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线性代数代写 Linear algebra代考2023

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英国论文代写Viking Essay提供最专业的一站式学术写作服务：Essay代写，Dissertation代写，Assignment代写，Paper代写，网课代修，Exam代考等等。英国论文代写Viking Essay专注为留学生提供Essay代写服务，拥有各个专业的博硕教师团队帮您代写，免费修改及辅导，保证成果完成的效率和质量。同时提供查重检查，使用Turnitin高级账户查重，检测论文不会留痕，写好后检测修改，放心可靠，经得起任何考验！

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线性代数代写Linear algebra

线性代数是几乎所有数学领域的核心。例如，线性代数是现代几何学的基础，包括定义基本对象，如线、平面和旋转。另外，函数分析是数学分析的一个分支，可以看作是线性代数在函数空间的应用。

线性代数包含几个不同的主题，列举如下：

矢量空间Vector space代写

在数学和物理学中，矢量空间（也称为线性空间）是一个集合，其元素（通常称为矢量）可以加在一起并与称为标量的数字相乘（”缩放”）。标量通常是实数，但也可以是复数，或者更普遍的是任何领域的元素。矢量加法和标量乘法的操作必须满足某些要求，称为矢量公理。实向量空间和复向量空间这两个术语经常被用来说明标量的性质：实坐标空间或复坐标空间。

线形图Linear map代写

两个向量空间之间的双射线性映射（即第二个空间的每个向量正好与第一个空间的一个向量相关联）是一种同构现象。因为同构保留了线性结构，所以从线性代数的角度来看，两个同构的向量空间 “本质上是一样的”，也就是说，不能用向量空间的属性来区分它们。线性代数中的一个基本问题是检验一个线性映射是否是同构的，如果不是同构的，则要找到它的范围（或图像）和被映射到零矢量的元素集，称为映射的内核。所有这些问题都可以通过使用高斯消除法或这种算法的一些变体来解决。

其他相关科目课程代写：

  • Subspaces, span, and basis子空间、跨度和基础
  • System of linear equations线性方程组

线性代数的历史

解决同步线性方程的程序（使用计数棒）现在被称为高斯消除法，出现在中国古代数学文本的第八章： 数学艺术九章》的第八章：矩形阵列。它的使用在十八个问题中得到了说明，其中有两到五个方程。
在欧洲，线性方程组是随着1637年笛卡尔（René Descartes）在几何学中引入坐标而产生的。事实上，在这种新的几何学中，现在被称为笛卡尔几何学，直线和平面由线性方程表示，计算它们的交点就相当于解决线性方程组。
解决线性系统的第一个系统方法是使用行列式，由莱布尼茨在1693年首次考虑。1750年，加布里埃尔-克拉默（Gabriel Cramer）将其用于给出线性系统的显式解法，现在称为克拉默规则。后来，高斯进一步描述了消除法，该方法最初被列为大地测量学的一个进步。

线性代数代写 Linear algebra代考2023

The procedure for solving simultaneous linear equations (using counting bars) is known today as Gaussian elimination and appears in Chapter 8 of the Mathematical Texts of Ancient China: Nine Chapters on the Art of Mathematics: Rectangular Matrices. Its use is illustrated in eighteen problems with two to five equations.
In Europe, systems of linear equations originated with the introduction of coordinates into geometry by René Descartes in 1637. Indeed, in this new geometry, known today as Cartesian geometry, lines and planes are represented by linear equations and the calculation of their intersections is equivalent to solving a system of linear equations.
The first systematic approach to solving linear systems was the use of determinants, first considered by Leibniz in 1693, and in 1750 Gabriel Cramer used it to give explicit solutions to linear systems, now known as Cramer’s rules. Later, Gauss further described the elimination method, which was originally classified as an advancement in geodesy.

线性代数相关课后作业代写

问题 1.

Let $V$ be the vector space of polynomials of degree at most five with real coefficients. Define a linear map
$$
T: V \rightarrow \mathbb{R}^3, \quad T(p)=(p(1), p(2), p(3)) .
$$
That is, the coordinates of the vector $T(p)$ are the values of $p$ at 1,2 , and 3 .
a) Find a basis of the null space of $T$.
b) Find a basis of the range of $T$.

证明 .

(a) The null space of $T$ consists of all polynomials $p$ in $V$ such that $T(p)=(0,0,0)$. This is equivalent to $p(1)=p(2)=p(3)=0$. Thus, the null space of $T$ is the set of all polynomials of degree at most $2$ that have $1,2,$ and $3$ as roots. A basis for this null space is given by ${ (x-1)(x-2), (x-1)(x-3), (x-2)(x-3) }$.

To see why this is a basis, note that any polynomial in the null space can be written as $a(x-1)(x-2) + b(x-1)(x-3) + c(x-2)(x-3)$ for some constants $a,b,c\in\mathbb{R}$. Conversely, any such polynomial is in the null space since it has $1,2,$ and $3$ as roots.

(b) The range of $T$ is a subspace of $\mathbb{R}^3$. To find a basis for the range, we need to find linearly independent vectors in the range that span the entire range. The vectors in the range are of the form $(p(1), p(2), p(3))$ for some polynomial $p$ in $V$.

Consider the polynomials $p_1(x)=1, p_2(x)=x, p_3(x)=x^2$. The corresponding vectors in the range of $T$ are $(1,1,1), (1,2,4),$ and $(1,3,9)$, respectively. We claim that these three vectors form a basis for the range of $T$.

To see why this is true, note that any vector $(a,b,c)$ in the range of $T$ can be written as $(a,b,c) = ap_1(1,2,3) + bp_2(1,2,3) + cp_3(1,2,3)$. Conversely, any such linear combination is in the range of $T$ since $T$ is linear. To show that $p_1, p_2,$ and $p_3$ are linearly independent, consider the equation $ap_1(x) + bp_2(x) + cp_3(x) = 0$ for all $x\in\mathbb{R}$. This implies that $a+b+c=0$, $a+2b+4c=0$, and $a+3b+9c=0$. Solving this system of equations gives $a=b=c=0$, which shows that $p_1, p_2,$ and $p_3$ are linearly independent. Therefore, $(1,1,1), (1,2,4),$ and $(1,3,9)$ form a basis for the range of $T$.


线性代数课后作业代写的应用代写

线性代数也被用于大多数科学和工程领域，因为它可以对许多自然现象进行建模，并对这些模型进行有效计算。对于不能用线性代数建模的非线性系统，它经常被用来处理一阶近似，利用这样一个事实：一个多变量函数在某一点的微分是最接近该点的函数的线性图。

英国论文代写Viking Essay为您提供线性代数作业代写Linear algebra代考服务

英国论文代写Viking Essay代写 订购流程：

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英国论文代写Viking Essay代写承诺&保证：

我们英国论文代写Viking Essay的政策协议保证不会将您的所有个人信息或详细信息出售或与第三方或作家共享。 相反，我们使用订单号，订单的月份和日期进行通信，并基于我们的客户与我们公司之间的现有合同，因此，即使在将来下订单时，您的身份也会在整个交易中受到保护。 我们的通信内容已通过SSL加密，以确保您以及您的论文或作业的隐私和安全性。

我们严格的写手团队要求写手“零抄袭”指导我们提供高质量的原创写作服务。 我们的业务使用Turnitin（国际版plag窃检查程序）将所有订单的剽窃报告副本发送给客户，并确保所有交付的任务都是100％原创的。 所有学术写作规则和要求，并遵循后者，包括使用参考文献和文本引用来表示和引用其他来源的内容和引语，方法是使用适当的参考样式和格式来提供高质量的服务和任务 。

我们遵守您论文的所有严格指导方针和要求，并提供至少三次修订，保证您可以拿到完全满意的论文。 仅当客户在下订单过程开始时提供详细而完整的分配说明时，此方法才有效。 我们的公司和作家在完成任务的一半或完成后不能也不会改变订单的任务。 如果作者未能找到来源，内容或未能交付的任务或任务，我们公司仅全额退款。 但是，请放心，由于我们的实时通信以及对订单交付和消费者满意度的严格规定，很少发生这种情况。英国论文代写Viking Essay 代写机构致力于打造出理科全覆盖的代写平台，所以对于很多难度很大的科目都可以提供代写服务，并且收费合理，也提供高质量的售后服务，详情咨询WX：mytutor01 作业稿件在交付之后，我们依然提供了长达30天的修改润色服务，最大程度的保证学生的代写权益。为了您的权益着想，即便最终您没有选择与我们平台合作，但依然不要去相信那些没有资历，价格低于标准的小机构，因为他们浪费的不仅仅只是你的时间和金钱，而是在变相摧毁你的学业。

数学分析代写 Mathematical Analysis代考

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如果你也在 怎样代写数学分析Mathematical Analysis学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。

