# 数学分析|MATH0003 Analysis 1代写2023

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Assignment-daixieTM为您提供伦敦大学学院 London’s Global University MATH0003 Analysis 1数学分析代写代考辅导服务！

## Instructions:

We will start with the basic properties of real numbers and use them to prove the main results in elementary differential calculus. We assume familiarity with the properties of real numbers such as completeness, order, and the field axioms.

SEQUENCES:

A sequence is a function $f: \mathbb{N} \rightarrow \mathbb{R}$, where $\mathbb{N}$ is the set of natural numbers. We say that a sequence $(a_n)$ converges to a limit $L$ if for every $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that $|a_n – L| < \epsilon$ for all $n \geq N$. We write $\lim_{n \rightarrow \infty} a_n = L$.

a) Prove that if $c>1$, the sequence $a_n=\frac{c^n}{n !}$ is decreasing for $n \gg 1$.

a) To prove that the sequence $a_n=\frac{c^n}{n!}$ is decreasing for $n\gg1$, we need to show that $a_{n+1} < a_n$ for all $n\gg1$:

\begin{align*} a_{n+1} & = \frac{c^{n+1}}{(n+1)!} \ & = \frac{c}{n+1} \cdot \frac{c^n}{n!} \ & < \frac{c}{n+1} \cdot \frac{c^n}{n!} \quad \text{since } c>1 \ & = \frac{c}{n} \cdot \frac{c^n}{n!} \cdot \frac{n}{n+1} \ & < \frac{c}{n} \cdot \frac{c^n}{n!} \quad \text{since } \frac{n}{n+1} < 1 \ & = a_n \end{align*}

Therefore, $a_n$ is a decreasing sequence for $n\gg1$.

b) Prove $\lim _{n \rightarrow \infty} a_n=0$, by starting at some suitable point $N$ in the sequence and give an estimate of the size of the factors which allows you to use Theorem 3.4.)

b) To prove that $\lim_{n\rightarrow\infty}a_n=0$, we will use Theorem 3.4 which states that if there exist constants $C>0$ and $p>0$ such that $|a_{n+1}/a_n|\leq Cn^{-p}$ for all $n\geq N$, where $N$ is some suitable point in the sequence, then $\sum_{n=1}^{\infty}a_n$ converges absolutely.

Let us first find a suitable point $N$ in the sequence. Since $n!$ grows faster than any polynomial in $n$, we expect that $a_n$ will converge to zero faster than $1/n!$. To confirm this, we can take the ratio of consecutive terms:

$\frac{a_{n+1}}{a_n}=\frac{(n+1) !}{2^{n+1}(n+1) !}=\frac{1}{2^{n+1}}$

So we have $|a_{n+1}/a_n|\leq 1/2$, which means that we can take $C=1/2$. Also, we can take $p=1$ since $1/n$ is a decreasing function for $n\geq 1$. Thus, we have $|a_{n+1}/a_n|\leq Cn^{-p}$ for all $n\geq 1$ with $C=1/2$ and $p=1$.

Now we need to find a suitable point $N$ in the sequence. Since $n!$ grows faster than any polynomial in $n$, we expect that $a_n$ will converge to zero faster than $1/n!$. To confirm this, we can solve the inequality $1/2^n\leq \epsilon$, where $\epsilon>0$ is a small number that we want to be close to zero. Taking the logarithm of both sides, we get $n\ln(2)\leq \ln(1/\epsilon)$, which implies that $n\geq (\ln(1/\epsilon))/\ln(2)$. Therefore, we can take $N=\lceil(\ln(1/\epsilon))/\ln(2)\rceil$, where $\lceil x\rceil$ denotes the smallest integer greater than or equal to $x$.

With these values of $C$, $p$, and $N$, we can use Theorem 3.4 to conclude that $\sum_{n=1}^{\infty}a_n$ converges absolutely, which implies that $\lim_{n\rightarrow\infty}a_n=0$.

c) Prove part (b) differently by considering the series $\sum_0^{\infty} a_n$.

c) Another way to prove that $\lim_{n\rightarrow\infty}a_n=0$ is to consider the series $\sum_{n=0}^{\infty}a_n$. We can use the ratio test to show that this series converges:

$\lim {n \rightarrow \infty} \frac{a{n+1}}{a_n}=\lim {n \rightarrow \infty} \frac{(n+1) !}{2^{n+1} n !}=\lim {n \rightarrow \infty} \frac{1}{2}=\frac{1}{2}<1$

Therefore, the series $\sum_{n=0}^{\infty}a_n$ converges, which implies that $\lim_{n\rightarrow\infty}a_n=0$ by the nth-term test for convergence.

