### 物理代写|理论力学代写theoretical mechanics代考|PHYS2201

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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|理论力学代写theoretical mechanics代考|Curl (Vortex Density)

The curl (rotation) of the vector field $\mathbf{a}(\mathbf{r})$ is the vector field
$$\operatorname{rot} \mathbf{a} \equiv \nabla \times \mathbf{a} \equiv \lim {V \rightarrow 0} \frac{1}{V} \int{(V)} \mathbf{d} \mathbf{f} \times \mathbf{a}$$
For the above-mentioned cube with the edges $\mathrm{d} x, \mathrm{~d} y, \mathrm{~d} z$, the $x$-component of the right-hand expression is equal to
$$\frac{1}{\mathrm{~d} x \mathrm{~d} y \mathrm{~d} z}\left[+\mathrm{d} z \mathrm{~d} x\left{a_{z}(x, y+\mathrm{d} y, z)-a_{z}(x, y, z)\right}\right.$$
$$\left.-\mathrm{d} x \mathrm{~d} y\left{a_{y}(x, y, z+\mathrm{d} z)-a_{y}(x, y, z)\right}\right]=\frac{\partial a_{z}}{\partial y}-\frac{\partial a_{y}}{\partial z} .$$
With $\partial_{i} \equiv 1 / \partial x_{i}$, we thus have
which is the vector product of the operators $\nabla$ and a. This explains the notation $\boldsymbol{\nabla} \times \mathbf{a}$. Moreover, we have
$$\int_{V} \mathrm{~d} V \nabla \times \mathbf{a}=\int_{(V)} \mathrm{d} \mathbf{f} \times \mathbf{a}$$
for all continuous vector fields, although they may become singular point-wise, and even along lines, as will become apparent shortly.
An important result is Stokes’s theorem
$$\int_{A} \mathrm{df} \cdot(\nabla \times \mathbf{a})=\int_{(A)} \mathrm{d} \mathbf{r} \cdot \mathbf{a}$$
where $\mathrm{df}$ is taken in the rotational sense on the edge $(A)$ and forms a right=hand screw.

## 物理代写|理论力学代写theoretical mechanics代考|Delta Function

In the following, we shall often use the Dirac delta function. Therefore, its properties are compiled here, even though it does not actually belong to vector analysis, but to general analysis (and in particular to integral calculus).
$$\delta_{i k}= \begin{cases}0 & \text { for } i \neq k \ 1 & \text { for } i=k\end{cases}$$
It is useful for many purposes. In particular we may use it to filter out the $k$ th element of a sequence $\left{f_{i}\right}$ :
$$f_{k}=\sum_{i} f_{i} \delta_{i k} .$$
Here, of course, within the sum, one of the $i$ has to take the value $k$. Now, if we make the transition from the countable (discrete) variables $i$ to a continuous quantity $x$, then we must also generalize the Kronecker symbol. This yields Dirac’s delta function $\delta\left(x-x^{\prime}\right)$. It is defined by the equation
$$f\left(x^{\prime}\right)=\int_{a}^{b} f(x) \delta\left(x-x^{\prime}\right) \mathrm{d} x \quad \text { for } a<x^{\prime}<b, \text { zero otherwise },$$
where $f(x)$ is an arbitrary continuous test function. If the variable $x$ (and hence also $\mathrm{d} x)$ is a physical quantity with unit $[x]$, the delta function has the unit $[x]^{-1}$.

Obviously, the delta function $\delta\left(x-x^{\prime}\right)$ is not an ordinary function, because it has to vanish for $x \neq x^{\prime}$ and it has to be singular for $x=x^{\prime}$, so that the integral becomes $\int \delta\left(x-x^{\prime}\right) \mathrm{d} x=1$. Consequently, we have to extend the concept of a function: $\delta\left(x-x^{\prime}\right)$ is a distribution, or generalized function, which makes sense only as a weight factor in an integrand, while an ordinary function $y=f(x)$ is a map $x \rightarrow y$. Every equation in which the delta function appears without an integral symbol is an equation between integrands: on both sides of the equation, the integral symbol and the test function have been left out.
The delta function is the derivative of the Heaviside step function:
$$\varepsilon\left(x-x^{\prime}\right)=\left{\begin{array}{ll} 0 & \text { for } xx^{\prime} \end{array} \quad \Longrightarrow \quad \delta(x)=\varepsilon^{\prime}(x)\right.$$