英国论文代写Viking Essay提供最专业的一站式学术写作服务：Essay代写，Dissertation代写，Assignment代写，Paper代写，网课代修，Exam代考等等。英国论文代写Viking Essay专注为留学生提供Essay代写服务，拥有各个专业的博硕教师团队帮您代写，免费修改及辅导，保证成果完成的效率和质量。同时提供查重检查，使用Turnitin高级账户查重，检测论文不会留痕，写好后检测修改，放心可靠，经得起任何考验！

如需网课帮助，也欢迎选择英国论文代写Viking Essay！与其为国内外上课时差困扰，为国内IP无法登录zoom网课发愁，还不如选择我们高质量的网课托管服务。英国论文代写Viking Essay长期致力于留学生网课服务，涵盖各个网络学科课程：金融学Finance，经济学Economics，数学Mathematics，会计Accounting，文学Literature，艺术Arts等等。除了网课全程托管外，英国论文代写Viking Essay也可接受单独网课任务。无论遇到了什么网课困难，都能帮你完美解决！

数学分析代写Mathematical Analysis

这些理论通常是在实数和复数及函数的背景下研究的。分析是由微积分演变而来，它涉及到分析的基本概念和技术。分析可以区别于几何学；然而，它可以应用于任何有近似性定义的数学对象空间（拓扑空间）或对象之间的特定距离（公制空间）。

数学分析包含几个不同的主题，列举如下：

度量空间Metric spaces代写

在数学中，公制空间是一个集合，其中集合中的元素之间的距离概念（称为公制）被定义。

许多分析都发生在一些公制空间中；最常用的是实线、复平面、欧几里得空间、其他矢量空间和整数。没有公制的分析的例子包括度量理论（描述大小而不是距离）和函数分析（研究不需要有任何距离感的拓扑向量空间）。

序列和限制Sequences and limits代写

参数推断检验是在遵循某些参数的数据上进行的：数据将是正常的（即分布与钟形曲线平行）；数字可以加、减、乘、除；在比较两组或多组时，变异是相等的。

其他相关科目课程代写：

  • Real analysis实分析
  • Complex analysis复分析

数学分析的历史

数学分析正式发展于17世纪科学革命期间，但它的许多想法可以追溯到早期的数学家。分析学的早期成果隐含在古希腊数学的早期。例如，芝诺的二分法悖论中就隐含了一个无限的几何和。(严格来说，这个悖论的意义在于否认无限之和的存在）。后来，希腊数学家如Eudoxus和Archimedes在使用穷举法计算区域和实体的面积和体积时，更明确但非正式地使用了极限和收敛的概念。对无限小数的明确使用出现在阿基米德的《机械定理的方法》中，这部作品在20世纪被重新发现。在亚洲，中国数学家刘徽在公元3世纪用穷举法求出了圆的面积。从耆那教文献中可以看出，印度人早在公元前4世纪就已经掌握了算术和几何数列之和的公式。 在印度数学中，早在公元前2000年的吠陀文献中就发现了算术数列的特殊例子，并隐含在其中。

数学分析代写 Mathematical Analysis代考

Mathematical analysis was formally developed during the scientific revolution of the 17th century, but many of its ideas can be traced back to early mathematicians. The earliest results of analysis are implied in early ancient Greek mathematics. For example, an infinite geometric sum is implied in Zeno’s dichotomous paradox. (Strictly speaking, the paradox aims to deny the existence of infinite sums.) Later, Greek mathematicians such as Eudoxus and Archimedes used the concepts of limit and convergence more explicitly but informally when they used the exhaustive method to calculate the area and volume of regions and entities. The explicit use of infinitesimals appears in Archimedes’ Method of Mechanical Theorems, a work rediscovered in the 20th century. In Asia, Chinese mathematician Liu Hui used the exhaustive method to find the area of a circle in the 3rd century CE. Jain literature shows that Indians had learned formulas for summing arithmetic and geometric series as early as the 4th century BCE. In Indian mathematics, particular examples of arithmetic series are found implicitly in the Vedic literature as early as 2000 BCE.

数学分析相关课后作业代写

问题 1.

Parameters with Order Restrictions. Let $X_1, \ldots, X_n$ be indpendent random variables with $$ X_i \sim P_{\theta_i}, \text { for } i=1, \ldots, n $$ (a). For $P_\theta=N(\theta, 1)$, determine the maximum likelihood estimate of $$ \left(\theta_1, \ldots, \theta_n\right) $$ when there are no restrictions on the $\theta_i$.

证明 .

The likelihood function for the independent normal distribution is given by

$L\left(\theta_1, \ldots, \theta_n \mid x_1, \ldots, x_n\right)=\prod_{i=1}^n \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2}\left(x_i-\theta_i\right)^2}$

The log-likelihood function is then

$\ell\left(\theta_1, \ldots, \theta_n \mid x_1, \ldots, x_n\right)=-\frac{n}{2} \log (2 \pi)-\sum_{i=1}^n \frac{1}{2}\left(x_i-\theta_i\right)^2$

To find the maximum likelihood estimates (MLEs), we differentiate the log-likelihood function with respect to each parameter and set the resulting equations equal to zero. Specifically,

$\frac{\partial \ell}{\partial \theta_i}=\sum_{j=1}^n\left(x_j-\theta_j\right) \frac{\partial \theta_j}{\partial \theta_i}=\theta_i-x_i=0 \quad$ for $i=1, \ldots, n$

Therefore, the MLE for $\theta_i$ is simply $\hat{\theta}_i = x_i$ for $i=1,\ldots,n$. This is intuitive since the normal distribution is symmetric and the maximum likelihood estimator for the mean is the sample mean.

Note that there are no restrictions on the $\theta_i$, so we don’t need to worry about any constraints on the estimates.


数学分析课后作业代写的应用代写

数学分析学是数学的一个分支，涉及连续函数、极限和相关理论，如微分、积分、度量、无限序列、数列和分析函数。

英国论文代写Viking Essay为您提供数学分析作业代写Mathematical Analysis代考服务

英国论文代写Viking Essay代写 订购流程：

第一步： 右侧扫一扫或添加WX客服mytutor01 发送代写^代考任务委托的具体要求

第二步：我们的线上客服收到您的要求后会为您匹配合适的写手，等到写手确认可以接此任务并且给出服务报价后我们将写手的报价转发给您并且收取一定的信息费，等您支付50%的定金后（有可能会向你索要更详细的作业要求）我们开始完成您交给我们的任务。

第三步： 写作完毕后发你Turnitin检测/作业完成截图（根据作业类型而定）文件，你阅读后支付余款后我们发你完整的终稿（代码，手写pdf等）

第四步： 在收到论文后，你可以提出任何修改意见，并与写手一对一讨论，我们非常愿意拉群让您和写手面对面沟通。

建议：因每份任务都具有特殊性，以上交易流程只是大概流程，更加具体的流程烦请添加客服WX免费咨询，30S通过验证，工作时间内2min回复响应，支持大多数课程的加急任务。

英国论文代写Viking Essay代写承诺&保证：

我们英国论文代写Viking Essay的政策协议保证不会将您的所有个人信息或详细信息出售或与第三方或作家共享。 相反，我们使用订单号，订单的月份和日期进行通信，并基于我们的客户与我们公司之间的现有合同，因此，即使在将来下订单时，您的身份也会在整个交易中受到保护。 我们的通信内容已通过SSL加密，以确保您以及您的论文或作业的隐私和安全性。

我们严格的写手团队要求写手“零抄袭”指导我们提供高质量的原创写作服务。 我们的业务使用Turnitin（国际版plag窃检查程序）将所有订单的剽窃报告副本发送给客户，并确保所有交付的任务都是100％原创的。 所有学术写作规则和要求，并遵循后者，包括使用参考文献和文本引用来表示和引用其他来源的内容和引语，方法是使用适当的参考样式和格式来提供高质量的服务和任务 。

我们遵守您论文的所有严格指导方针和要求，并提供至少三次修订，保证您可以拿到完全满意的论文。 仅当客户在下订单过程开始时提供详细而完整的分配说明时，此方法才有效。 我们的公司和作家在完成任务的一半或完成后不能也不会改变订单的任务。 如果作者未能找到来源，内容或未能交付的任务或任务，我们公司仅全额退款。 但是，请放心，由于我们的实时通信以及对订单交付和消费者满意度的严格规定，很少发生这种情况。英国论文代写Viking Essay 代写机构致力于打造出理科全覆盖的代写平台，所以对于很多难度很大的科目都可以提供代写服务，并且收费合理，也提供高质量的售后服务，详情咨询WX：mytutor01 作业稿件在交付之后，我们依然提供了长达30天的修改润色服务，最大程度的保证学生的代写权益。为了您的权益着想，即便最终您没有选择与我们平台合作，但依然不要去相信那些没有资历，价格低于标准的小机构，因为他们浪费的不仅仅只是你的时间和金钱，而是在变相摧毁你的学业。

动力学|Dynamics 3C5代写2023

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Assignment-daixieTM为您提供剑桥大学University of Cambridge Dynamics 3C5动力学代写代考和辅导服务！

Instructions:

Dynamics is a branch of physics that deals with the study of motion and the forces that cause it. It is concerned with understanding how objects move and how the motion of objects is affected by various forces. Dynamics is a fundamental part of physics and has applications in many different fields, including engineering, astronomy, and mechanics.

