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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|数值分析代写numerical analysis代考|Effects of Finite Precision Arithmetic

In developing a numerical method for a given problem we begin with mathematical or physical intuition about the particular problem and use that intuition to devise a scheme that, we hope, will converge to a solution of the problem. Since most methods do not converge in a finite number of iterations we include convergence criteria that indicate when we have achieved practical convergence, that is, when we have found sufficiently accurate values for our purposes (or at least, as accurate as we can reasonably hope to find).

Whether a method does or does not converge in the usual mathematical sense, as a limit, and how rapidly it does so, is the first issue. The second issue is stability, by which we mean the sensitivity of the method to small errors such as round-off errors. If a method converges only when infinite precision arithmetic is used but not on a computer then it will not be of much interest to us.

Here’s an example of a method that is not convergent: Consider the problem of finding a fixed point of $f(x)=3 \cos (x)$ near $x=1$ (see the MATLAB subsection of Sec. 1.3). The direct application of fixed point iteration gives
$$x_{k+1}=3 \cos \left(x_k\right)$$ (see Fig. 1). There is a solution at $x^* \doteq 1.17$; however, since $f^{\prime}\left(x^\right)=-3 \sin \left(x^\right) \approx$ $-2.76$ the fixed point is not attracting (i.e., it is repelling) and the method cannot converge for any $x_0$ in a deleted neighborhood of $x^*$, even using infinite precision arithmetic. The method is not convergent in the usual sense of the calculus; Newton’s method, of course, would be convergent for the corresponding root-finding problem $f(x)-x=0$.

Lack of stability is a different issue, however. To explore it, let’s consider the two-step recurrence relation
$$x_{k+1}=x_k+x_{k-1}$$
with the initial conditions $x_0=1, x_1=(1-\sqrt{5}) / 2 \doteq-0.6180$.

## 数学代写|数值分析代写numerical analysis代考|Polynomial Roots

In general it is not reasonable to ask that a numerical method find all roots of even a continuous function of a single variable. There may be infinitely many such roots; there may be closely spaced roots (consider $\sin (1 / x)$ near the origin); there may be a root at such a large value of $x$ that the method simply never ends up looking for it there; the root may be a blip between two adjacent machine numbers with the function positive at all máchine numbers; and so on. Even for well-behaved functions, a multiple root is hard to handle. If the multiplicity is even, as for $f(x)=x^2$, then the root is hard to see numerically, and whether it is even or odd, the fact that $f^{\prime}(x) \approx 0$ in a neighborhood of the root means that the function is flat there, so points near it will seem like roots too.

Further, experience shows that methods like Newton’s method sometimes have “favorite” zeroes, that is, if we find an approximate root $x_a$ of $f(x)=0$ starting from some initial guess $x_0$, then choose a new initial guess $z_0$, it is not uncommon for the method to find the same approximate root $x_a$ from this new initial guess even if there are other roots to be found. One way to defeat this is to eliminate the newly found zero by dividing it out of the function, giving a new function $$g(x)=\frac{f(x)}{x-x^}$$ which has zeroes exactly where $f$ does, save for $x^$ (assuming $x^$ is an exact root of $f(x)=0)$. This process is known as deflation. In practice $f\left(x_a\right)$ is not exactly zero and so $$g(x)=\frac{f(x)}{x-x_a}$$ has a pole, that is, an infinitely large spike, next to a zero. Often the use of deflation is enough to drive the method away from the known zero near $x^$, but not always. The pole-near-a-zero phenomenon can itself cause difficulties, and so the cure can be as bad as the disease in some cases. Still, repeated deflation of the form
$$g(x)=\frac{f(x)}{\left(x-x_a\right)\left(x-x_b\right)\left(x-x_c\right)}$$
(after three different approximate roots $x_a, x_b$, and $x_c$ have been found) can be a useful tool. When $f(x)$ is a polynomial of known degree $n$, we know we can stop after we have found $n$ zeroes of it.

# 数值分析代考

## 有限元方法代写

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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