From: Christophe Guyeux Date: Fri, 10 Oct 2014 13:20:12 +0000 (+0200) Subject: Avancées dans la preuve par induction X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/GMRES2stage.git/commitdiff_plain/c21f0db90e14920174026f446c338c0a88e35262 Avancées dans la preuve par induction --- diff --git a/paper.tex b/paper.tex index 436909a..d9880d1 100644 --- a/paper.tex +++ b/paper.tex @@ -759,11 +759,17 @@ where $\alpha$ and $\beta$ are defined as in Proposition~\ref{prop:saad}. \begin{proof} Let $r_k = b-Ax_k$, where $x_k$ is the approximation of the solution after the $k$-th iterate of TSIRM. -We will prove that $r_k \rightarrow 0$ when $k \rightarrow +\infty$. +We will prove by a mathematical induction that, for each $k \in \mathbb{N}^\ast$, +$||r_m|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_0||.$ -Each step of the TSIRM algorithm \\ +The base case is obvious, as for $k=1$, the TSIRM algorithm simply consists in applying GMRES($m$) once, leading to a new residual $r_1$ which follows the inductive hypothesis due to Proposition~\ref{prop:saad}. -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,\\ +Suppose now that the claim holds for all $m=1, 2, \hdots, k-1$, that is, $\forall m \in \{1,2,\hdots, k-1\}$, $||r_m|| \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{m}{2}} ||r_0||$. +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 + +\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,\\ $\min_{\alpha \in \mathbb{R}^s} ||b-R\alpha ||_2 = \min_{\alpha \in \mathbb{R}^s} ||b-AS\alpha ||_2$ $\begin{array}{ll} @@ -773,6 +779,7 @@ $\begin{array}{ll} & \leqslant \min_{\lambda \in \mathbb{R}} ||b-\lambda Ax_{k-1} ||_2\\ & \leqslant ||b-Ax_{k-1}||_2 . \end{array}$ +\end{itemize} \end{proof} We can remark that, at each iterate, the residue of the TSIRM algorithm is lower