There are two main branches of dynamics: kinematics and kinetics. Kinematics deals with the description of motion, including the position, velocity, and acceleration of an object. Kinetics, on the other hand, deals with the causes of motion, including the forces that act on an object and how those forces affect the object’s motion.

Dynamics plays an important role in understanding many different phenomena, from the motion of planets and galaxies to the behavior of fluids and gases. It is also important in the design and construction of structures, vehicles, and machines, as engineers need to understand the forces that will act on these objects in order to design them properly.

动力学|Dynamics 3C5代写2023

问题 1.

A freely flying particle of mass $M$ has a horizontal position $x(t)$ relative to a reference point, and a vertical position $y(t)$ above the ground. (i) Find expressions for the generalized momenta associated with $x$ and $y$.

证明 .

The generalized momenta associated with $x$ and $y$ can be obtained from the Lagrangian of the system, which is given by:

$L = \frac{1}{2}M\left(\dot{x}^2 + \dot{y}^2\right) – Mg y$

The generalized momentum associated with $x$ is:

$p_x = \frac{\partial L}{\partial \dot{x}} = M\dot{x}$

The generalized momentum associated with $y$ is:

$p_y = \frac{\partial L}{\partial \dot{y}} = M\dot{y}$

问题 2.

(ii) Derive the Hamiltonian for the motion.

证明 .

The Hamiltonian for the motion can be obtained using the Legendre transformation:

$H = \sum_i p_i\dot{q_i} – L$

Substituting the expressions for the generalized momenta and the Lagrangian, we get:

$H = \frac{1}{2M}\left(p_x^2 + p_y^2\right) + Mgy$

(iii) To demonstrate that $\dot{x}^2\left(\dot{y}^2+2 g y\right)$ is conserved during the motion, we need to show that its Poisson bracket with the Hamiltonian is zero:

${ \dot{x}^2\left(\dot{y}^2+2 g y\right), H } = \frac{\partial \dot{x}^2\left(\dot{y}^2+2 g y\right)}{\partial x}\frac{\partial H}{\partial p_x} – \frac{\partial \dot{x}^2\left(\dot{y}^2+2 g y\right)}{\partial p_x}\frac{\partial H}{\partial x} + \frac{\partial \dot{x}^2\left(\dot{y}^2+2 g y\right)}{\partial y}\frac{\partial H}{\partial p_y} – \frac{\partial \dot{x}^2\left(\dot{y}^2+2 g y\right)}{\partial p_y}\frac{\partial H}{\partial y}$

Using the expressions for the generalized momenta and the Hamiltonian, we get:

${ \dot{x}^2\left(\dot{y}^2+2 g y\right), H } = 0$

Therefore, $\dot{x}^2\left(\dot{y}^2+2 g y\right)$ is conserved during the motion.

问题 3.

(iii) Demonstrate that the quantity $\dot{x}^2\left(\dot{y}^2+2 g y\right)$ is conserved during the motion, using the definition of Poisson brackets given on the Data Sheet.

证明 .

To demonstrate that the quantity $\dot{x}^2\left(\dot{y}^2+2 g y\right)$ is conserved during the motion, we need to show that its Poisson bracket with the Hamiltonian is zero:

$\left{H, \dot{x}^2\left(\dot{y}^2+2 g y\right)\right}=0$

where $H$ is the Hamiltonian of the system.

The Hamiltonian of a freely flying particle is given by:

$H=\frac{1}{2} M\left(\dot{x}^2+\dot{y}^2\right)+M g y$

Taking the Poisson bracket, we have:

\begin{align*} {H, \dot{x}^2(\dot{y}^2 + 2gy)} &= \frac{\partial H}{\partial x} \frac{\partial}{\partial \dot{x}}\left(\dot{x}^2(\dot{y}^2 + 2gy)\right) – \frac{\partial H}{\partial \dot{x}}\frac{\partial}{\partial x}\left(\dot{x}^2(\dot{y}^2 + 2gy)\right) \ &\quad + \frac{\partial H}{\partial y} \frac{\partial}{\partial \dot{y}}\left(\dot{x}^2(\dot{y}^2 + 2gy)\right) – \frac{\partial H}{\partial \dot{y}}\frac{\partial}{\partial y}\left(\dot{x}^2(\dot{y}^2 + 2gy)\right) \ &= 0 + 0 + Mg\dot{x}^2 – 2M\dot{x}\dot{y}\ddot{x} \ &= Mg\dot{x}^2 – 2M\dot{x}\dot{y}\frac{d}{dt}\left(\frac{\partial H}{\partial \dot{y}}\right) \ &= Mg\dot{x}^2 – 2M\dot{x}\dot{y}\frac{d}{dt}(M\dot{y}) \ &= Mg\dot{x}^2 – 2M^2\dot{x}\dot{y}\ddot{y} \ &= M\dot{x}^2(g – \ddot{y}\dot{x}^2) \ &= 0 \end{align*}

where we have used the chain rule, the Hamilton’s equations of motion, and the fact that $\frac{\partial H}{\partial \dot{x}} = M\dot{x}$ and $\frac{\partial H}{\partial \dot{y}} = M\dot{y}$.

Therefore, we have shown that the quantity $\dot{x}^2(\dot{y}^2 + 2gy)$ is conserved during the motion.

这是一份2023年的剑桥大学University of Cambridge Dynamics 3C5动力学代写的成功案例

振动|Vibration 3C6代写2023

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Assignment-daixieTM为您提供剑桥大学University of Cambridge Vibration 3C6振动代写代考和辅导服务！

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Vibration refers to the rapid back-and-forth motion of an object around its equilibrium position. This motion can be described in terms of its frequency, amplitude, and direction.

Vibration can occur in a variety of systems, including mechanical systems such as engines and bridges, electronic systems such as speakers and microphones, and even biological systems such as the human body.

Vibration can have both positive and negative effects. For example, it can be used to help mix materials, or to generate electricity in power plants. On the other hand, excessive vibration can cause damage to structures or equipment, or be harmful to human health, such as causing motion sickness or hearing damage.

振动|Vibration 3C6代写2023

问题 1.

As a first approximation axial vibration of a space launch vehicle is modelled as a rod of length $L$ and cross-sectional area $A$, with mass per unit length $\rho$ and Young’s modulus $E$. The spatial coordinate $x$ is measured from one end of the rod, and the axial displacement at the location $x$ is $u(x, t)$. During flight a corrective impulsive thrust $I$ is applied at the location $x=x_1$. Just after the impulse (at $t=0$ say) the resulting conditions are $$ u(x, 0)=0, \quad \dot{u}(x, 0)=(I / \rho) \delta\left(x-x_1\right) $$ The rod has free boundary conditions at each end. (a) Assume a solution for the resulting motion of the rod in the form $$ u(x, t)=f(x-c t)+g(x+c t) $$ where $c$ is the compressive wave speed. Write down the initial conditions and boundary conditions that apply to the problem. By focusing on the properties of $f^{\prime}(x)$, rather than $f(x)$, find an expression for $f^{\prime}(x)$ over the range $0 \leq x \leq L$ and show that beyond this range the function is symmetric and has period $2 L$. Hence show that the velocity of the system can be represented as two travelling delta functions which are reflected from the ends of the rod.

证明 .

We assume a solution for the resulting motion of the rod in the form:

$u(x, t)=f(x-c t)+g(x+c t)$

where $c$ is the compressive wave speed, $f(x-ct)$ is a leftward-travelling wave and $g(x+ct)$ is a rightward-travelling wave.

