From 093b40c0c2a74ebfa6e3366e75a49bac1d4f59f2 Mon Sep 17 00:00:00 2001 From: Christophe Guyeux Date: Fri, 10 Oct 2014 15:49:20 +0200 Subject: [PATCH] =?utf8?q?avanc=C3=A9es=20dans=20la=20preuve?= MIME-Version: 1.0 Content-Type: text/plain; charset=utf8 Content-Transfer-Encoding: 8bit --- paper.tex | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) 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,\\ $\min_{\alpha \in \mathbb{R}^s} ||b-R\alpha ||_2 = \min_{\alpha \in \mathbb{R}^s} ||b-AS\alpha ||_2$ $\begin{array}{ll} -- 2.39.5