热力学代写 Thermodynamics代考2023

0
Posted on by

如果你也在 怎样代写热力学Thermodynamics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！

热力学代写Thermodynamics

热力学是物理学的一个分支，涉及热、功和温度，以及它们与能量、熵以及物质和辐射的物理特性的关系。这些数量的行为受热力学四大定律的制约，这些定律使用可测量的宏观物理量来传达定量描述，但可以用统计力学的微观成分来解释。热力学适用于科学和工程的各种主题，特别是物理化学、生物化学、化学工程和机械工程，但也适用于其他复杂的领域，如气象学。

热力学包含几个不同的主题，列举如下：

经典热力学Classical thermodynamics代写

经典热力学是对处于近平衡状态的热力学系统状态的描述，它使用宏观的、可测量的属性。它被用来模拟基于热力学定律的能量、功和热的交换。修饰语 “经典 “反映了这样一个事实，即它代表了对这一主题在19世纪发展的第一层理解，并以宏观经验（大尺度和可测量的）参数来描述一个系统的变化。这些概念的微观解释后来由统计力学的发展提供。

统计力学Statistical mechanics代写

统计力学，又称统计热力学，是随着19世纪末20世纪初原子和分子理论的发展而出现的，并以对单个粒子或量子力学状态之间的微观相互作用的解释补充了经典热力学。这一领域将单个原子和分子的微观属性与人类尺度上可以观察到的材料的宏观、批量属性联系起来，从而将经典热力学解释为统计学、经典力学和量子理论在微观层面的自然结果。

其他相关科目课程代写：

  • Chemical thermodynamics化学热力学
  • Equilibrium thermodynamics平衡热力学

热力学的历史

热力学是经典物理学和化学的一个分支，它研究和描述在一个热力学系统中，在涉及状态变量温度和能量变化的过程中，从热到功所引起的热力学转化。

经典热力学是基于宏观系统的概念，即在物理上或概念上与外部环境分离的一部分质量，为方便起见，通常假设它不会受到与系统的能量交换的干扰（孤立系统）：处于平衡状态的宏观系统的状态是由被称为热力学变量或状态函数的量来规定的，如温度、压力、体积和化学成分。化学热力学的主要符号已经由IUPAC建立。

热力学代写 Thermodynamics代考2023

 

Thermodynamics is a branch of classical physics and chemistry that studies and describes the thermodynamic transformations induced from heat to work in a thermodynamic system in processes involving changes in the state variables temperature and energy.

Classical thermodynamics is based on the concept of a macroscopic system, i.e. a part of the mass physically or conceptually separated from the external environment, which for convenience is usually assumed not to be disturbed by the exchange of energy with the system (isolated system): the state of a macroscopic system in equilibrium is specified by quantities called thermodynamic variables or state functions, such as temperature, pressure, volume and chemical composition. The main notations of chemical thermodynamics have been established by IUPAC.

热力学相关课后作业代写

问题 1.

A state function for a Van der Waals gas is given by an equation between thermodynamic variables that depend on model parameters $A, B$, and a physical constant $R$ :
$$
\left(P+\frac{A N^2}{V^2}\right)(V-N B)=N R T
$$
where $A N^2 / V^2$ is referred to as the internal pressure due to the attraction between molecules and $N B$ is an extra volume, sometimes associated with the the volume per molecule.

Write out a differential expression for $d N$ in terms of differentials of the thermodynamic variables.

证明 .

A state function for a Van der Waals gas is given by an equation between thermodynamic variables that depend on model parameters $A, B$, and a physical constant $R$ :
$$
\left(P+\frac{A N^2}{V^2}\right)(V-N B)=N R T
$$
where $A N^2 / V^2$ is referred to as the internal pressure due to the attraction between molecules and $N B$ is an extra volume, sometimes associated with the volume per molecule.

Write out a differential expression for $d N$ in terms of differentials of the thermodynamic variables.

The solution is pretty straightforward. One way is to differentiate the entire expression and group the terms corresponding to $d N, d P, d T$, and $d V$. Another way to do is by implicit differentiation. The real gas equation can be rewritten such that,
$$
\begin{aligned}
N & =N(T, V, P) \
d N & =\left(\frac{\partial N}{\partial P}\right){T, V} d P+\left(\frac{\partial N}{\partial V}\right){T, P} d V+\left(\frac{\partial N}{\partial T}\right){P, V} d T \end{aligned} $$ In an equivalent way, you could have written the function $P=P(V, T, N)$ and extract $d N$ from the following. $$ d P=\left(\frac{\partial P}{\partial N}\right){T, V} d N+\left(\frac{\partial P}{\partial V}\right){T, N} d V+\left(\frac{\partial P}{\partial T}\right){N, V} d T
$$
For instance, the first term $\left(\frac{\partial P}{\partial N}\right){T, V}$ can be evaluated as $$ \left(\frac{\partial P}{\partial T}\right){P, V}=\frac{N R}{V-N B}
$$
Using any one of the methods you would get
$$
d N=\frac{(V-B N) d P+\left(P-\left(A N^2 / V^2\right)+\left(2 A B N^3 / V^3\right)\right) d V-R N d T}{B P+R T+\left(3 A B N^2 / V^2\right)-(2 A N / V)}
$$