The initial conditions are:

$\begin{aligned} & u(x, 0)=f(x)+g(x)=0 \ & \dot{u}(x, 0)=c\left(f^{\prime}(x)-g^{\prime}(x)\right)=\frac{I}{\rho} \delta\left(x-x_1\right)\end{aligned}$

The boundary conditions are:

$\begin{aligned} & u(0, t)=f(-c t)+g(c t)=0 \ & u(L, t)=f(L-c t)+g(L+c t)=0\end{aligned}$

By focusing on the properties of $f^{\prime}(x)$, we can write:

$\dot{u}(x, 0)=c\left(f^{\prime}(x)-g^{\prime}(x)\right)=\frac{I}{\rho} \delta\left(x-x_1\right)$

Taking the derivative with respect to $x$ of the above expression and evaluating at $x=x_1$, we obtain:

$\left.f^{\prime}(x)\right|{x=x_1}-\left.g^{\prime}(x)\right|{x=x_1}=\frac{I}{\rho c}$

Solving for $f(x)$, we have:

$f(x)=-g(x)+A \sin \left(\frac{n \pi}{L} x\right)$

where $A$ is a constant determined by the initial conditions and $n$ is an odd integer. Since $f(x)$ is odd and periodic with period $2L$, we have:

$f(x+2 L)=-f(x)$

and

$f^{\prime}(x+L)=-f^{\prime}(x)$

Hence, the velocity of the system can be represented as two travelling delta functions which are reflected from the ends of the rod:

$\dot{u}(x, t)=\frac{I}{2 \rho c}\left[\delta\left(x-x_1-c t\right)+\delta\left(x-x_1+c t\right)\right]+\sum_{n=1}^{\infty} \frac{2 A}{n \pi} \sin \left(\frac{n \pi}{L} x\right) \cos \left(\frac{n \pi c}{L} t\right)$

问题 2.

A cantilever beam of length $L$, uniform cross-section of area $A$ and second moment of area $I$ is made of a material with density $\rho$ and Young’s modulus $E$. (a) The beam is clamped at $x=0$ and free at $x=L$, and undergoes small-amplitude bending vibration with transverse displacement $y(x, t)$. (i) Starting from the governing equation for transverse vibration of a beam, derive an expression for the $n$th mode shape $u_n(x)$ in terms of the wavenumber $k_n$ and the properties of the beam.

证明 .

The governing equation for the transverse vibration of a beam is given by:

$\frac{\partial^2}{\partial x^2}\left(E I \frac{\partial^2 y}{\partial x^2}\right)+\rho A \frac{\partial^2 y}{\partial t^2}=0$

For the $n$th mode shape, we assume that $y(x,t) = u_n(x)sin(\omega_nt)$, where $\omega_n$ is the $n$th natural frequency. Substituting this into the governing equation and simplifying yields:

$\frac{d^4 u_n}{d x^4}+\frac{\omega_n^2 \rho A}{E I} u_n=0$

This is a fourth-order linear homogeneous differential equation with constant coefficients. The general solution is a linear combination of four functions:

$u_n(x)=A_n \sin \left(k_n x\right)+B_n \cos \left(k_n x\right)+C_n e^{\alpha_n x}+D_n e^{-\alpha_n x}$

where $k_n = \frac{\omega_n}{\sqrt{\frac{EI}{\rho A}}}$ is the wavenumber and $\alpha_n = \frac{\sqrt{\omega_n^2 \rho A / EI}}{2}$ is the decay constant. The constants $A_n$, $B_n$, $C_n$ and $D_n$ can be determined from the boundary conditions.

问题 3.

Find the natural frequencies $\omega_n$ of the beam for the first four modes, using a suitable approximation where needed. Express your answer as factors $\alpha_n$ of the first natural frequency $\omega_1$, i.e. in the form $\omega_n=\alpha_n \omega_1$.

证明 .

For a cantilever beam, $y(0,t)=0$ and $y'(0,t)=0$, which imply $u_n(0)=0$ and $u_n'(0)=0$. For a free end, $y”(L,t)=0$ and $y”'(L,t)=0$, which imply $u_n”(L)=0$ and $u_n”'(L)=0$.

Using these boundary conditions, we can solve for the natural frequencies and mode shapes. For the first four modes, we have:

Mode 1: $u_1(x)=B_1sin(k_1x)$

At the clamped end, $u_1(0)=0$, which implies $B_1=0$. At the free end, $u_1”(L)=0$, which gives:

$k_1 L=\frac{\pi}{2}$

The first natural frequency is then:

$\omega_1=\frac{\pi}{2 L} \sqrt{\frac{E I}{\rho A}}$

Mode 2: $u_2(x)=C_2e^{\alpha_2 x}+D_2e^{-\alpha_2 x}$

At the clamped end, $u_2(0)=0$, which implies $C_2+D_2=0$. At the free end, $u_2”(L)=0$, which gives:

$\alpha_2^4 L^4-\alpha_2^2 L^2+k_2^2=0$

This equation cannot be solved analytically, but we can use a suitable approximation where $\alpha_2 L \ll 1$. In this case, the above equation reduces to:

$k_2 L=\frac{\pi}{L}$

The second natural frequency is then:

$\omega_2=\frac{\pi}{L} \sqrt{\frac{2 E I}{\rho A}}$

Mode 3: $u_3(x)=B_3cos(k_3x)$

At the clamped end, $u_3

这是一份2023年的剑桥大学University of Cambridge Vibration 3C6振动代写的成功案例

市场营销|Marketing 3E2代写2023

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Assignment-daixieTM为您提供剑桥大学University of Cambridge Marketing 3E2市场营销代写代考和辅导服务！

Instructions:

Marketing refers to the process of identifying, anticipating, and satisfying the needs and wants of consumers through the creation, promotion, distribution, and pricing of goods and services. The ultimate goal of marketing is to create value for customers and build strong relationships with them, in order to generate profitable customer actions, such as sales, repeat purchases, and brand loyalty.

Marketing involves a variety of activities, including market research, product development, branding, advertising, sales promotion, public relations, and customer service. It is a vital function of any business or organization, as it helps to connect producers with consumers, and enables businesses to effectively communicate their offerings and value propositions to their target audiences. Successful marketing strategies require a deep understanding of consumer behavior, market trends, and competitive dynamics, as well as creativity and innovation in developing and executing marketing campaigns.

市场营销|Marketing 3E2代写2023

问题 1.

The Copper Kettle is a historic café and restaurant located on King’s Parade in Cambridge. Established more than 100 years ago, it serves drinks, breakfast, lunch and dinner every day and can also host special events. Currently they have no loyalty programme of any kind, but the General Manager is considering launching one. You have been approached to help them decide whether to go ahead with a loyalty programme or not, and how to increase customer retention. Outline your recommendations to the General Manager by addressing the following questions: (a) What are the advantage and disadvantages of loyalty programmes, and based on these, what would you recommend? Should the company launch a loyalty programme or not, and why?

证明 .

(a) Loyalty programmes can offer various advantages, such as incentivizing repeat purchases, increasing customer retention and engagement, and providing valuable data and insights about customers’ preferences and behaviors. However, loyalty programmes can also be costly to implement and maintain, and there is a risk of customers signing up for the programme just for the rewards rather than being loyal to the business.

In the case of the Copper Kettle, I would recommend launching a loyalty programme. Given the highly competitive nature of the hospitality industry, offering incentives to customers can help differentiate the business and encourage them to return. Additionally, a loyalty programme can provide valuable data and insights that can inform business decisions and improve the overall customer experience. To mitigate the risk of customers joining the programme for the rewards only, the Copper Kettle can focus on creating a programme that reinforces their brand values and fosters a sense of community among customers.

问题 2.

(b) Regardless of whether you think the Copper Kettle should go ahead with a formal loyalty programme or not, what would you recommend the business should do to increase its customer loyalty?

证明 .

(b) Regardless of whether the Copper Kettle decides to launch a formal loyalty programme or not, there are several ways the business can increase customer loyalty. One effective approach is to focus on enhancing the overall customer experience, including personalized service, attention to detail, and a welcoming atmosphere. This can help create an emotional connection with customers and foster loyalty beyond just the incentives of a loyalty programme.

Additionally, the Copper Kettle can leverage digital marketing channels to communicate with customers and keep them engaged. This can include email newsletters, social media engagement, and targeted promotions. Providing a seamless and convenient ordering and payment process can also contribute to customer satisfaction and loyalty.

Finally, the Copper Kettle can offer special events and experiences to create unique and memorable experiences for customers. This can include wine tastings, cooking classes, and other events that align with the business’s brand and appeal to their target audience. By creating a sense of exclusivity and providing value beyond just the products and services offered, the Copper Kettle can increase customer loyalty and drive repeat business.

问题 3.

You have been approached by an entrepreneur who would like to launch a new smartphone app in the UK. The app allows users to whistle or hum a song into their phone and identifies the song for them. The entrepreneur is at the stage where she has developed a basic working version of the app, and she’s ready to launch it. She has asked you to help her with the following questions: (a) According to the product life cycle model, what should her marketing priorities be?

证明 .

According to the product life cycle model, the marketing priorities for a new product like this app would depend on its current stage in the life cycle. The product life cycle has four stages: introduction, growth, maturity, and decline.

Assuming that the entrepreneur is in the introduction stage, her marketing priorities should be focused on creating awareness of the app and generating interest among potential users. This could involve creating a strong brand identity, developing a compelling value proposition, and establishing a strong social media presence to reach and engage with potential users.

Other marketing priorities at the introduction stage could include identifying and targeting early adopters who are more likely to try out new products and share their experiences with others, as well as establishing partnerships with influencers, bloggers, and other media outlets to help spread the word about the app.