## 物理代写|理论力学代写theoretical mechanics代考|Fourier Transform

If the region of definition is infinite on both sides, we use
$$f(x)=\int_{-\infty}^{\infty} g(k, x) f(k) \mathrm{d} k, \quad f(k)=\int_{-\infty}^{\infty} g^{}(k, x) f(x) \mathrm{d} x$$ with $g(k, x)=1 / \sqrt{2 \pi} \exp (\mathrm{i} k x)$ : \begin{aligned} &f(x)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \exp (+\mathrm{i} k x) f(k) \mathrm{d} k \ &f(k)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \exp (-\mathrm{i} k x) f(x) \mathrm{d} x \end{aligned} Generally, $f(x)$ and $f(k)$ are different functions of their arguments, but we would like to distinguish them only through their argument. [The less symmetric notation $f(x)=\int \exp (\mathrm{i} k x) F(k) \mathrm{d} k$ with $F(k)=f(k) / \sqrt{2 \pi}$ is often used. This avoids the square root factor with the agreement that $(2 \pi)^{-1}$ always appears with $\mathrm{d} x$.] Instead of the pair of variables $x \leftrightarrow k$, the pair $t \leftrightarrow \omega$ is also often used. Important properties of the Fourier transform are $f(x)=f^{}(x) \quad \Longleftrightarrow \quad f(k)=f^{*}(-k)$,
$f(x)=g(x) h(x) \Longleftrightarrow f(k)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} g\left(k-k^{\prime}\right) h\left(k^{\prime}\right) \mathrm{d} k^{\prime}$,
$f(x)=g\left(x-x^{\prime}\right) \quad \Longleftrightarrow \quad f(k)=\exp \left(-\mathrm{i} k x^{\prime}\right) g(k) .$
For a periodic function $f(x)=f(x-l)$ the last relation leads to the condition $k_{n}=$ $2 \pi n / l$ with $n \in{0, \pm 1, \pm 2, \ldots}$, thus to a Fourier series instead of the integral. In addition, by Fourier transform, all convolution integrals $\int g\left(x-x^{\prime}\right) h\left(x^{\prime}\right) \mathrm{d} x^{\prime}$ can clearly be turned into products $\sqrt{2 \pi} g(k) h(k)$ (Problem 3.9), which are much easier to handle.

## 物理代写|理论力学代写theoretical mechanics代考|Curl (Vortex Density)

$$\operatorname{rot} \mathbf{a} \equiv \nabla \times \mathbf{a} \equiv \lim V \rightarrow 0 \frac{1}{V} \int(V) \mathbf{d} \mathbf{f} \times \mathbf{a}$$

$\backslash$ frac ${1} \backslash \backslash$ mathrm ${\sim \mathrm{d}} \times \backslash m a t h r m{\sim \mathrm{d}}$ y $\backslash$ mathrm ${\sim \mathrm{d}}$ z $} \backslash$ left $\left[+\backslash m a t h r m{\mathrm{~d}} \mathrm{z} \backslash \mathrm{mathrm}{\sim \mathrm{d}} \mathrm{x} \backslash\right.$ left $\left{\mathrm{a}{-}{\mathrm{z}}(\mathrm{x}, \mathrm{y}+\backslash \mathrm{mathrm}{\mathrm{d}}\right.$ y, $\mathrm{z})-\mathrm{a}{-}{$
Veft.-\mathrm{d} x \mathrm{ ${\mathrm{d}} \mathrm{~ y ~ \ l e f t}$

$$\int_{V} \mathrm{~d} V \nabla \times \mathbf{a}=\int_{(V)} \mathrm{d} \mathbf{f} \times \mathbf{a}$$

$$\int_{A} \mathrm{df} \cdot(\nabla \times \mathbf{a})=\int_{(A)} \mathrm{d} \mathbf{r} \cdot \mathbf{a}$$

## 物理代写|理论力学代写theoretical mechanics代考|Delta Function

$$\delta_{i k}={0 \quad \text { for } i \neq k 1 \quad \text { for } i=k$$

$$f_{k}=\sum_{i} f_{i} \delta_{i k} .$$

$$f\left(x^{\prime}\right)=\int_{a}^{b} f(x) \delta\left(x-x^{\prime}\right) \mathrm{d} x \quad \text { for } a<x^{\prime}<b, \text { zero otherwise }$$

delta 函数是 Heaviside 阶跃函数的导数：
$\$ \$$Ivarepsilon \backslash left (x x \wedge{\backslash prime} \backslash right )=\backslash \operatorname{left}{ 0 for x x^{\prime} Iquad ILongrightarrow \quad Idelta (x)= Ivarepsilon^{1prime }(x) \backslash right. \ \$$

## 物理代写|理论力学代写theoretical mechanics代考|Fourier Transform

$$f(x)=\int_{-\infty}^{\infty} g(k, x) f(k) \mathrm{d} k, \quad f(k)=\int_{-\infty}^{\infty} g(k, x) f(x) \mathrm{d} x$$

$$f(x)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \exp (+\mathrm{i} k x) f(k) \mathrm{d} k \quad f(k)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \exp (-\mathrm{i} k x) f(x) \mathrm{d} x$$

$f(x)=f(x) \Longleftrightarrow f(k)=f^{*}(-k)$,
$f(x)=g(x) h(x) \Longleftrightarrow f(k)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} g\left(k-k^{\prime}\right) h\left(k^{\prime}\right) \mathrm{d} k^{\prime}$
$f(x)=g\left(x-x^{\prime}\right) \Longleftrightarrow f(k)=\exp \left(-\mathrm{i} k x^{\prime}\right) g(k)$

$\sqrt{2 \pi} g(k) h(k)$ (问题 3.9），这更容易处理。

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## MATLAB代写

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