热力学课后作业代写的应用代写

一个 “热力学系统 “是宇宙的任何部分，人们对其作为调查对象感兴趣（宇宙的其余部分被称为环境）。这部分空间与宇宙的其余部分，即外部环境，被一个控制面（真实的或想象的表面，刚性的或可变形的）分开，是内部转变和与外部环境的物质或能量交换的场所。因此，这些交换本身导致了系统的转变，因为它从一个起始条件到一个不同的条件。在实践中，当一个系统从最初的平衡状态到最终的平衡状态时，它就被转化了。另一方面，环境保持近似 “不变”，因为系统相对于它是如此之小，以至于能量或物质的交换对于环境中相同的整体来说是不相关的，否则我们就不是在谈论环境，而是在谈论另一个系统（根据定义，环境并不对应于它）。

英国论文代写Viking Essay为您提供热力学作业代写Thermodynamics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。

实分析|MATH0051 Analysis 4: Real Analysis代写2023

0
Posted on by

Assignment-daixieTM为您提供伦敦大学学院 London’s Global University MATH0051 Analysis 4: Real Analysis实分析代写代考和辅导服务！

Instructions:

实分析|MATH0051 Analysis 4: Real Analysis代写2023

问题 1.

Let $R \in \mathbb{R}$, and let $f:[R, \infty) \rightarrow \mathbb{R}$. We say that $f(x)$ converges to $L$ as $x \rightarrow \infty$ if for every $\epsilon>0$ there exists $M \geq R$ such that for all $x \geq M$ we have $|f(x)-L|<\epsilon$. We write $f(x) \rightarrow L$ as $x \rightarrow \infty$ or $$ \lim _{x \rightarrow \infty} f(x)=L $$ [A similar definition can be formulated for limits as $x \rightarrow-\infty$ but we will not do so here.]

(a) Prove that $$ \lim _{x \rightarrow \infty} \frac{x^2}{x^2+1}=1 $$

证明 .

(a) To prove that $\lim_{x \rightarrow \infty} \frac{x^2}{x^2+1} = 1$, we need to show that for every $\epsilon > 0$, there exists $M \geq R$ such that $|f(x)-1| < \epsilon$ for all $x \geq M$.

Let $\epsilon > 0$ be given. We want to find $M \geq R$ such that $\left| \frac{x^2}{x^2+1} – 1 \right| = \frac{1}{x^2+1} < \epsilon$ for all $x \geq M$. This is equivalent to $\frac{1}{\epsilon} – 1 < x^2+1$, or $x^2 > \frac{1-\epsilon}{\epsilon}$. So we can choose $M$ to be any real number greater than $\sqrt{\frac{1-\epsilon}{\epsilon}}$. Then for all $x \geq M$, we have $\left| \frac{x^2}{x^2+1} – 1 \right| < \epsilon$, which proves that $\lim_{x \rightarrow \infty} \frac{x^2}{x^2+1} = 1$.

问题 2.

(b) Prove that $$ \lim _{x \rightarrow \infty} \sin x $$ does not exist.

证明 .

(b) To prove that $\lim_{x \rightarrow \infty} \sin x$ does not exist, we need to show that there exist two different values $L_1$ and $L_2$ such that for every $M \geq R$, there exist $x_1, x_2 \geq M$ such that $|\sin x_1 – L_1| \geq \epsilon$ and $|\sin x_2 – L_2| \geq \epsilon$ for some $\epsilon > 0$.

Let $L_1 = 1$ and $L_2 = -1$. Let $\epsilon = \frac{1}{2}$ be given. For any $M \geq R$, we can choose $x_1 = \left(2k+\frac{1}{4}\right)\pi$ and $x_2 = \left(2k+\frac{3}{4}\right)\pi$ for some integer $k$ such that $x_1, x_2 \geq M$. Then we have $\sin x_1 = 1$ and $\sin x_2 = -1$, so $|\sin x_1 – L_1| = |\sin x_2 – L_2| = 0.5 \geq \epsilon$. Therefore, $\lim_{x \rightarrow \infty} \sin x$ does not exist.

问题 3.

Let $S \subset \mathbb{R}$. We say that $f: S \rightarrow \mathbb{R}$ is Lipschitz continuous on $S$ if there exists $L \geq 0$ such that for all $x, y \in S$
$$
|f(x)-f(y)| \leq L|x-y|
$$
Prove that if $f: S \rightarrow \mathbb{R}$ is Lipschitz continuous on $S$ then $f$ is uniformly continuous on $S$.