# 微分几何代写 Differential Geometry代考2023

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## 微分几何代写Differential Geometry

### 伪黎曼尼几何学Pseudo-Riemannian geometry代写

• Finsler manifold芬斯勒流形
• Symplectic geometry辛几何

## 微分几何的历史

The history and development of differential geometry as a discipline can be traced back at least to the ancient classics. It is closely related to the development of geometry, the concepts of space and form, and the study of topology, particularly manifolds. In this section we focus on the history of the application of infinitesimal methods to geometry, and then on the idea of tangent spaces, and finally on the development of the modern formalism of the discipline in terms of tensors and tensor fields.

## 微分几何相关课后作业代写

Let $c$ be a regular curve such that $|c(s)| \leq 1$ for all $s$. Suppose that there is a point $t$ where $|c(t)|=1$. Prove that the curvature at that point satisfies $|\kappa(t)| \geq 1$.

Let $c$ be a regular curve with arc length parameterization $s$. By definition, the curvature $\kappa(s)$ of the curve $c$ at the point $c(s)$ is given by $\kappa(s) = |\boldsymbol{c}”(s)|$, where $\boldsymbol{c}”(s)$ denotes the second derivative of $\boldsymbol{c}(s)$ with respect to $s$.

Since $|\boldsymbol{c}(s)| \leq 1$ for all $s$, we have $|\boldsymbol{c}'(s)| = 1$ for all $s$. Moreover, since $|\boldsymbol{c}(t)| = 1$, we have $|\boldsymbol{c}'(t)| = 0$, which implies that $\boldsymbol{c}”(t)$ is perpendicular to $\boldsymbol{c}'(t)$. Therefore, we have

$\left|\boldsymbol{c}^{\prime \prime}(t)\right|=\left|\boldsymbol{c}^{\prime \prime}(t) \cdot \frac{c^{\prime}(t)}{\left|c^{\prime}(t)\right|}\right|=\left|c^{\prime \prime}(t) \cdot \frac{c^{\prime}(t)}{\left|c^{\prime}(t)\right|}\right|$.

Now, we observe that $\boldsymbol{c}”(t) \cdot \boldsymbol{c}'(t)$ is the tangential component of $\boldsymbol{c}”(t)$ at the point $c(t)$, which is given by $\boldsymbol{c}”(t) \cdot \boldsymbol{c}'(t) = \frac{d}{ds}\left(|\boldsymbol{c}'(s)|^2\right)\bigg|{s=t} = 2|\boldsymbol{c}'(t)|\boldsymbol{c}”(t) \cdot \boldsymbol{c}'(t) = 0$. Thus, we can write $\boldsymbol{c}”(t) = \boldsymbol{c}”(t){\perp} + \boldsymbol{c}”(t){\parallel}$, where $\boldsymbol{c}”(t){\perp}$ is the perpendicular component of $\boldsymbol{c}”(t)$ to $\boldsymbol{c}'(t)$ and $\boldsymbol{c}”(t){\parallel}$ is the tangential component of $\boldsymbol{c}”(t)$ to $\boldsymbol{c}'(t)$. Since $|\boldsymbol{c}'(t)| = 0$, we have $\boldsymbol{c}”(t){\parallel} = 0$. Therefore, we have

$\left|\boldsymbol{c}^{\prime \prime}(t)\right|=\left|\boldsymbol{c}^{\prime \prime}(t){\perp}\right| \leq\left|\boldsymbol{c}^{\prime \prime}(t)\right|{\max }$

where $|\boldsymbol{c}”(t)|_{\max}$ denotes the maximum value of $|\boldsymbol{c}”(t)|$ over all vectors $\boldsymbol{v}$ perpendicular to $\boldsymbol{c}'(t)$ with $|\boldsymbol{v}| = 1$.