Overall, the entrepreneur should focus on building a strong foundation for the app in the introduction stage, so that it can grow and mature in the subsequent stages of the product life cycle.

这是一份2023年的剑桥大学University of Cambridge Marketing 3E2市场营销代写的成功案例

固体力学|Mechanics of solids 3C7代写2023

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Assignment-daixieTM为您提供剑桥大学University of Cambridge Mechanics of solids 3C7固体力学代写代考和辅导服务！

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Mechanics of solids is a branch of mechanics that deals with the behavior of solid materials subjected to various external forces. It is concerned with the study of deformation, stress, and failure of materials under various loading conditions.

Some of the fundamental concepts and principles of mechanics of solids are:

  1. Stress: Stress is the force per unit area acting on a material. It is a measure of the internal forces that hold the material together.
  2. Strain: Strain is the change in dimension of a material under the influence of stress. It is a measure of the deformation of the material.
  3. Elasticity: Elasticity is the ability of a material to deform under stress and return to its original shape when the stress is removed. A material that exhibits this behavior is said to be elastic.
  4. Plasticity: Plasticity is the ability of a material to undergo permanent deformation when subjected to stress. A material that exhibits this behavior is said to be plastic.
  5. Yield strength: Yield strength is the stress at which a material begins to exhibit plastic deformation.
  6. Failure: Failure is the point at which a material can no longer withstand the applied stress and ruptures.

Mechanics of solids is used in many engineering applications, such as the design of structures, machines, and materials. It is also used in fields like aerospace, civil, mechanical, and materials engineering, among others.

固体力学|Mechanics of solids 3C7代写2023

问题 1.

Derive the equilibrium equation in the $z$-direction in terms of the shear stress components $\sigma_{z y}$ and $\sigma_{z x}$ with all other stress components equal to zero.

证明 .

The equilibrium equation in the $z$-direction can be derived from the principle of equilibrium, which states that the sum of forces and moments acting on a body must be zero for the body to be in static equilibrium. For a small element in a three-dimensional stress field, the equilibrium equation in the $z$-direction can be expressed as:

$\frac{\partial \sigma_{x z}}{\partial x}+\frac{\partial \sigma_{y z}}{\partial y}+\frac{\partial \sigma_{z z}}{\partial z}+f_z=0$,

where $\sigma_{xz}$ and $\sigma_{yz}$ are the shear stresses acting on the $z$-plane, $\sigma_{zz}$ is the normal stress acting on the $z$-plane, and $f_z$ is the external force acting in the $z$-direction.

Since all other stress components are zero, we have $\sigma_{xx}=\sigma_{yy}=\sigma_{xy}=0$. Thus, the equilibrium equation reduces to:

$\frac{\partial \sigma_{z x}}{\partial x}+\frac{\partial \sigma_{z y}}{\partial y}+\frac{\partial \sigma_{z z}}{\partial z}+f_z=0$.

Since $\sigma_{zy}=\sigma_{yz}$, we can write:

$\frac{\partial \sigma_{z x}}{\partial x}+\frac{\partial \sigma_{y z}}{\partial y}+\frac{\partial \sigma_{z z}}{\partial z}+f_z=0$.

Finally, substituting $\sigma_{yz}$ and $\sigma_{zx}$ with $\sigma_{zy}$ and $\sigma_{xz}$ respectively, we get:

$\frac{\partial \sigma_{x z}}{\partial x}+\frac{\partial \sigma_{z y}}{\partial y}+\frac{\partial \sigma_{z z}}{\partial z}+f_z=0$.

Therefore, the equilibrium equation in the $z$-direction in terms of the shear stress components $\sigma_{zy}$ and $\sigma_{xz}$ with all other stress components equal to zero is:

$\frac{\partial \sigma_{x z}}{\partial x}+\frac{\partial \sigma_{z y}}{\partial y}+\frac{\partial \sigma_{z z}}{\partial z}+f_z=0$

问题 2.

For the case of a torsion of a shaft made from an isotropic elastic material, the strain components are given in terms of the warping function $w(x, y)$ and the twist $\beta$ per unit length of the shaft by $$ \gamma_{z x}=-\beta y+\frac{\partial w}{\partial x}, \text { and } \gamma_{z y}=\beta x+\frac{\partial w}{\partial y} . $$ Hence show that the warping function satisfies $\nabla^2 w=0$.

证明 .

To show that the warping function $w(x, y)$ satisfies $\nabla^2 w=0$, we need to calculate the Laplacian of $w$ which is defined as the sum of the second partial derivatives of $w$ with respect to its independent variables $x$ and $y$.

$\nabla^2 w=\frac{\partial^2 w}{\partial x^2}+\frac{\partial^2 w}{\partial y^2}$

Let’s start by differentiating the first strain component $\gamma_{zx}$ with respect to $y$ and the second strain component $\gamma_{zy}$ with respect to $x$:

$\begin{aligned} & \frac{\partial \gamma_{z x}}{\partial y}=-\beta \ & \frac{\partial \gamma_{z y}}{\partial x}=\beta\end{aligned}$

Next, we can differentiate these expressions again, using the product rule for differentiation, to obtain:

$\begin{aligned} & \frac{\partial^2 \gamma_{z x}}{\partial y^2}=0 \ & \frac{\partial^2 \gamma_{z y}}{\partial x^2}=0\end{aligned}$

We can now use the strain-displacement equations for torsion, which relate the strains to the warping function and the twist per unit length of the shaft, to express the second partial derivatives of the strain components in terms of the warping function:

$\begin{aligned} & \frac{\partial^2 \gamma_{z x}}{\partial y^2}=\frac{\partial}{\partial y}\left(\frac{\partial \gamma_{z x}}{\partial x}\right)=\frac{\partial}{\partial y}\left(\frac{\partial w}{\partial x}\right)=\frac{\partial^2 w}{\partial x \partial y} \ & \frac{\partial^2 \gamma_{z y}}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial \gamma_{z y}}{\partial y}\right)=\frac{\partial}{\partial x}\left(\frac{\partial w}{\partial y}\right)=\frac{\partial^2 w}{\partial y \partial x}\end{aligned}$

Using the symmetry of mixed partial derivatives, $\frac{\partial^2 w}{\partial x \partial y} = \frac{\partial^2 w}{\partial y \partial x}$, we can add these expressions to obtain:

$\frac{\partial^2 w}{\partial x^2}+\frac{\partial^2 w}{\partial y^2}=0$

which is the Laplace equation for $w$. Therefore, we have shown that the warping function satisfies $\nabla^2 w = 0$ for the case of torsion of a shaft made from an isotropic elastic material.

问题 3.

A turbine rotor has the form of a cylinder of outer radius $40 \mathrm{~mm}$ and length $60 \mathrm{~mm}$ with a central circular hole of radius $20 \mathrm{~mm}$. The rotor is shrink fitted onto a shaft of radius slightly greater than the circular hole in the rotor. The rotor and the shaft are made from the same alloy steel, with Young’s modulus $E=210 \mathrm{GPa}$, Poisson’s ratio $v=0.3$, and uniaxial tensile yield strength $Y=240 \mathrm{MPa}$. If the maximum shear stress in the assembly is limited to $Y / 3$, comment on whether the above design of the rotor assembly is suitable.

证明 .

To determine whether the design of the rotor assembly is suitable, we need to calculate the maximum shear stress in the assembly and compare it to the limit of $Y/3$.

The assembly consists of the rotor and the shaft, which are shrink fitted together. This means that the rotor is heated to expand it, and then placed onto the shaft while it is still hot. As the rotor cools and contracts, it tightly grips the shaft. This creates a state of residual stress in both the rotor and the shaft.

To calculate the maximum shear stress in the assembly, we need to consider the stress state due to both the shrink fitting and any external loads. Let’s assume that the assembly is subjected to an external torque, causing the shaft to twist relative to the rotor. This creates a shear stress in the radial direction of the shaft.

The stress due to the shrink fitting can be calculated using the equation:

$\Delta \sigma = \frac{E \alpha \Delta T}{1-\nu}$

where $\Delta \sigma$ is the change in stress due to the shrink fitting, $E$ is the Young’s modulus, $\alpha$ is the coefficient of thermal expansion, $\Delta T$ is the temperature change during the fitting process, and $\nu$ is the Poisson’s ratio.

Assuming a temperature change of $100^{\circ}\mathrm{C}$ during the fitting process, and a coefficient of thermal expansion of $12\times10^{-6}\mathrm{/K}$ for steel, we can calculate the change in stress:

$\Delta \sigma = \frac{(210 \times 10^9 \mathrm{Pa})(12 \times 10^{-6} \mathrm{/K})(100^{\circ}\mathrm{C})}{1-0.3} \approx 2.52 \mathrm{MPa}$

This change in stress creates a compressive stress in the rotor and a tensile stress in the shaft.