证明 .

To prove that $f$ is uniformly continuous on $S$, we need to show that for any $\epsilon>0$, there exists a $\delta>0$ such that for all $x,y \in S$ with $|x-y|<\delta$, we have $|f(x)-f(y)|<\epsilon$.

Since $f$ is Lipschitz continuous on $S$, we know that there exists $L\geq 0$ such that $|f(x)-f(y)|\leq L|x-y|$ for all $x,y\in S$.

Now, let $\epsilon>0$ be given. Choose $\delta=\frac{\epsilon}{L}$. Then, for any $x,y\in S$ with $|x-y|<\delta$, we have

$|f(x)-f(y)| \leq L|x-y|<L \cdot \frac{\epsilon}{L}=\epsilon$.

Thus, we have shown that for any $\epsilon>0$, there exists a $\delta>0$ such that for all $x,y\in S$ with $|x-y|<\delta$, we have $|f(x)-f(y)|<\epsilon$. Therefore, $f$ is uniformly continuous on $S$.

这是一份2023年的伦敦大学学院 MATH0051 Analysis 4: Real Analysis实分析代写的成功案例

数学方法|MATH0016 Mathematical Methods 3代写2023

0
Posted on by

Assignment-daixieTM为您提供伦敦大学学院 London’s Global University MATH0016 Mathematical Methods 3数学方法代写代考和辅导服务！

Instructions:

To summarize, the course covers the following topics:

(a) Fourier theory: This topic deals with decomposing a periodic function into a sum of sine and cosine functions, which can help analyze and understand the behavior of the function over time. The Fourier transform is a powerful tool used in signal processing, image analysis, and many other fields.

(b) The calculus of variations: This is a branch of mathematics that deals with finding functions that maximize or minimize an integral, subject to certain conditions. This topic has applications in many fields, including physics, engineering, and economics.

(c) Partial differential equations: These are equations that involve partial derivatives of a function, and they are commonly used to describe physical phenomena such as fluid dynamics and heat transfer. The course will cover both linear and quasilinear partial differential equations of the first and second order.

(d) Vector calculus: This topic deals with the manipulation of vector fields, including the divergence and curl operations. These operations are useful in many areas of physics and engineering, including electromagnetism and fluid dynamics. The course will also cover the divergence and Stokes’ theorem.

I hope this helps clarify what you’ll be studying in the course! Let me know if you have any questions or if there’s anything else I can help you with.

数学方法|MATH0016 Mathematical Methods 3代写2023

问题 1.

Solve the initial value problem
$$
y^{\prime} y^{\prime \prime}-t=0, \quad y(1)=2, \quad y^{\prime}(1)=1 .
$$

证明 .

We can begin by using the method of integration by parts to integrate $y’y”$. Let $u = y’$ and $v’ = y”$, so that $u’ = y”$ and $v = y’$. Then we have

$y^{\prime} y^{\prime \prime}=u v^{\prime}=\frac{1}{2}\left(u^2\right)^{\prime}$.

Using this, we can rewrite the differential equation as

$\frac{1}{2}\left(y^{\prime 2}\right)^{\prime}-t=0$

Integrating once with respect to $t$, we get

$\frac{1}{2} y^{\prime 2}-\frac{1}{2} t^2=C_1$,

where $C_1$ is a constant of integration. We can then solve for $y’$ to get

$y^{\prime}= \pm \sqrt{t^2+2 C_1}$.

Using the initial condition $y(1) = 2$, we get

$y^{\prime}(1)= \pm \sqrt{1+2 C_1}=1$

Solving for $C_1$, we find that $C_1 = 0$. Thus, we have

$y^{\prime}= \pm \sqrt{t^2}, \quad y(1)=2$.

Since $y$ is increasing, we take the positive sign, giving

$y^{\prime}=t, \quad y(1)=2$.

Integrating with respect to $t$, we get

$y=\frac{1}{2} t^2+C_2$.

Using the initial condition $y(1) = 2$, we get $C_2 = \frac{3}{2}$, so the solution to the initial value problem is

$y=\frac{1}{2} t^2+\frac{3}{2}$

问题 2.

Consider the differential equation $y^{\prime}=y(5-y)(y-4)^2$. (a)Determine the critical points (stationary solutions).

证明 .

To determine the critical points, we need to find the values of $y$ for which $y’=0$. Thus, we solve the equation:

$y^{\prime}=y(5-y)(y-4)^2=0$

This equation is satisfied when $y=0,5$, or $4$. Therefore, the critical points are $y=0,5,$ and $4$.

问题 3.

(b)Sketch the graph of $f(y)=y(5-y)(y-4)^2$.

证明 .

To sketch the graph of $f(y)=y(5-y)(y-4)^2$, we can use the critical points found in part (a) and the behavior of $f(y)$ as $y$ approaches positive or negative infinity.