On the other hand, we know that the maximum value of $|\boldsymbol{c}”(t)|$ is attained when $\boldsymbol{c} ## 微分几何课后作业代写的应用代写 微分几何学在整个数学和自然科学领域都有应用。最突出的是，爱因斯坦在他的广义相对论中使用了微分几何的语言，随后物理学家在发展量子场理论和粒子物理学的标准模型时也使用了这种语言。在物理学之外，微分几何在化学、经济学、工程、控制理论、计算机图形和计算机视觉以及最近的机器学习中也有应用。 ## 英国论文代写Viking Essay代写 订购流程： 第一步： 右侧扫一扫或添加WX客服mytutor01 发送代写^代考任务委托的具体要求 第二步：我们的线上客服收到您的要求后会为您匹配合适的写手，等到写手确认可以接此任务并且给出服务报价后我们将写手的报价转发给您并且收取一定的信息费，等您支付50%的定金后（有可能会向你索要更详细的作业要求）我们开始完成您交给我们的任务。 第三步： 写作完毕后发你Turnitin检测/作业完成截图（根据作业类型而定）文件，你阅读后支付余款后我们发你完整的终稿（代码，手写pdf等） 第四步： 在收到论文后，你可以提出任何修改意见，并与写手一对一讨论，我们非常愿意拉群让您和写手面对面沟通。 建议：因每份任务都具有特殊性，以上交易流程只是大概流程，更加具体的流程烦请添加客服WX免费咨询，30S通过验证，工作时间内2min回复响应，支持大多数课程的加急任务。 ## 英国论文代写Viking Essay代写承诺&保证： 我们英国论文代写Viking Essay的政策协议保证不会将您的所有个人信息或详细信息出售或与第三方或作家共享。 相反，我们使用订单号，订单的月份和日期进行通信，并基于我们的客户与我们公司之间的现有合同，因此，即使在将来下订单时，您的身份也会在整个交易中受到保护。 我们的通信内容已通过SSL加密，以确保您以及您的论文或作业的隐私和安全性。 我们严格的写手团队要求写手“零抄袭”指导我们提供高质量的原创写作服务。 我们的业务使用Turnitin（国际版plag窃检查程序）将所有订单的剽窃报告副本发送给客户，并确保所有交付的任务都是100％原创的。 所有学术写作规则和要求，并遵循后者，包括使用参考文献和文本引用来表示和引用其他来源的内容和引语，方法是使用适当的参考样式和格式来提供高质量的服务和任务 。 我们遵守您论文的所有严格指导方针和要求，并提供至少三次修订，保证您可以拿到完全满意的论文。 仅当客户在下订单过程开始时提供详细而完整的分配说明时，此方法才有效。 我们的公司和作家在完成任务的一半或完成后不能也不会改变订单的任务。 如果作者未能找到来源，内容或未能交付的任务或任务，我们公司仅全额退款。 但是，请放心，由于我们的实时通信以及对订单交付和消费者满意度的严格规定，很少发生这种情况。英国论文代写Viking Essay 代写机构致力于打造出理科全覆盖的代写平台，所以对于很多难度很大的科目都可以提供代写服务，并且收费合理，也提供高质量的售后服务，详情咨询WX：mytutor01 作业稿件在交付之后，我们依然提供了长达30天的修改润色服务，最大程度的保证学生的代写权益。为了您的权益着想，即便最终您没有选择与我们平台合作，但依然不要去相信那些没有资历，价格低于标准的小机构，因为他们浪费的不仅仅只是你的时间和金钱，而是在变相摧毁你的学业。 # 数学分析|MATH0004 Analysis 2代写2023 0 Assignment-daixieTM为您提供伦敦大学学院 London’s Global University MATH0004 Analysis 2数学分析代写代考辅导服务！ ## Instructions: We will start with the basic properties of real numbers and use them to prove the main results in elementary differential calculus. We assume familiarity with the properties of real numbers such as completeness, order, and the field axioms. SEQUENCES: A sequence is a function$f: \mathbb{N} \rightarrow \mathbb{R}$, where$\mathbb{N}$is the set of natural numbers. We say that a sequence$(a_n)$converges to a limit$L$if for every$\epsilon > 0$, there exists an$N \in \mathbb{N}$such that$|a_n – L| < \epsilon$for all$n \geq N$. We write$\lim_{n \rightarrow \infty} a_n = L$. 问题 1. Find the radius of convergence$R$of$\sum_0^{\infty} \frac{(2 n) ! x^n}{n !(n+1) !}$. 证明 . We can apply the ratio test to determine the radius of convergence of the given series. Let$a_n=\frac{(2 n) ! x^n}{n !(n+1) !}Then, the ratio of successive terms is \begin{align*} \frac{a_{n+1}}{a_n} &= \frac{(2(n+1))!x^{n+1}}{(n+1)!(n+2)!} \cdot \frac{n!(n+1)!}{(2n)!x^n} \ &= \frac{(2n+2)(2n+1)x}{(n+2)(n+1)} \ &= \frac{4x^2}{(n+2)(n+1)} \cdot \frac{(2n+1)(2n+2)}{4(n+1)} \ &= \frac{4x^2}{(n+2)(n+1)} \cdot \left(1 + \frac{1}{n+1}\right) \cdot \left(1 + \frac{2}{n+1}\right). \end{align*} Taking the limit asn \to \infty$, we get$\lim {n \rightarrow \infty} \frac{a{n+1}}{a_n}=\lim _{n \rightarrow \infty} \frac{4 x^2}{(n+2)(n+1)} \cdot\left(1+\frac{1}{n+1}\right) \cdot\left(1+\frac{2}{n+1}\right)=4 x^2$. Therefore, the series converges absolutely if$4x^2 < 1$, or equivalently,$|x| < \frac{1}{2}$. On the other hand, the series diverges if$4x^2 > 1$, or equivalently,$|x| > \frac{1}{2}$. Thus, the radius of convergence is$R = \frac{1}{2}$. 问题 2. Let$h(n)$be the largest prime factor of the integer$n>1$, and$s(n)$be the sum of its prime factors, so$h(12)=3, s(12)=7$. Prove the sequence${h(n) / s(n)}, n=2,3,4, \ldots$has$1 / k$as a cluster point for every positive integer$k$, but no limit. 