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How Many Steps Still Left to $x$*?
SIAM Review ( IF 10.2 ) Pub Date : 2021-08-05 , DOI: 10.1137/19m1244858
Emil Cătinaş

SIAM Review, Volume 63, Issue 3, Page 585-624, January 2021.
The high speed of $x_{k}\rightarrow x^\ast\in{\mathbb R}$ is usually measured using the $C$-, $Q$-, or $R$-orders: \[ łim \tfrac |x^\ast - x_k+1||x^\ast - x_k|^p_0\in(0,+\infty), łim \tfrac łn |x^\ast - x_k+1|łn |x^\ast - x_k| =q_0, or łim \bigěrtłn |x^\ast - x_k| \big ěrt ^\frac1k =r_0. \] By connecting them to the natural, term-by-term comparison of the errors of two sequences, we find that the $C$-orders---including (sub)linear---are in agreement. Weird relations may appear though for the $Q$-orders: we expect $|x^\ast - x_k|=\mathcal{O}(|x^\ast - y_k|^\alpha)$ $\forall\alpha>1$ to imply “$\geq$" for the $Q$-orders of $\{x_{k}\}$ vs. $\{y_{k}\}$; the contrary is shown by an example providing no vs. infinite $Q$-orders. The $R$-orders appear to be even worse: an $\{x_{k}\}$ with infinite $R$-order may have unbounded nonmonotone errors: $|x^\ast - x_{k+1}|/|x^\ast - x_k|\rightarrow +\infty$. Such aspects motivate the study of equivalent definitions, computational variants, and so on. These orders are also the perspective from which we analyze the three basic iterative methods for nonlinear equations in $\mathbb R$. The Newton method, widely known for its quadratic convergence, may in fact attain any $C$-order from $[1,+\infty]$ (including sublinear); we briefly recall such convergence results, along with connected aspects (such as historical notes, known asymptotic constants, floating point arithmetic issues, and radius of attraction balls), and provide examples. This approach leads to similar results for the successive approximations method, while the secant method exhibits different behavior: it may not have high $C$-orders, but only $Q$-orders.


中文翻译:

$x$* 还剩下多少步骤?

SIAM 评论,第 63 卷,第 3 期,第 585-624 页,2021 年 1 月。
$x_{k}\rightarrow x^\ast\in{\mathbb R}$ 的高速通常使用 $C$-、$Q$- 或 $R$-orders 来衡量: \[ łim \tfrac |x^\ast - x_k+1||x^\ast - x_k|^p_0\in(0,+\infty), łim \tfrac łn |x^\ast - x_k+1|łn |x^\ast - x_k| =q_0,或 łim \bigěrtłn |x^\ast - x_k| \big ěrt ^\frac1k =r_0. \] 通过将它们与两个序列的错误的自然、逐项比较,我们发现 $C$-orders---包括(sub)linear---是一致的。但是对于 $Q$-orders 可能会出现奇怪的关系:我们期望 $|x^\ast - x_k|=\mathcal{O}(|x^\ast - y_k|^\alpha)$ $\forall\alpha> 1$ 表示 $\{x_{k}\}$ 与 $\{y_{k}\}$ 的 $Q$-orders 的“$\geq$”;相反的例子显示为不提供与无限的 $Q$-orders. $R$-orders 似乎更糟:具有无限 $R$-order 的 $\{x_{k}\}$ 可能具有无限的非单调错误:$|x^\ast - x_{k+1}|/|x^\ast - x_k|\rightarrow +\infty$。这些方面激发了对等效定义、计算变体等的研究。这些阶数也是我们分析 $\mathbb R$ 中非线性方程的三种基本迭代方法的视角。Newton 方法以其二次收敛而广为人知,实际上可以从 $[1,+\infty]$(包括次线性)获得任何 $C$ 阶;我们简要回顾了这样的收敛结果,以及相关的方面(例如历史记录、已知的渐近常数、浮点算术问题和吸引力球的半径),并提供了示例。这种方法对逐次逼近方法产生了类似的结果,而割线方法表现出不同的行为:它可能没有高 $C$-orders,但只有 $Q$-orders。
更新日期:2021-08-07
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