First, we note that $f(y)$ changes sign at $y=0,4,$ and $5$. We can determine the sign of $f(y)$ on each interval by testing a point in each interval. For example, for $y<0$, we can test $y=-1$, and we get:

$f(-1)=(-1)(5-(-1))(4-(-1))^2=-120<0$.

Therefore, $f(y)$ is negative on $(-\infty,0)$.

Next, we can consider the behavior of $f(y)$ as $y$ approaches infinity or negative infinity. Since the highest power of $y$ in $f(y)$ is $y^4$, we know that $f(y)$ approaches infinity as $y$ approaches positive or negative infinity.

Putting all of this together, we can sketch the graph of $f(y)$ as follows:

The graph has a local maximum at $y=4$ and a local minimum at $y=5$. The critical point $y=0$ is a double root.

这是一份2023年的伦敦大学学院 MATH0016 Mathematical Methods 3数学方法代写的成功案例

进一步线性代数|MATH0014 Algebra 3: Further Linear Algebra代写2023

0
Posted on by

Assignment-daixieTM为您提供伦敦大学学院 London’s Global University MATH0014 Algebra 3: Further Linear Algebra进一步线性代数代写代考和辅导服务！

Instructions:

Linear algebra is a fundamental subject in mathematics that has many applications in other fields such as physics, engineering, computer science, and economics.

In this course, you will deepen your understanding of linear algebra by exploring polynomial rings over fields, which are important in algebraic geometry and coding theory. You will also study matrix diagonalizability and the Jordan normal form, which are important for understanding the behavior of linear transformations and systems of linear equations.

In addition, you will learn about linear and bilinear forms, which are functions that take vectors as inputs and produce scalars as outputs. These concepts are important in the study of linear transformations and inner product spaces, which are vector spaces equipped with an additional structure that allows you to measure angles and distances between vectors.

Overall, this course will provide you with a deeper understanding of the key concepts of linear algebra and their applications, which will be useful in a wide range of fields.

进一步线性代数|MATH0014 Algebra 3: Further Linear Algebra代写2023

问题 1.

Suppose $A$ is the matrix
$$
A=\left[\begin{array}{llll}
0 & 1 & 2 & 2 \
0 & 3 & 8 & 7 \
0 & 0 & 4 & 2
\end{array}\right]
$$
(a) Find all special solutions to $A x=0$ and describe in words the whole nullspace of $A$.

证明 .

(a) Find all special solutions to $A x=0$ and describe in words the whole nullspace of $A$.
Solution: First, by row reduction
$$
\left[\begin{array}{llll}
0 & 1 & 2 & 2 \
0 & 3 & 8 & 7 \
0 & 0 & 4 & 2
\end{array}\right] \rightarrow\left[\begin{array}{llll}
0 & 1 & 2 & 2 \
0 & 0 & 2 & 1 \
0 & 0 & 4 & 2
\end{array}\right] \rightarrow\left[\begin{array}{llll}
0 & 1 & 0 & 1 \
0 & 0 & 2 & 1 \
0 & 0 & 0 & 0
\end{array}\right] \rightarrow\left[\begin{array}{llll}
0 & 1 & 0 & 1 \
0 & 0 & 1 & \frac{1}{2} \
0 & 0 & 0 & 0
\end{array}\right]
$$
so the special solutions are
$$
s_1=\left[\begin{array}{l}
1 \
0 \
0 \
0
\end{array}\right], s_2=\left[\begin{array}{c}
0 \
-1 \
-\frac{1}{2} \
1
\end{array}\right]
$$
Thus, $N(A)$ is a plane in $\mathbb{R}^4$ given by all linear combinations of the special solutions.

问题 2.

(b) Describe the column space of this particular matrix $A$. “All combinations of the four columns” is not a sufficient answer.

证明 .

(b) Describe the column space of this particular matrix $A$. “All combinations of the four columns” is not a sufficient answer.

Solution: $C(A)$ is a plane in $\mathbb{R}^3$ given by all combinations of the pivot columns, namely
$$
c_1\left[\begin{array}{l}
1 \
3 \
0
\end{array}\right]+c_2\left[\begin{array}{l}
2 \
8 \
4
\end{array}\right]
$$

问题 3.

(c) What is the reduced row echelon form $R^*=\operatorname{rref}(B)$ when $B$ is the 6 by 8 block matrix
$$
B=\left[\begin{array}{cc}
A & A \
A & A
\end{array}\right] \text { using the same } A \text { ? }
$$

证明 .