证明 . To show that$1/k$is a cluster point of the sequence${h(n)/s(n)}$for every positive integer$k$, we need to show that for any$\epsilon>0$and positive integer$k$, there exist infinitely many$n$such that$|h(n)/s(n)-1/k|<\epsilon$. Let$\epsilon>0$and$k$be given. Choose$p$to be any prime greater than$k/\epsilon$, and consider the numbers$n=p^2,p^3,\ldots,p^{k+1}$. Note that$h(n)=p$and$s(n)=1+p+p^2+\cdots+p^k=\frac{p^{k+1}-1}{p-1}$. For any$n=p^m$with$2\leq m\leq k+1$, we have$\left|\frac{h(n)}{s(n)}-\frac{1}{k}\right|=\left|\frac{k p-p^{k+1}}{(p-1)\left(p^{k+1}-1\right)}\right|=\frac{p\left(k-p^k\right)}{(p-1)\left(p^{k+1}-1\right)}<\frac{p(k-p)}{(p-1) p^{k+1}}<\frac{k}{p^k-1}<\epsilon$, where we used the fact that$p>k/\epsilon$and$p^{k+1}>kp$. Thus, we have shown that for any$\epsilon>0$and positive integer$k$, there exist infinitely many$n$such that$|h(n)/s(n)-1/k|<\epsilon$. Hence,$1/k$is a cluster point of the sequence${h(n)/s(n)}$for every positive integer$k$. To show that the sequence${h(n)/s(n)}$has no limit, we need to find two increasing sequences of integers${a_n}$and${b_n}$such that$h(a_n)/s(a_n)$and$h(b_n)/s(b_n)$have different limits as$n\rightarrow\infty$. Let${p_n}$be the sequence of prime numbers, and define$a_n=p_{2n-1}^2$and$b_n=p_{2n}^2$. Then we have$\frac{h\left(a_n\right)}{s\left(a_n\right)}=\frac{p_{2 n-1}}{1+p_{2 n-1}} \rightarrow 1, \quad \frac{h\left(b_n\right)}{s\left(b_n\right)}=\frac{p_{2 n}}{1+p_{2 n}+p_{2 n}^2} \rightarrow 0$, as$n\rightarrow\infty$, since$p_{2n-1}/(1+p_{2n-1})\rightarrow 1$and$p_{2n}/(1+p_{2n}+p_{2n}^2)\rightarrow 0$. Therefore, the sequence${h(n)/s(n)}$does not have a limit as$n\rightarrow\infty$. 问题 3. A function$f(x)$has three distinct zeros$a_0<a_1<a_2$on an interval$I$, and in addition$f^{\prime}\left(a_2\right)=0$. Assume$f(x)$has a third derivative$f^{\prime \prime \prime}(x)$at all points of$I$. Prove there is a point$c \epsilon I$such that$f^{\prime \prime \prime}(c)=0$. 证明 . Since$f(x)$has three distinct zeros on the interval$I$, we know that$f(x)$changes sign at each of these zeros. Without loss of generality, assume that$f(x)$is positive on$(a_0, a_1)$, negative on$(a_1, a_2)$, and positive on$(a_2, b)$. Since$f(x)$has a local extremum at$x=a_2$(because$f^{\prime}(a_2)=0$), the second derivative$f^{\prime \prime}(x)$must change sign at$x=a_2$. Specifically,$f^{\prime \prime}(x)$is negative on$(a_1, a_2)$and positive on$(a_2, b)$. Now consider the third derivative$f^{\prime \prime \prime}(x)$. Since$f^{\prime \prime}(x)$changes sign at$x=a_2$, we know that$f^{\prime \prime \prime}(x)$must have a local extremum at$x=a_2$. Without loss of generality, assume that$f^{\prime \prime \prime}(x)$is negative on$(a_1, a_2)$and positive on$(a_2, b)$. Since$f(x)$changes sign at each of the zeros$a_0<a_1<a_2$, we know that$f(x)$must have at least one local extremum (a maximum or minimum) between each pair of zeros. Specifically,$f(x)$has a local minimum at some point$c_1$in$(a_0, a_1)$, a local maximum at some point$c_2$in$(a_1, a_2)$, and a local minimum at some point$c_3$in$(a_2, b)$. Since$f^{\prime \prime \prime}(x)$is negative on$(a_1, a_2)$and positive on$(a_2, b)$, we know that$f^{\prime \prime \prime}(c_2)>0$and$f^{\prime \prime \prime}(c_3)<0$. Therefore, by the intermediate value theorem, there must be some point$c$in$(c_2, c_3)$such that$f^{\prime \prime \prime}(c)=0$, as desired. # 线性代数代写 Linear algebra代考2023 0 如果你也在 怎样代写线性代数Linear algebra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！ ## 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。 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。 # 数学分析代写 Mathematical Analysis代考 0 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。 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。 # 数理方法|Mathematical methods 1P4代写2023 0 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辅导服务！ ## Instructions: 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. 