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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Fixed-point iteration ： ウィキペディア英語版 Fixed-point iteration (詳細はnumerical analysis, fixed-point iteration is a method of computing fixed points of iterated functions. More specifically, given a function $f$ defined on the real numbers with real values and given a point $x\_0$ in the domain of $f$, the fixed point iteration is :$x\_=f(x\_n),\; \backslash ,\; n=0,\; 1,\; 2,\; \backslash dots$ which gives rise to the sequence $x\_0,\; x\_1,\; x\_2,\; \backslash dots$ which is hoped to converge to a point $x$. If $f$ is continuous, then one can prove that the obtained $x$ is a fixed point of $f$, i.e., :$f(x)=x$. More generally, the function $f$ can be defined on any metric space with values in that same space. ==Examples== * A first simple and useful example is the Babylonian method for computing the square root of ''a''>0, which consists in taking $f(x)=\backslash frac\; 12\backslash left(\backslash frac\; ax\; +\; x\backslash right)$, i.e. the mean value of ''x'' and ''a/x'', to approach the limit $x\; =\; \backslash sqrt\; a$ (from whatever starting point $x\_0\; \backslash gg\; 0$). This is a special case of Newton's method quoted below. * The fixed-point iteration $x\_=\backslash cos\; x\_n\backslash ,$ converges to the unique fixed point of the function $f(x)=\backslash cos\; x\backslash ,$ for any starting point $x\_0.$ This example does satisfy the assumptions of the Banach fixed point theorem. Hence, the error after n steps satisfies $|x\_n-x\_0|\; \backslash leq\; |\; x\_1\; -\; x\_0\; |\; =\; C\; q^n$ (where we can take $q\; =\; 0.85$, if we start from $x\_0=1$.) When the error is less than a multiple of $q^n$ for some constant ''q'', we say that we have linear convergence. The Banach fixed-point theorem allows one to obtain fixed-point iterations with linear convergence. * The fixed-point iteration $x\_=2x\_n\backslash ,$ will diverge unless $x\_0=0$. We say that the fixed point of $f(x)=2x\backslash ,$ is repelling. * The requirement that ''f'' is continuous is important, as the following example shows. The iteration :$x\_\; =$ \begin \frac, & x_n \ne 0\\ 1, & x_n=0 \end converges to 0 for all values of $x\_0$. However, 0 is ''not'' a fixed point of the function :$f(x)\; =$ \begin \frac, & x \ne 0\\ 1, & x = 0 \end as this function is ''not'' continuous at $x=0$, and in fact has no fixed points. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Fixed-point iteration」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.