(c) What is the reduced row echelon form $R^*=\operatorname{rref}(B)$ when $B$ is the 6 by 8 block matrix
$$
B=\left[\begin{array}{cc}
A & A \
A & A
\end{array}\right] \text { using the same } A \text { ? }
$$
Solution: Note that $B$ immediately reduces to
$$
B=\left[\begin{array}{cc}
A & A \
0 & 0
\end{array}\right]
$$
We reduced $A$ above: the row reduced echelon form of of $B$ is thus
$$
B=\left[\begin{array}{rr}
\operatorname{rref}(A) & \operatorname{rref}(A) \
0 & 0
\end{array}\right], \operatorname{rref}(A)=\left[\begin{array}{llll}
0 & 1 & 0 & 1 \
0 & 0 & 1 & \frac{1}{2} \
0 & 0 & 0 & 0
\end{array}\right]
$$

这是一份2023年的伦敦大学学院 MATH0014 Algebra 3: Further Linear Algebra进一步线性代数代写的成功案例

流体力学|MATH0015 Fluid Mechanics代写2023

0
Posted on by

Assignment-daixieTM为您提供伦敦大学学院 London’s Global University MATH0015 Fluid Mechanics流体力学代写代考和辅导服务！

Instructions:

Fluid Mechanics is the study of fluids, which can be either liquids or gases, and their behavior when subjected to various forces and conditions. It is a branch of physics and engineering that deals with the motion and equilibrium of fluids, as well as their interactions with solid objects.

Fluid mechanics involves the study of topics such as fluid statics, which deals with fluids at rest, and fluid dynamics, which deals with fluids in motion. Fluid mechanics also encompasses topics such as viscosity, turbulence, and boundary layers, which are important in understanding the behavior of fluids.

Applications of fluid mechanics are found in various fields, including aerospace engineering, civil engineering, chemical engineering, mechanical engineering, and biomedical engineering. Understanding fluid mechanics is important for the design and analysis of many devices and systems, such as pumps, turbines, pipes, and blood flow in the human body.

Overall, fluid mechanics plays an important role in understanding the behavior of fluids and their applications in various fields of engineering and science.

流体力学|MATH0015 Fluid Mechanics代写2023

问题 1.

Consider a sinusoidal traveling wave of small amplitude $(\mathrm{Ak}<<1)$ such that the free surface is given by: $$ \eta=A \cos (\mathrm{kx}-\omega t) $$ The corresponding velocities are given by: $$ \mathrm{u}=\mathrm{A} \omega \mathrm{e}^{\mathrm{ky}} \cos (\mathrm{kx}-\omega t) $$ and $$ v=A \omega e^{k y} \sin (k x-\omega t) $$ ( $\mathrm{y}$ is the vertical coordinate and is negative in the direction of $\mathrm{g} ; \mathrm{x}$ is the horizontal coordinate) a) Sketch the streamlines at $\mathrm{t}=\mathrm{o}$ that pass through: $\mathrm{x}=o, \mathrm{y}=o ; \mathrm{x}=0, \mathrm{y}=\frac{-2 \pi}{k}$

证明 .

a) To sketch the streamlines at $t=0$ that pass through $(x,y)=(0,0)$ and $(x,y)=(0,-\frac{2\pi}{k})$, we need to plot a few representative particle trajectories.

The particle trajectory can be obtained by integrating the velocity vector along the path. For small amplitude waves, we can assume that the fluid motion is irrotational, and thus the streamlines are parallel to the velocity vector. Therefore, the particle trajectory will be perpendicular to the streamline at each point.

Let’s start with the point $(0,0)$. The velocity vector at this point is given by $(u,v)=(A\omega,0)$, so the streamline will be a horizontal line. The particle trajectory passing through $(0,0)$ will be a vertical line, as shown below:

Now, let’s consider the point $(0,-\frac{2\pi}{k})$. The velocity vector at this point is given by $(u,v)=(0,A\omega e^{-2\pi})$. Since $v$ is negative, the particle will move downwards along the streamline. The streamline itself can be obtained by integrating the velocity vector, which gives:

$\frac{d y}{d x}=\frac{v}{u}=-\frac{e^{-2 \pi}}{k}$

Integrating this expression yields:

$y=-\frac{e^{-2 \pi}}{k} x+C$

where $C$ is a constant. We can determine $C$ by setting $x=0$ and $y=-\frac{2\pi}{k}$, which gives $C=-\frac{2\pi}{k}$. Therefore, the streamline passing through $(0,-\frac{2\pi}{k})$ is given by:

$y=-\frac{e^{-2 \pi}}{k} x-\frac{2 \pi}{k}$

The particle trajectory passing through $(0,-\frac{2\pi}{k})$ will be perpendicular to this streamline and can be obtained by integrating the velocity vector along the streamline. The resulting particle trajectories are shown below:

问题 2.

b) Sketch the particle line at $\mathrm{t}=\frac{2 \pi}{\omega}$ for the particle which was at $\mathrm{x}=\mathrm{o}, \mathrm{y}=\mathrm{o}$ at $\mathrm{t}=\mathrm{o}$

证明 .

b) To sketch the particle line at $t=\frac{2\pi}{\omega}$ for the particle which was at $(x,y)=(0,0)$ at $t=0$, we need to find the position of the particle at $t=\frac{2\pi}{\omega}$.