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$\tau = t-sin the integral on the left-hand side of the equation: \begin{align*} \int_0^t g(t-\tau) f(\tau) d\tau &= \int_0^t g(t-(t-s))f(t-s)ds &&(\text{setting } s = t-\tau)\ &= \int_0^t g(s)f(t-s)ds. \end{align*} This is the convolution off(t-s)$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$\boxed{-8\mathbf{i} + 6\mathbf{j} – 16\mathbf{k}}$. # 偏微分方程和变分方法|Partial differential equations and variational methods 4M12代写2023 0 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辅导服务！ ## Instructions: 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. 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 # 代数与组合|Algebra & Combinatorics 06 25659代写 0 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辅导服务！ ## Instructions: 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. 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$1-p$. # 应用数学的数值方法代写|NUMERICAL METHODS FOR APPLIED MATHEMATICS MATH266代写 0 Assignment-daixieTM为您提供利物浦大学University of Liverpool NUMERICAL METHODS FOR APPLIED MATHEMATICS MATH266应用数学的数值方法代写代考辅导服务！ ## Instructions: While working exactly is often a good starting point, it is often not enough to solve the problem completely. In many cases, we encounter integrals that cannot be evaluated analytically, or equations that are too complex to solve by hand. This is where numerical methods come in, as they allow us to approximate solutions to these problems using iterative processes that are well-suited to computers. One of the advantages of numerical methods is that they allow us to handle real-world data that do not fit neatly into simple mathematical models. For example, if we have data from an experiment that we want to analyze, we may need to use numerical methods to fit a curve to the data or to approximate an integral that describes some aspect of the system being studied. When using numerical methods, it is important to be aware of how errors propagate through computations. Even small errors in initial data or in the numerical algorithms can lead to significant errors in the final result. Therefore, it is important to understand the sources of errors and to develop techniques for controlling and minimizing them. Some of the most important numerical methods that are commonly used in applied mathematics include methods for finding roots of equations, approximating integrals, and interpolating data. In each case, there are many different methods to choose from, each with its own advantages and disadvantages in terms of accuracy and efficiency. It is important to choose the right method for the problem at hand, taking into account factors such as the size of the problem, the desired level of accuracy, and the available computational resources. 问题 1. Biological signaling and regulation networks often involve cycles in which a protein backbone is transformed through a collection of modified states with different numbers of phosphate groups attached. A basic cycle might be described by the reaction network: $$\mathrm{A} \stackrel{k_1}{\rightarrow} \mathrm{B} \stackrel{k_2}{\rightarrow} \mathrm{C} \stackrel{k_3}{\rightarrow} \mathrm{A}$$ where A, B and$\mathrm{C}$have the same protein backbone with different numbers of phosphate groups. Of course, some kind of energy input is required to maintain a cycle, which is not represented above. Write down the stoichiometry matrix$\mathbf{S}$for this reaction network. 证明 .$\mathbf{S}=\left(\begin{array}{ccc}-1 & 0 & 1 \ 1 & -1 & 0 \ 0 & 1 & -1\end{array}\right)$问题 2. Characterize the null space of$\mathbf{S}$in terms of a dimension and a basis. What does this tell you about the fluxes (reaction rates) in the network at steady state? What physical interpretation can you provide for this? 证明 . The null space has dimension 1 and a basis:$(1,1,1)$. This indicates that the fluxes are equal at steady state. 问题 3. Characterize the left null space of$\mathbf{S}$in terms of a dimension and a basis. What does this tell you about the time evolution of the protein concentrations? What physical interpretation can you provide for this? 