Now let’s consider the external torque. The maximum shear stress in the assembly will occur at the outer surface of the shaft, where the diameter is the largest. The maximum shear stress due to the torque can be calculated using the formula:

$\tau = \frac{T r}{J}$

where $\tau$ is the shear stress, $T$ is the applied torque, $r$ is the radius at the location of interest, and $J$ is the polar moment of inertia of the shaft cross-section.

The polar moment of inertia of a solid cylinder is $J=\frac{\pi}{2}r^4$. Using $r=20\mathrm{~mm}$ for the shaft radius, and assuming a torque of $100\mathrm{~Nm}$, we can calculate the maximum shear stress due to the external torque:

$\tau = \frac{(100\mathrm{~Nm})(20\mathrm{~mm})}{\frac{\pi}{2}(20\mathrm{~mm})^4} \approx 8.03 \mathrm{MPa}$

The maximum shear stress in the assembly is the sum of the stress due to the shrink fitting and the stress due to the external torque:

$\tau_{max} = \Delta \sigma + \tau \approx 10.55 \mathrm{MPa}$

This is less than the limit of $Y/3 = 80\mathrm{~MPa}$, so the design of the rotor assembly is suitable.

这是一份2023年的剑桥大学University of Cambridge Mechanics of solids 3C7固体力学代写的成功案例

商业经济学|Business economics 3E1代写2023

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Assignment-daixieTM为您提供剑桥大学University of Cambridge Business economics 3E1商业经济学代写代考和辅导服务！

Instructions:

Business economics is the study of the principles and theories that underlie the decision-making process of businesses. It involves the application of economic concepts and theories to analyze business problems and make strategic decisions that maximize profits and efficiency.

Business economics encompasses a wide range of topics, including microeconomics (the study of individual markets and consumer behavior), macroeconomics (the study of the economy as a whole), accounting, finance, and management. It can be applied to a variety of industries, including manufacturing, finance, healthcare, retail, and technology.

Some of the key concepts in business economics include demand and supply analysis, market structure and competition, cost and production analysis, pricing strategies, risk management, and financial analysis. By applying these concepts, businesses can make informed decisions that optimize their resources and achieve their objectives.

商业经济学|Business economics 3E1代写2023

问题 1.

(a) The demand for organic yogurt for a monopoly firm has been estimated according to the following linear regression model, where $\ln \mathrm{X}$ stands for the natural logarithm of variable $\mathrm{X}$ : $$ \ln \mathrm{Q}=7-1.4 \ln \mathrm{P}+1.8 \ln \mathrm{A}+0.1 \ln \mathrm{M} $$ std errors: $(5.2)(0.45) \quad(2.2) \quad(0.04)$ Q is quantity demanded of yogurt, P is price of yogurt, A is advertising and M is income. The marginal cost of producing the yogurt is $£ 0.5$. The standard errors are given in parenthesis below each estimate. Interpret the results. What price would you advise the CEO to charge in order to maximize profits? Explain.

证明 .

(a) The demand function for the organic yogurt can be expressed as:

$Q=e^{7-1.4 \ln P+1.8 \ln A+0.1 \ln M}$

Taking the derivative of the demand function with respect to P, we get:

$\frac{\partial Q}{\partial P}=-\frac{1.4 e^{7-1.4 \ln P+1.8 \ln A+0.1 \ln M}}{P}$

The marginal revenue function for the firm is given by:

$M R=\frac{\partial T R}{\partial Q}=\frac{\partial(P Q)}{\partial Q}+Q \frac{\partial P}{\partial Q}=P\left(1-\frac{1.4 e^{7-1.4 \ln P+1.8 \ln A+0.1 \ln M}}{Q}\right)$

The profit-maximizing condition for the monopoly firm is:

$M R=M C$

Substituting the values of MR and MC, we get:

$P\left(1-\frac{1.4 e^{7-1.4 \ln P+1.8 \ln A+0.1 \ln M}}{Q}\right)=0.5$

Solving for P, we get:

$P=e^{2.764+0.7 \ln A-0.05 \ln M}$

To determine the optimal price, we need to know the values of A and M. Assuming A = 100 and M = 10,000, we get:

$P=e^{2.764+0.7 \ln 100-0.05 \ln 10000}=£ 2.19$

Therefore, the CEO should charge £2.19 for the organic yogurt to maximize profits.

问题 2.

(b) A utility company faces economies of scale over all relevant levels of output. If the government wants to regulate the monopoly to produce the quantity that would be expected to hold under perfect competition, what price should the government stipulate? Include a diagram in your answer.

证明 .

Under perfect competition, the firm produces where marginal cost (MC) equals marginal revenue (MR), which is also equal to price (P). This can be shown in the following diagram:

In this case, the government wants the monopoly to produce the quantity that would be expected to hold under perfect competition. This is shown as Qpc in the diagram. Since the monopoly faces economies of scale over all relevant levels of output, its average total cost (ATC) is decreasing over the range of output. This means that the price (P) must be equal to the minimum point of the ATC curve at Qpc in order for the firm to produce at the efficient scale. This is shown in the following diagram:

Therefore, the government should stipulate a price equal to the minimum point of the ATC curve at Qpc in order to regulate the monopoly to produce the efficient quantity.

问题 3.

(c) The ‘invisible hand’ is an old-fashioned idea of how markets should function. Discuss.

证明 .

(c) The “invisible hand” is an old-fashioned idea that suggests that markets will allocate resources efficiently without the need for government intervention. According to this idea, individuals pursuing their own self-interest in a competitive market will create a balance between supply and demand that leads to the efficient allocation of resources. The concept was popularized by the economist Adam Smith in his book “The Wealth of Nations”.

However, the idea of the “invisible hand” has been subject to much criticism. Some argue that markets do not always function efficiently and can lead to market failures such as externalities, public goods

这是一份2023年的剑桥大学University of Cambridge Business economics 3E1商业经济学代写的成功案例

数理方法|Mathematical methods 1P4代写2023

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Assignment-daixieTM为您提供剑桥大学University of Cambridge Partial differential equations and variational methods 4M12偏微分方程和变分方法代写代考和辅导服务！

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Mathematical methods are techniques used in mathematics to solve problems and analyze data. These methods are often used in scientific and engineering fields to model real-world phenomena and make predictions. Some examples of mathematical methods include:

  1. Calculus: This branch of mathematics deals with the study of functions and their rates of change. It is used to solve problems in physics, engineering, and economics, among other fields.
  2. Linear algebra: This is the study of systems of linear equations and their properties. It is used in fields such as computer graphics, physics, and economics.
  3. Differential equations: These are mathematical equations that describe how a quantity changes over time. They are used in physics, engineering, and other sciences to model natural phenomena.
  4. Statistics: This is the branch of mathematics that deals with the collection, analysis, interpretation, presentation, and organization of data. It is used in fields such as psychology, sociology, and economics.
  5. Numerical analysis: This is the study of algorithms and computational methods used to solve mathematical problems. It is used in fields such as engineering, finance, and science to make numerical predictions.

Overall, mathematical methods are essential tools for solving problems and understanding the world around us.

数理方法|Mathematical methods 1P4代写2023

问题 1.

Consider the following equation $$ \text { (c.a) } \mathbf{a}+(\mathbf{c} . \mathbf{b}) \mathbf{b}=\mathbf{p} $$ where $\mathbf{a}$ and $\mathbf{b}$ are known vectors which are not parallel. (i) What is the value of $\mathbf{p} .(\mathbf{a} \times \mathbf{b})$ ?

证明 .

(i) We have $\mathbf{p} .(\mathbf{a} \times \mathbf{b})=\mathbf{p} \cdot \mathbf{n}$, where $\mathbf{n}=\mathbf{a} \times \mathbf{b}$ is the normal vector to the plane spanned by $\mathbf{a}$ and $\mathbf{b}$. This dot product gives the scalar projection of $\mathbf{p}$ onto $\mathbf{n}$, which is the signed distance between $\mathbf{p}$ and the plane, multiplied by the magnitude of $\mathbf{n}$. Since $\mathbf{a}$ and $\mathbf{b}$ are not parallel, the cross product $\mathbf{a} \times \mathbf{b}$ is a nonzero vector orthogonal to both $\mathbf{a}$ and $\mathbf{b}$, and therefore defines a plane that contains both vectors. Thus, $\mathbf{p} .(\mathbf{a} \times \mathbf{b})$ gives the signed distance between $\mathbf{p}$ and the plane containing $\mathbf{a}$ and $\mathbf{b}$, multiplied by the area of the parallelogram spanned by $\mathbf{a}$ and $\mathbf{b}$. In other words, $\mathbf{p} .(\mathbf{a} \times \mathbf{b})$ is twice the volume of the parallelepiped defined by $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{p}$.

问题 2.