The particle position can be obtained by integrating the velocity vector along the particle trajectory. For a particle starting at $(0,0)$, the particle trajectory is a vertical line, as shown in part (a). The velocity vector along this trajectory is given by $(u,v)=(A\omega e^{ky},0)$. Integrating this expression from $y=0$ to $y=-\frac{2\pi}{k}$ and from $t=0$ to $t=\frac{2\pi}{\omega}$ gives:

$x=0 \quad$ and $\quad y=-\frac{2 \pi}{k}$

Therefore, the particle that was at $(x,y)=(0,0)$ at $t=0$ will be at $(x,y)=(0,-\frac{2\pi}{k})$ at $t=\frac{2\pi}{\omega}$

问题 3.

For an ideal rectilinear vortex, the velocity profile is given by $\mathrm{v}_{\mathrm{r}}=\mathrm{v}_{\mathrm{z}}=0 ; \mathrm{v}_\theta=\mathrm{K} / \mathrm{r}$. $A$ qualitatively similar result is associated with an ideal smoke ring. a) Why does a smoke ring propel itself? Why does the velocity of translation diminish with time?

证明 .

A smoke ring is a toroidal vortex ring that propagates through the air. It is created by a sudden release of fluid, such as smoke, from a circular orifice. The smoke ring propels itself because it is a closed loop of fluid that has rotational motion. The fluid inside the smoke ring rotates in the same direction as the ring itself, creating a low-pressure area in the center of the ring.

As the smoke ring moves forward, the low-pressure area at the center of the ring pulls in air from behind, which helps to maintain the shape of the ring. This process of entraining air into the ring creates a drag force, which slows down the velocity of the smoke ring over time.

The velocity of translation of the smoke ring diminishes with time because of the loss of energy due to drag forces. As the smoke ring moves through the air, it creates turbulence and friction with the surrounding air, which causes the fluid within the ring to lose energy. This loss of energy translates into a decrease in the velocity of translation of the smoke ring over time. Eventually, the smoke ring will come to a stop as all of its kinetic energy is dissipated through friction with the air.

这是一份2023年的伦敦大学学院 MATH0015 Fluid Mechanics流体力学代写的成功案例

光学代写 optics代考2023

0
Posted on by

如果你也在 怎样代写热力学Thermodynamics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！

光学代写optics

光学是研究光的行为和属性的物理学分支，包括它与物质的相互作用以及使用或探测它的仪器的构造。光学通常描述可见光、紫外光和红外光的行为。因为光是一种电磁波，其他形式的电磁辐射，如X射线、微波和无线电波也表现出类似的特性。

光学包含几个不同的主题，列举如下：

几何光学Geometrical optics代写

几何光学，或称射线光学，是一种用射线来描述光的传播的光学模型。几何光学中的光线是一个抽象的概念，有助于近似地描述光线在某些情况下的传播路径。

量子光学Quantum optics代写

量子光学是原子、分子和光学物理学的一个分支，处理单个光量子（称为光子）如何与原子和分子互动的问题。它包括研究光子的类似粒子的特性。光子已被用来测试量子力学的许多反直觉预测，如纠缠和远程传输，并且是量子信息处理的有用资源。

其他相关科目课程代写：

  • 反射（物理学）Reflection (physics)
  • 折射Refraction

光学的历史

光学的历史是科学史的一部分。光学一词来自古希腊语的τα ὀπτικά。它最初是指与眼睛有关的一切科学。希腊人将光学从dioptrics和catoptrics中区分出来。我们可能会把第一种科学称为视觉，第二种科学称为镜片，第三种科学称为镜子。希腊光学的伟大名字是欧几里德、亚历山大的赫伦和托勒密。

自古代以来，光学经历了许多发展。这个词的含义已经发生了变化，从对视觉的研究到对光的研究，经历了几个阶段，最近才被纳入到更广泛的物理学体系中。

光学代写 optics代考2023

 

The history of optics is a part of the history of science. The term optics comes from the ancient Greek τα ὀπτικά. It is originally the science of everything related to the eye. The Greeks distinguish optics from dioptrics and catoptrics. We would probably call the first science of vision, the second science of lenses and the third science of mirrors. The great names of Greek optics are Euclid, Heron of Alexandria and Ptolemy.

Since antiquity, optics has undergone many developments. The very meaning of the word has varied and from the study of vision, it has passed in several stages to the study of light, before being incorporated recently into a broader body of physics.

光学相关课后作业代写

问题 1.

A glass plate is sprayed with uniform opaque particles. When a distant point source of light is observed looking through the plate, a diffuse halo is seen whose angular width is about $2^{\circ}$. Estimate the size of the particles. (Hint: consider Fraunhoffer diffraction through random gratings, and use Babinet’s principle)

证明 .

