\alpha ||_2 = \min_{\alpha \in \mathbb{R}^s} ||b-AS\alpha ||_2$
 
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