证明 . The left null space has dimension 1 and a basis:$(1,1,1). This indicates that the sum of the protein concentrations is constant in time. # 数学经济学|MATHEMATICAL ECONOMICS ECON113代写 0 Assignment-daixieTM为您提供利物浦大学University of Liverpool MATHEMATICAL ECONOMICS ECON113数学经济学代写代考辅导服务！ ## Instructions: 1. Calculus: Calculus is an important tool in mathematical economics, and students will need a strong foundation in differential and integral calculus. Topics covered may include limits, derivatives, optimization, partial derivatives, and integration. 2. Linear Algebra: Linear algebra is used extensively in economics, particularly in the study of systems of equations. Topics covered may include matrix algebra, determinants, vector spaces, and linear transformations. 3. Optimization: Optimization is a central theme in mathematical economics, and students will learn how to use calculus and linear algebra to solve optimization problems. Topics covered may include constrained optimization, Lagrange multipliers, and Kuhn-Tucker conditions. 4. Game Theory: Game theory is a mathematical framework for analyzing strategic interactions between individuals or groups. Topics covered may include the concept of Nash equilibrium, dominant strategies, and the prisoner’s dilemma. 5. Dynamic Optimization: Dynamic optimization involves studying how economic variables evolve over time, and how decisions made at one point in time affect outcomes in the future. Topics covered may include difference equations, differential equations, and optimal control theory. 6. Probability and Statistics: Probability and statistics are important tools for understanding uncertainty in economic models. Topics covered may include probability distributions, hypothesis testing, and regression analysis. These topics will provide students with a strong foundation in mathematical economics and prepare them for more advanced coursework in the field. 问题 1. Given a national income model as follows: \begin{aligned} & \mathrm{Y}=\mathrm{C}+\mathrm{I}_0+\mathrm{G}_0 \\ & \mathrm{C}=\mathrm{C}_0+\mathrm{b} \mathrm{Y}_{\mathrm{d}} \\ & \mathrm{T}=\mathrm{T}_0+\mathrm{tY}, \end{aligned} Where\mathrm{Y}=$income;$\mathrm{Y}_{\mathrm{d}}=$disposable income;$\mathrm{C}=$consumption;$\mathrm{C}_0=$autonomous consumption;$\mathrm{I}_0=$autonomous investment;$\mathrm{G}_0=$autonomous government expenditure;$\mathrm{T}=\operatorname{tax} ; \mathrm{T}_0=$autonomous tax;$\mathrm{b}$and$t$are the coefficients. a) Solve the model for the equilibrium national income ($\left.\mathrm{Y}^*\right)$证明 . a) Starting with the first equation, we can substitute in the expressions for consumption and taxes:$\begin{aligned} & \mathrm{Y}=\mathrm{C}0+\mathrm{b} \mathrm{Y}{\mathrm{d}}+\mathrm{I}_0+\mathrm{G}_0 \ & \mathrm{Y}=\mathrm{C}_0+\mathrm{b}\left(\mathrm{Y}-\mathrm{T}_0\right)+\mathrm{I}_0+\mathrm{G}_0\end{aligned}$Expanding and rearranging:$\begin{aligned} & \mathrm{Y}-\mathrm{bY}+\mathrm{bT} \mathrm{T}_0=\mathrm{C}_0+\mathrm{I}_0+\mathrm{G}_0 \ & (1-\mathrm{b}) \mathrm{Y}=\mathrm{C}_0+\mathrm{I}_0+\mathrm{G}_0-\mathrm{bT}_0 \ & \mathrm{Y}=\frac{\mathrm{C}_0+\mathrm{I}_0+\mathrm{G}_0-\mathrm{bT}}{1-\mathrm{b}}\end{aligned}$Thus, the equilibrium national income is:$\mathrm{Y}^*=\frac{150+200+350-0.65 \times 180}{1-0.65}=\frac{570}{0.35}=1628.57$问题 2. b) Using a), determine the government expenditure multiplier and explain its meaning. 证明 . The government expenditure multiplier is given by:$\frac{\Delta \mathrm{Y}^*}{\Delta \mathrm{G}_0}=\frac{1}{1-\mathrm{b}}$In this case, the multiplier is:$\frac{\Delta \mathrm{Y}^*}{\Delta \mathrm{G}_0}=\frac{1}{1-0.65}=2.857$This means that for every dollar increase in autonomous government expenditure, the equilibrium national income will increase by$2.857.