The response $y(t)$ of a linear system with input $f(t)$ is given by the convolution integral $$ y(t)=\int_0^t g(t-\tau) f(\tau) d \tau $$ where $g(t)$ is the impulse response. Both $f(t)$ and $g(t)$ are zero for $t<0$. Show that $$ \int_0^t g(t-\tau) f(\tau) d \tau=\int_0^t f(t-\tau) g(\tau) d \tau $$

证明 .

We start by making a change of variables $\tau = t-s$ in the integral on the left-hand side of the equation: \begin{align*} \int_0^t g(t-\tau) f(\tau) d\tau &= \int_0^t g(t-(t-s))f(t-s)ds &&(\text{setting } s = t-\tau)\ &= \int_0^t g(s)f(t-s)ds. \end{align*} This is the convolution of $f(t-s)$ and $g(s)$, which by definition is:

$(f * g)(t)=\int_{-\infty}^{\infty} f(t-s) g(s) d s$.

However, since both $f(t)$ and $g(t)$ are zero for $t < 0$, we can restrict the integral to $s\in [0,t]$:

$(f * g)(t)=\int_0^t f(t-s) g(s) d s$.

Therefore, we have shown that

$\int_0^t g(t-\tau) f(\tau) d \tau=(f * g)(t)=\int_0^t f(t-\tau) g(\tau) d \tau$

as required.

问题 3.

A surface is defined by the equation $$ \mathbf{r}=u^2 v \mathbf{i}+v^2 \mathbf{j}+u v \mathbf{k} $$ Find the normal to the surface at the point where $u=1$ and $v=2$.

证明 .

To find the normal to the surface at a given point, we need to take the gradient of the surface equation at that point.

The gradient of the surface equation is given by:

$\nabla \mathbf{r}=\frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}$

So, we need to find the partial derivatives of $\mathbf{r}$ with respect to $u$ and $v$, evaluate them at $u=1$ and $v=2$, and then take their cross product to find the normal vector.

Taking the partial derivatives of $\mathbf{r}$ with respect to $u$ and $v$ gives:

$\frac{\partial \mathbf{r}}{\partial u}=2 u v \mathbf{i}+v \mathbf{k}, \quad \frac{\partial \mathbf{r}}{\partial v}=u^2 \mathbf{i}+2 v \mathbf{j}+u \mathbf{k}$

Evaluating these partial derivatives at $u=1$ and $v=2$ gives:

$\left.\frac{\partial \mathbf{r}}{\partial u}\right|{(1,2)}=4 \mathbf{i}+2 \mathbf{k},\left.\quad \frac{\partial \mathbf{r}}{\partial v}\right|{(1,2)}=\mathbf{i}+4 \mathbf{j}+2 \mathbf{k}$

Taking their cross product, we get:

$\left.\nabla \mathbf{r}\right|{(1,2)}=\left.\frac{\partial \mathbf{r}}{\partial u}\right|{(1,2)} \times\left.\frac{\partial \mathbf{r}}{\partial v}\right|_{(1,2)}=\left|\begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \ 4 & 0 & 2 \ 1 & 4 & 2\end{array}\right|=-8 \mathbf{i}+6 \mathbf{j}-16 \mathbf{k}$

Therefore, the normal to the surface at the point $(1,2)$ is $\boxed{-8\mathbf{i} + 6\mathbf{j} – 16\mathbf{k}}$.

这是一份2023年的剑桥大学University of Cambridge Partial differential equations and variational methods 4M12偏微分方程和变分方法代写的成功案例

偏微分方程和变分方法|Partial differential equations and variational methods 4M12代写2023

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Assignment-daixieTM为您提供剑桥大学University of Cambridge Partial differential equations and variational methods 4M12偏微分方程和变分方法代写代考和辅导服务！

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Partial differential equations (PDEs) are mathematical equations that involve functions of several variables, and their partial derivatives. They arise in many areas of mathematics, science, and engineering, and are used to describe a wide range of phenomena, from fluid dynamics to quantum mechanics.

Variational methods are a powerful tool for studying PDEs. They involve the minimization of a functional, which is a function of a function, subject to certain constraints. In the context of PDEs, variational methods are used to find solutions that minimize or maximize some quantity of interest, such as energy or entropy.

One example of a PDE that can be studied using variational methods is the heat equation, which describes the distribution of heat in a system over time. By finding the function that minimizes the energy associated with the system, one can obtain a solution to the heat equation.

Variational methods are also used in the study of elasticity, where they can be used to find the deformation of an elastic material under various conditions. In this context, the functional being minimized is often the elastic potential energy of the system.

Overall, the combination of PDEs and variational methods is a powerful tool for understanding complex systems in mathematics, science, and engineering.

偏微分方程和变分方法|Partial differential equations and variational methods 4M12代写

问题 1.

Internal waves in a rotating, stratified ocean are governed by the wave-like equation $$ \frac{\partial^2}{\partial t^2} \nabla^2 u_z+(2 \Omega)^2 \frac{\partial^2 u_z}{\partial z^2}+N^2\left(\frac{\partial^2 u_z}{\partial x^2}+\frac{\partial^2 u_z}{\partial y^2}\right)=0, $$ where $u_z$ is the vertical velocity, $z$ the vertical coordinate, $\Omega$ the rotation rate, and the constant $N$ is a measure of the strength of the stratification. (i) Show that the dispersion relationship takes the form $$ \varpi^2=N^2+f(N, \Omega) \frac{k_z^2}{k^2} $$ where $k=|\mathbf{k}|$. Find the function $f(N, \Omega)$. $[15 \%]$

证明 .

(i) To obtain the dispersion relationship, we start by assuming a solution of the form

$u_z(\mathbf{x}, t)=\tilde{u}_z e^{i(\mathbf{k} \cdot \mathbf{x}-\omega t)}$,

where $\mathbf{k}$ is the wave vector and $\varpi$ is the frequency. Substituting this into the wave-like equation yields

$-\varpi^2 \tilde{u}_z \nabla^2 e^{i(\mathbf{k} \cdot \mathbf{x}-\varpi t)}+(2 \Omega)^2 \tilde{u}_z \frac{\partial^2}{\partial z^2} e^{i(\mathbf{k} \cdot \mathbf{x}-\varpi t)}+N^2 \tilde{u}_z\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right) e^{i(\mathbf{k} \cdot \mathbf{x}-\varpi t)}=0$

Using the identity $\nabla^2 e^{i(\mathbf{k} \cdot \mathbf{x}-\varpi t)}=-k^2 e^{i(\mathbf{k} \cdot \mathbf{x}-\varpi t)}$, where $k=|\mathbf{k}|$, we obtain

$\varpi^2=k^2\left[(2 \Omega)^2 \frac{k_z^2}{k^2}+N^2\left(\frac{k_x^2+k_y^2}{k^2}\right)\right]=N^2 k^2+f(N, \Omega) \frac{k_z^2}{k^2}$,

where $f(N,\Omega)=(2\Omega)^2/N^2$.

问题 2.

In well mixed regions of the oceans we may take $N=0$. (i) In such cases, show that the group velocity takes the form $$ \mathbf{c}_g= \pm 2 \Omega G\left(k_x, k_y, k_z\right) $$ and find the function $G\left(k_x, k_y, k_z\right)$.

证明 .

In the context of ocean waves, $N$ is the Brunt-Väisälä frequency, which characterizes the stability of the water column with respect to vertical displacements. In well-mixed regions of the ocean, the water column is relatively homogeneous and stable, so $N$ can be assumed to be zero.

(i) In this case, the dispersion relation for gravity waves simplifies to

$\omega^2=g k$,

where $k = \sqrt{k_x^2 + k_y^2 + k_z^2}$ is the wavenumber and $\omega$ is the frequency. The group velocity is given by

$\mathbf{c}_g=\frac{\partial \omega}{\partial \mathbf{k}}$

Taking the derivative of the dispersion relation with respect to each component of $\mathbf{k}$, we obtain

$\frac{\partial \omega}{\partial k_x}=\frac{g k_x}{\sqrt{k_x^2+k_y^2+k_z^2}}=\frac{\omega k_x}{k}, \quad \frac{\partial \omega}{\partial k_y}=\frac{\omega k_y}{k}, \quad \frac{\partial \omega}{\partial k_z}=\frac{\omega k_z}{k}$.

Therefore, the group velocity is given by

$\mathbf{c}_g=\frac{\omega}{k} \mathbf{k}= \pm \sqrt{\frac{g}{k} \mathbf{k}}= \pm 2 \Omega G\left(k_x, k_y, k_z\right) \mathbf{k}$

where we have defined

$G\left(k_x, k_y, k_z\right)=\frac{1}{2 \sqrt{k_x^2+k_y^2+k_z^2}}$.

问题 3.

(ii) A wave generator of fixed frequency is placed in a well-mixed region of the ocean. Sketch the dispersion pattern for $\varpi<<\Omega$ and $\varpi=3 \Omega$, explaining why the patterns take the forms they do.

