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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|广义相对论代写General relativity代考|First-order displacement along a curve

Let us consider a manifold $\mathbf{M}$ with a coordinate system $\left(x^{1}, \ldots, x^{n}\right)$ and a path joining two nearby points $\mathbf{p}$ and $\mathbf{p}^{\prime}$. Let $\lambda$ be a parameter chosen on the path and $\left{x^{i}(\lambda)\right}$ and $\left{x^{i}(\lambda+\Delta \lambda)\right}$ the coordinates of the two points. The coordinate separation between $\mathbf{p}$ and $\mathbf{p}^{\prime}$ can be expanded in $\Delta \lambda$ as follows
$$x^{i}(\lambda+\Delta \lambda)-x^{i}(\lambda)=\frac{d x^{i}}{d \lambda} \Delta \lambda+\mathcal{O}\left(\Delta \lambda^{2}\right)=\delta x^{i}+\mathcal{O}\left(\Delta \lambda^{2}\right) \quad(i=1, \ldots, n),$$
where
$$\delta x^{i}=\frac{d x^{i}}{d \lambda} \Delta \lambda$$
are the components of a vector $\overrightarrow{\delta x}$, which is the first-order displacement between $\mathbf{p}$ and $\mathbf{p}^{\prime}$. Note that the vector $\overrightarrow{\delta x}$ does not depend on the parametrization of the path joining the two points. If the terms $\mathcal{O}\left(\Delta \lambda^{2}\right)$ can be neglected, $\overrightarrow{\delta x}$ is the infinitesimal displacement from $\mathbf{p}$ to $\mathbf{p}^{\prime}$. If we apply the basis one-forms $d x^{i}$ to $\overrightarrow{\delta x}$, we get
$$d x^{i}(\overrightarrow{\delta x})=\delta x^{i}$$
thus, the basis one-forms $d x^{i}$ can be considered as the components of the infinitesimal displacement along a generic direction.

## 物理代写|广义相对论代写General relativity代考|Vector fields and one-form fields

Vectors and one-forms are defined at a point $\mathbf{p}$ of the manifold, and belong to the vector spaces $\mathbf{T}{\mathbf{p}}$ and $\mathbf{T}{\mathbf{p}}^{*}$, respectively, which also are defined in $\mathbf{p}$; to make this explicit, we could denote a vector in $\mathbf{p}$ as $\vec{V}{\mathbf{p}}$, a one-form in $\mathbf{p}$ as $\tilde{q}{\mathbf{p}}$, but we shall often leave the point implicit in order to simplify the notation. We shall now define vector fields and one-form fields.
Given an open set $\mathbf{S}$ of a differentiable manifold $\mathbf{M}$, we define the vector spaces
\begin{aligned} &\mathbf{T}{\mathbf{S}} \equiv \bigcup{\mathbf{p} \in \mathbf{S}} \mathbf{T}{\mathbf{p}} \ &\mathbf{T}{\mathbf{S}}^{} \equiv \bigcup_{\mathbf{p} \in \mathbf{S}} \mathbf{T}{\mathbf{p}}^{} \end{aligned}
i.e., the union of the tangent spaces in the points $\mathbf{p} \in \mathbf{S}$, and the union of the cotangent spaces in the points $\mathbf{p} \in \mathbf{S}$. A vector field $\vec{V}$ is a mapping
\begin{aligned} \vec{V}: \mathbf{S} & \rightarrow \mathbf{T}{\mathbf{S}} \ \mathbf{p} & \mapsto \vec{V}{\mathbf{p}} \end{aligned} which associates to every point $\mathbf{p} \in \mathbf{S}$, a vector $\vec{V}{\mathbf{p}}$ defined on the tangent space in $\mathbf{p}, \mathbf{T} \mathbf{p}$. A one-form field $\tilde{q}$ is a mapping
\begin{aligned} \tilde{q}: \mathbf{S} & \rightarrow \mathbf{T}{\mathbf{S}}^{} \ \mathbf{p} & \mapsto \tilde{q}{\mathbf{p}} \end{aligned}
which associates to every point $\mathbf{p} \in \mathbf{S}$, a one-form $\tilde{q}{\mathbf{p}}$ defined on the cotangent space in $\mathbf{p}$, $\mathbf{T}{\mathbf{p}}^{}$. If a coordinate system (a chart) $\left{x^{i}\right}$ is defined on $\mathbf{S}$, we can indicate the vector field and the one-form field as $\vec{V}(x), \tilde{q}(x)$, respectively.

In the following, we will mainly consider vector fields and one-form fields; however, for brevity of notation, we will usually refer to them as vectors and one-forms.

# 广义相对论代考

## 物理代写|广义相对论代写General relativity代考|First-order displacement along a curve

$$x^{i}(\lambda+\Delta \lambda)-x^{i}(\lambda)=\frac{d x^{i}}{d \lambda} \Delta \lambda+\mathcal{O}\left(\Delta \lambda^{2}\right)=\delta x^{i}+\mathcal{O}\left(\Delta \lambda^{2}\right) \quad(i=1, \ldots, n),$$

$$\delta x^{i}=\frac{d x^{i}}{d \lambda} \Delta \lambda$$
$$d x^{i}(\overrightarrow{\delta x})=\delta x^{i}$$

## 物理代写|广义相对论代写General relativity代考|Vector fields and one-form fields

$$\mathbf{T S} \equiv \bigcup \mathbf{p} \in \mathbf{S T} \mathbf{p} \quad \mathbf{T S} \equiv \bigcup_{\mathbf{p} \in \mathbf{S}} \mathbf{T} \mathbf{p}$$

$$\vec{V}: \mathbf{S} \rightarrow \mathbf{T S} \mathbf{p} \quad \mapsto \vec{V} \mathbf{p}$$

$$\tilde{q}: \mathbf{S} \rightarrow \mathbf{T} \mathbf{p}_{\mathbf{p}} \quad \mapsto \tilde{q} \mathbf{p}$$

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assignmentutor™您的专属作业导师
assignmentutor™您的专属作业导师