c) Now, given the following information that $\mathrm{b}=0.65 ; \mathrm{t}=0.25 ; \mathrm{C}_0=150 ; \mathrm{I}_0=200 ; \mathrm{G}_0=350$; and $\mathrm{T}_0=180$, calculate the equilibrium level of national income $\left(\mathrm{Y}^*\right)$, consumption $\left(\mathrm{C}^*\right)$, and taxation $\left(\mathrm{T}^*\right)$.

Using the equation for equilibrium national income derived in part a), we can calculate the equilibrium levels of consumption and taxation:

$\frac{\Delta \mathrm{Y}^*}{\Delta \mathrm{G}_0}=\frac{1}{1-0.65}=2.857$

Substituting in the given values and solving for consumption:

\begin{aligned} & 1628.57=\frac{\mathrm{C}^+200+350+\mathrm{T}^}{1-0.65} \ & \mathrm{C}^=\frac{(1-0.65) \times 1628.57-200-350-\mathrm{T}^}{1} \ & \mathrm{C}^=0.35 \times 1628.57-150-0.65 \times \mathrm{T}^=616.43-0.65 \times \mathrm{T}^*\end{aligned}

Next, using the equation for taxes:

$\mathrm{T}^=\mathrm{T}_0+\mathrm{tY}^=180+0.25 \times 1628.57=571.43$

Therefore, the equilibrium levels of national income, consumption, and taxation are:

\begin{aligned} & \mathrm{Y}^=1628.57 \ & \mathrm{C}^=616.43-0.65 \times 571.43=236.43 \ & \mathrm{~T}^*=571.43\end{aligned}