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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

## 物理代写|统计力学代写Statistical mechanics代考|The assumption of weak interactions

We take a composite system (system $A$ interacting with its surroundings $B$; see Fig. 1.9) and consider it an isolated system of total energy $E$. Let $\Gamma$ denote its phase space with canonical coordinates $\left{p_{i}, q_{i}\right}, i=1, \cdots, 3 N$, with $A$ having coordinates $\left{p_{i}, q_{i}\right}$ for $i=1, \cdots, 3 n$ and $B$ having coordinates $\left{p_{i}, q_{i}\right}$ for $i=3 n+1, \cdots, 3 N$, where $N \gg n$. We can write the Hamiltonian of the composite system in the form $H(A, B)=H_{A}(A)+H_{B}(B)+V(A, B)$, where $H_{A}\left(H_{B}\right)$ is a function of the canonical coordinates of system $A(B)$, and $V(A, B)$ describes the interactions between $A$ and $B$ involving both sets of coordinates. The energies $E_{A} \equiv H_{A}, E_{B} \equiv H_{B}$ far exceed the energy of interaction $V(A, B)$ because $E_{A}$ and $E_{B}$ are proportional to the volumes of $A$ and $B$, whereas $V(A, B)$ is proportional to the surface area of contact between them (for short-range forces). For macroscopic systems, $|V(A, B)|$ is negligible in comparison with $E_{A}, E_{B}$. Thus, we take
$$E=E_{A}+E_{B},$$
the assumption of weak interaction between $A$ and $B$ (even though “no interaction” might seem more apt). We can’t take $V(A, B) \equiv 0$ because $A$ and $B$ would then be isolated systems. Equilibrium is established and maintained through a continual process of energy transfers between system and environment; taking $V(A, B)=0$ would exclude that possibility. For systems featuring shortrange interatomic forces, we can approximate $E \approx E_{A}+E_{B}$ when the surface area of contact between $A$ and $B$ does not increase too rapidly in relation to bulk volume (more than $V^{2 / 3}$ is too fast). ${ }^{3}$ No matter how small in relative terms the energy of interaction $V(A, B)$ might be, it’s required to estahlish equilihrium hetween $A$ and $R$. We’re not concerned (in equilihrium statistical mechanics) with how a system comes to be in equilibrium (in particular how much time is required). We assume that, in equilibrium, $E_{A}, E_{B} \gg|V(A, B)|$, with Eq. (4.4) as a consequence.

## 物理代写|统计力学代写Statistical mechanics代考|Partition function: Laplace transform of the density of states function

The canonical distribution requires that we know the density-of-state functions of the system and the surroundings. Density-of-states functions satisfy a convolution relation, Eq. (4.9). By the convolution theorem, ${ }^{6}$ the integral transform $T$ of a convolution integral is equal to the product of the transforms of the functions appearing in the convolution, $T(f * g)=T(f) T(g)$. The Laplace transform of $\Omega(x)$ is known as the partition function: ${ }^{7}$
$$Z(\alpha) \equiv \int_{0}^{\infty} \mathrm{e}^{-\alpha x} \Omega(x) \mathrm{d} x,$$
where $\alpha>0$. The Laplace transform is the natural choice of transform (as opposed to Fourier) because $\Omega(x \leq 0)=0$. We’ll soon give $\alpha$ a physical interpretation, but for now we treat it as a mathematical parameter. ${ }^{8}$

Partition functions obey a simple composition law. By taking the Laplace transform of Eq. (4.9), we find for a system composed of $n$ subsystems $\left(Z_{i}(\alpha)\right.$ is the Laplace transform of $\Omega_{i}(x)$ ),
$$Z(\alpha)=\prod_{i=1}^{n} Z_{i}(\alpha),$$
where the parameter $\alpha$ applies to all subsystems. ${ }^{9}$ Equation (4.16) implies a useful feature of partition functions. We can consider each molecule of a system as a subsystem! For a gas of $N$ identical molecules, if $z(\alpha)$ is the partition function of a single molecule (Laplace transform of its density of states function), $Z(\alpha)=[z(\alpha)]^{N}$.

# 统计力学代考

## 物理代写|统计力学代写Statistical mechanics代考|The assumption of weak interactions

\eft 的分隔符缺失或无法识别，和 $A$ 有坐标 $\backslash$ left 的分隔符缺失或无法识别
\eft 的分隔符缺失或无法识别 为了 $i=3 n+1, \cdots, 3 N$ ，在哪里 $N \gg n$. 我们可以将复合系统的哈密顿量写成以下形式
$H(A, B)=H_{A}(A)+H_{B}(B)+V(A, B)$ ， 在哪里 $H_{A}\left(H_{B}\right)$ 是系统规范坐标的函数 $A(B)$ ，和 $V(A, B)$ 描述了之间的相互作用 $A$ 和 $B$ 涉及两组坐标。能量
$E_{A} \equiv H_{A}, E_{B} \equiv H_{B}$ 远远超过相互作用的能量 $V(A, B)$ 因为 $E_{A}$ 和 $E_{B}$ 与体积成正比 $A$ 和 $B$ ，然而 $V(A, B)$ 与它们之间的接触表面积成正比 (对于短程力)。对于宏 观系统， $|V(A, B)|$ 相比起来可以忽略不计 $E_{A}, E_{B}$. 因此，我们取
$$E=E_{A}+E_{B}$$

## 物理代写|统计力学代写Statistical mechanics代考|Partition function: Laplace transform of the density of states function

$$Z(\alpha) \equiv \int_{0}^{\infty} \mathrm{e}^{-\alpha x} \Omega(x) \mathrm{d} x$$

$$Z(\alpha)=\prod_{i=1}^{n} Z_{i}(\alpha)$$

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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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