The diffraction pattern of an opaque circular particle is complementary to that due to circular apertures of the same size in an otherwise opaque screen.
Under the Fraunhofer condition $\left(\frac{k\left(x^{\prime 2}+y^{\prime 2}\right)}{2 z} \ll 1, \frac{k\left(x^2+y^2\right)}{2 z} \ll 1\right)$
$$
\begin{aligned}
& E\left(x^{\prime}, y^{\prime}\right) \approx \frac{1}{z} \iint \exp \left(-i k\left(\theta_{x^{\prime}} x+\theta_{y^{\prime}} y\right)\right) t(x, y) E(x, y) d x d y \
& \text { Where } \theta_{x^{\prime}} \approx \frac{x \prime}{z}, \theta_{y^{\prime}} \approx \frac{y \prime}{z} \
&
\end{aligned}
$$
For the given problem, we may further assume $\mathrm{E}(\mathrm{x}, \mathrm{y})$ is a plane wave at normal incidence, and the transmission function $t(x, y)$ for a single can be expressed as:
$$
t(x, y)=1-\operatorname{circ}\left(\frac{\sqrt{x^2+y^2}}{R}\right)
$$
Where $R$ is the radius of the opaque particles.
$$
\begin{gathered}
E\left(x^{\prime}, y^{\prime}\right) \approx \frac{1}{z} \iint \exp \left(-i k\left(\theta_{x^{\prime}} x+\theta_{y^{\prime}} y\right)\right)\left[1-\operatorname{circ}\left(\frac{\sqrt{x^2+y^2}}{R}\right)\right] d x d y \
E\left(x^{\prime}, y^{\prime}\right) \approx \frac{1}{z} \mathcal{F}\left[1-\operatorname{circ}\left(\frac{\sqrt{x^2+y^2}}{R}\right)\right] \
\text { With } x^{\prime}=\frac{z}{k} k_x, y^{\prime}=\frac{z}{k} k_y \
E\left(k_x, k_y\right) \approx \frac{1}{z}\left[\delta\left(\sqrt{k_x{ }^2+k_y{ }^2}\right)-|R|^2 \frac{2 \pi J_1\left(R \sqrt{k_x^2+k_y{ }^2}\right)}{R \sqrt{k_x{ }^2+k_y{ }^2}}\right]
\end{gathered}
$$

Where $\gamma=R \sqrt{k_x^2+k_y^2}=\frac{2 \pi}{\lambda} R \theta$
From the above table,
$$
\frac{2 \pi}{\lambda} R \Delta \theta=7.106-3.832=3.274
$$
Taking central wavelength at visible frequency, $\lambda=500 \mathrm{~nm}$ and given $\Delta \theta=2^{\circ}$, we find the radius of the particle:
$$
R=\lambda \frac{3.274}{(2 \pi)^2\left(\frac{\Delta \theta}{360}\right)}=500 \mathrm{~nm} \times\left(\frac{3.274}{(2 \pi)^2 \frac{2}{360}}\right)=7463 \mathrm{~nm}=7.4 \mu \mathrm{m}
$$


光学课后作业代写的应用代写

大多数的光学现象都可以用经典的光的电磁描述来解释，然而完整的光的电磁描述往往很难在实践中应用。实用光学通常使用简化模型。其中最常见的是几何光学，它将光视为直线传播的光线的集合，当它们通过或从表面反射时，会发生弯曲。物理光学是一个更全面的光的模型，它包括波的效应，如衍射和干涉，这些效应在几何光学中是无法解释的。从历史上看，基于射线的光的模型首先被开发出来，然后是光的波模型。19世纪电磁理论的进步使人们发现光波实际上是电磁辐射。

英国论文代写Viking Essay为您提供光学作业代写optics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。

电磁学代写 Electromagnetism代考2023

0
Posted on by

如果你也在 怎样代写电磁学Electromagnetism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！

电磁学代写Electromagnetism

在物理学中，电磁力是具有电荷的粒子之间通过电磁场发生的一种相互作用。电磁力是自然界的四种基本力之一。它是原子和分子相互作用中的主导力量。电磁学可以被认为是静电和磁力的结合，这是两种不同但又紧密相连的现象。电磁力发生在任何两个带电粒子之间，造成带相反电荷的粒子之间的吸引和带相同电荷的粒子之间的排斥，而磁力是专门发生在相对运动的带电粒子之间的互动。这两种效应结合在一起，在带电粒子附近产生电磁场，它可以通过洛伦兹力加速其他带电粒子。在高能量下，弱力和电磁力被统一为单一的弱电力。

电磁学包含几个不同的主题，列举如下：

基本相互作用Fundamental interaction代写

在物理学中，基本相互作用，也被称为基本力，是指似乎不能还原为更基本的相互作用。目前已知存在四种基本相互作用：引力和电磁相互作用，它们产生重要的长程力，其影响可以在日常生活中直接看到；强相互作用和弱相互作用，它们在微小的亚原子距离上产生力，并支配核相互作用。一些科学家假设可能存在第五种力，但这些假设仍然是推测性的。