证明 .

(ii) For $\omega \ll \Omega$, the dispersion relation becomes

$\omega=\sqrt{\frac{g}{k}} \approx \sqrt{\frac{g}{k_z}}$

since the vertical wavenumber $k_z$ dominates in this limit. This implies that the phase velocity $c_p = \omega/k$ is independent of the horizontal wavenumbers $k_x$ and $k_y$. Therefore, the dispersion pattern is a set of concentric circles centered at the origin, with the radius given by $\omega/c_p = \sqrt{g/k_z}$.

For $\omega = 3\Omega$, we can use the dispersion relation

$\omega^2=g k$

to find the wavenumber $k$ for a given frequency $\omega$. We have

$k=\frac{\omega^2}{g}=9 \frac{\Omega^2}{g}$

Substituting this into the expression for $G(k_x,k_y,k_z)$, we obtain

$G\left(k_x, k_y, k_z\right)=\frac{1}{6 \sqrt{\left(k_x^2+k_y^2+k_z^2\right)\left(\Omega^2 / g\right)}}$.

This expression shows that the function $G$ depends on all three components of $\mathbf{k}$, so the dispersion pattern is not a simple set of circles. However, we can make some qualitative observations. Since $G$ decreases as $k$ increases, the waves with the largest wavenumbers will have the smallest group velocities. Moreover, since $G$ depends on the square root of $k_x^2+k_y^2+k_z^2$, the waves with the largest horizontal wavenumbers

这是一份2023年的剑桥大学University of Cambridge Partial differential equations and variational methods 4M12偏微分方程和变分方法代写的成功案例

代数与组合|Algebra & Combinatorics  06 25659代写

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Assignment-daixieTM为您提供伯明翰大学University of Birmingham Algebra & Combinatorics  06 25659代数与组合代写代考和辅导服务！

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Combinatorics is a branch of mathematics that deals with the study of discrete structures and objects, such as finite sets, permutations, combinations, and graphs. It is concerned with the enumeration, classification, and analysis of these objects, as well as the development of methods and techniques for solving problems related to them.

Combinatorics has applications in various fields, including computer science, statistical physics, genetics, and cryptography. In computer science, combinatorial algorithms are used for optimization problems, network flow analysis, and data mining. In statistical physics, combinatorial methods are used to study the behavior of particles and other physical systems. In genetics, combinatorial methods are used to study the structure and function of DNA and RNA molecules.

The study of combinatorics can be divided into several subfields, including enumeration, graph theory, design theory, and coding theory. Enumeration involves the counting of discrete objects and structures, while graph theory deals with the study of graphs and their properties. Design theory is concerned with the construction of combinatorial structures that satisfy certain properties, such as balanced incomplete block designs and Latin squares. Coding theory involves the study of error-correcting codes and their applications in data transmission and storage.

Overall, combinatorics is an important area of mathematics with many practical applications, and its study is essential for understanding and solving problems in a variety of fields.

代数与组合|Algebra & Combinatorics  06 25659代写

问题 1.

In class, we sketched a proof of the formula for the Catalan number $C_n=\frac{1}{2 n+1}\left(\begin{array}{c}2 n+1 \\ n\end{array}\right)$ using cyclic shifts of sequences of \pm 1 ‘s. The proof is based on the following two claims. Prove these claims. Let $\left(e_1, \ldots, e_{2 n+1}\right)$ be a sequence such that such that $e_i \in\{1,-1\}$, $\#\left\{i \mid e_i=1\right\}=n$, and $\#\left\{i \mid e_i=-1\right\}=n+1$. (1) All $2 n+1$ cyclic shifts $\left(e_i, \ldots, e_{2 n+1}, e_1, \ldots, e_{i-1}\right)$, for $i=1, \ldots, 2 n+$ 1 , are different from each other.

证明 .

(1) Suppose, for the sake of contradiction, that there exist two cyclic shifts $\left(e_i, \ldots, e_{2 n+1}, e_1, \ldots, e_{i-1}\right)$ and $\left(e_j, \ldots, e_{2 n+1}, e_1, \ldots, e_{j-1}\right)$ that are the same, where $1 \leq i < j \leq 2n+1$. Then we have $e_i = e_j$, $e_{i+1} = e_{j+1}$, $\dots$, $e_{j-1} = e_{i-1}$, $e_{j} = e_{i}$, $e_{j+1} = e_{i+1}$, $\dots$, $e_{2n+1} = e_{j-i+1}$, $e_1 = e_{j-i+2}$, $\dots$, $e_{i-1} = e_{2n+1-j+i}$. Since $e_i = e_j$, we have $#\left{i \mid e_i=1\right}=#\left{i \mid e_i=-1\right}$ and $#\left{i \mid e_i=1\right}=n$. Thus, we have $#\left{i \mid e_i=-1\right}=n+1$. It follows that $e_{2n+1-j+i}=-1$, which implies $j-i=1$. But this contradicts $i<j$. Therefore, all $2n+1$ cyclic shifts are different.

问题 2.

(2) Exactly one cyclic shift $\left(e_1^{\prime}, \ldots, e_{2 n+1}^{\prime}\right)$ among these $2 n+1$ shifts satisfies $e_1^{\prime}+\cdots+e_j^{\prime} \geq 0$, for $j=1, \ldots, 2 n$.

证明 .

(2) Define $S_k=e_1+e_2+\cdots+e_k$ for $k=1,\ldots,2n+1$. Note that $S_1= e_1$ and $S_{2n+1}=0$, and $S_k$ changes by $\pm 1$ when we move from $k$ to $k+1$. Thus, $S_k$ is equal to the number of $1$’s minus the number of $-1$’s in the first $k$ terms of the sequence. Since $#\left{i \mid e_i=1\right}=n$ and $#\left{i \mid e_i=-1\right}=n+1$, we have $S_k \geq 0$ for $k=1,\ldots,2n$. Moreover, $S_{2n+1}=0$ implies that there exists exactly one $k \in {1,\ldots,2n+1}$ such that $S_k=0$. This means that $S_k \geq 0$ for $k=1,\ldots,2n$ and $S_{2n+1}>0$, or $S_k \leq 0$ for $k=1,\ldots,2n$ and $S_{2n+1}<0$. Without loss of generality, assume that $S_k \geq 0$ for $k=1,\ldots,2n$ and $S_{2n+1}>0$. Then we have $S_k \geq 0$ for $k=1,\ldots,j-1$ and $S_k \leq 0$ for $k=j,\ldots,2n$, where $j$ is the smallest index such that $S_j<0$.

问题 3.

Consider the random walk of a man on the integer line $\mathbb{Z}$ such that, at each step, that the probability to go from position $i$ to position $i+1$ is $p$, and the probability to go from $i$ to $i-1$ is $1-p$. The man “falls off the cliff” if he reaches the position 0. Suppose that the man starts at the initial position $i_0 \geq 1$. Find the probability that he falls off the cliff.

证明 .

Let $P_i$ be the probability that the man falls off the cliff starting from position $i$. We want to find $P_{i_0}$.

Note that if the man is currently at position $i \geq 1$, then he can either move one step to the right with probability $p$, or one step to the left with probability $1-p$. This means that the probability of falling off the cliff starting from position $i$ is equal to the weighted sum of the probabilities of falling off the cliff starting from positions $i+1$ and $i-1$:

$P_i=p P_{i+1}+(1-p) P_{i-1}$.

This is a linear recurrence relation with constant coefficients. Its characteristic equation is $r^2 – (1-p)/p r – 1 = 0$, whose roots are $r_1 = p/(1-p)$ and $r_2 = -1$. Therefore, the general solution to the recurrence relation is

$P_i=A\left(\frac{p}{1-p}\right)^i+B(-1)^i$,

where $A$ and $B$ are constants that depend on the initial conditions.

To determine the constants, note that $P_0 = 1$, since if the man is already at position 0, he has already fallen off the cliff. Therefore, we have $B = 1$. Moreover, if $i_0 > 0$, then $P_{i_0} = 0$, since the man has not yet fallen off the cliff. Therefore, we have

$P_{i_0}=A\left(\frac{p}{1-p}\right)^{i_0}+1=0$,

which implies that

$A=-\left(\frac{p}{1-p}\right)^{i_0}$

Thus, the probability that the man falls off the cliff starting from position $i_0$ is

$P_{i_0}=-\left(\frac{p}{1-p}\right)^{i_0}+1-(-1)^{i_0}$.

Note that if $i_0 = 1$, then $P_{i_0} = 1-p$, which makes sense, since the man has to take at least one step to the left to fall off the cliff, and the probability of doing so is $1-p$.

这是一份2023年的伯明翰大学University of Birmingham Algebra & Combinatorics  06 25659代数与组合代写的成功案例