From 0501cb56e1c4b2968b9f435cc1f53e348a8bd434 Mon Sep 17 00:00:00 2001 From: Christophe Guyeux Date: Fri, 10 Oct 2014 14:51:15 +0200 Subject: [PATCH] i8n de la preuve --- paper.tex | 12 +++++++----- 1 file changed, 7 insertions(+), 5 deletions(-) diff --git a/paper.tex b/paper.tex index 4e40730..4a8bc4d 100644 --- a/paper.tex +++ b/paper.tex @@ -757,13 +757,16 @@ $k$-th iterate of TSIRM. We will prove that $r_k \rightarrow 0$ when $k \rightarrow +\infty$. Each step of the TSIRM algorithm \\ + +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 vectors $S$. So,\\ $\min_{\alpha \in \mathbb{R}^s} ||b-R\alpha ||_2 = \min_{\alpha \in \mathbb{R}^s} ||b-AS\alpha ||_2$ $\begin{array}{ll} -& = \min_{x \in Vect\left(S_{k-s}, S_{k-s+1}, \hdots, S_{k-1} \right)} ||b-AS\alpha ||_2\\ -& = \min_{x \in Vect\left(x_{k-s}, x_{k-s}+1, \hdots, x_{k-1} \right)} ||b-AS\alpha ||_2\\ -& \leqslant \min_{x \in Vect\left( x_{k-1} \right)} ||b-Ax ||_2\\ -& \leqslant ||b-Ax_{k-1}|| +& = \min_{x \in span\left(S_{k-s}, S_{k-s+1}, \hdots, S_{k-1} \right)} ||b-AS\alpha ||_2\\ +& = \min_{x \in span\left(x_{k-s}, x_{k-s}+1, \hdots, x_{k-1} \right)} ||b-AS\alpha ||_2\\ +& \leqslant \min_{x \in span\left( x_{k-1} \right)} ||b-Ax ||_2\\ +& \leqslant \min_{\lambda \in \mathbb{R}} ||b-\lambda Ax_{k-1} ||_2\\ +& \leqslant ||b-Ax_{k-1}||_2 . \end{array}$ \end{proof} @@ -1069,4 +1072,3 @@ Curie and Juqueen respectively based in France and Germany. % that's all folks \end{document} - -- 2.39.5