经典电磁学Classical electrodynamics代写

1600年，威廉-吉尔伯特在他的De Magnete中提出，电和磁虽然都能引起物体的吸引和排斥，但却是不同的效果。海员们注意到，雷击有能力干扰罗盘针。直到本杰明-富兰克林在1752年提出的实验，法国的托马斯-弗朗索瓦-达利巴德在1752年5月10日用一根40英尺高（12米）的铁棒代替风筝进行了实验，他成功地从云中提取了电火花，这才证实了闪电和电力之间的联系。

其他相关科目课程代写：

  • Nonlinear system非线性系统
  • Magnetohydrodynamics磁流体力学

电磁学的历史

对这一现象最早的研究可能要追溯到希腊哲学家泰勒斯（公元前600年），他研究了琥珀的电学特性，琥珀是一种化石树脂，在摩擦时能吸引其他物质：它的希腊名字是elektron（ἤλεκτρον），而 “电 “这个词就来源于此。古希腊人意识到，琥珀能够吸引轻的物体，如头发，而且反复摩擦琥珀本身甚至可以产生火花。

电磁学代写 Electromagnetism代考2023

The earliest study of this phenomenon probably goes back to the Greek philosopher Thales (600 BC), who studied the electrical properties of amber, a fossil resin that attracts other substances when rubbed: its Greek name is elektron (ἤλεκτρον), from which the word ‘electricity’ is derived. The ancient Greeks realised that amber could attract light objects, such as hair, and that repeated rubbing of the amber itself could even produce sparks.

电磁学相关课后作业代写

问题 1.

Show that $S^4$ has no symplectic structure. Show that $S^2 \times S^4$ has no symplectic structure.

证明 .

To show that $S^4$ has no symplectic structure, we will use the following fact from symplectic geometry: a compact symplectic manifold has even dimension.

Suppose that $S^4$ has a symplectic structure. Then $S^4$ is a compact symplectic manifold, so its dimension must be even. However, the dimension of $S^4$ is $4$, which is not even. Therefore, $S^4$ cannot have a symplectic structure.

To show that $S^2 \times S^4$ has no symplectic structure, we will use the following fact: the product of two symplectic manifolds is symplectic if and only if both factors have even dimension.

Suppose that $S^2 \times S^4$ has a symplectic structure. Then both $S^2$ and $S^4$ are symplectic manifolds, so their dimensions must both be even. However, the dimension of $S^2$ is $2$, which is not even. Therefore, $S^2 \times S^4$ cannot have a symplectic structure.


电磁学课后作业代写的应用代写

电磁力是日常生活中观察到的许多化学和物理现象的原因。原子核和其电子之间的静电吸引力将原子固定在一起。电力还允许不同的原子结合成分子，包括构成生命基础的大分子，如蛋白质。同时，电子的自旋和角动量磁矩之间的磁相互作用也在化学反应性中起作用；自旋化学中研究了这种关系。电磁学在现代技术中也起着至关重要的作用：电能的生产、转化和分配，光、热和声音的产生和检测，光纤和无线通信，传感器，计算，电解，电镀和机械马达和执行器。

英国论文代写Viking Essay为您提供电磁学作业代写Electromagnetism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。

复分析|MATH0013 Analysis 3: Complex Analysis代写2023

0
Posted on by

Assignment-daixieTM为您提供伦敦大学学院 London’s Global University MATH0013 Analysis 3: Complex Analysis复分析代写代考和辅导服务！

Instructions:

Complex analysis is a branch of mathematics that studies complex functions, which are functions that have complex numbers as their inputs and outputs. It is an important area of study because complex functions have many interesting and useful properties that are not shared by real-valued functions.

Starting from complex numbers is a good way to introduce the subject, as it allows you to develop the algebraic and geometric properties of complex numbers that are essential for understanding complex functions. Cauchy’s theorem and Cauchy’s integral formulae are fundamental results in complex analysis that relate the values of a complex function inside a closed contour to its values on the boundary of the contour. These theorems have many important applications, such as evaluating definite integrals and finding the number of zeros of a complex polynomial in a region.

The theorem of residues is another important result in complex analysis that allows you to compute certain types of integrals using the residues of a function. This theorem is particularly useful for evaluating integrals involving trigonometric and logarithmic functions.

Laurent’s theorem is a generalization of Taylor’s theorem for complex functions, which allows you to express a complex function as a power series with both positive and negative powers of the complex variable. This theorem is useful for studying the singularities of a complex function, which are points where the function is not defined or behaves in an unusual way.

Overall, the course on complex functions seems to cover a lot of important topics in complex analysis that have many applications in mathematics and other fields.

数学方法|MATH0010 Mathematical Methods 1代写2023

问题 1.

Let $G$ be an undirected graph with $n$ vertices. Then a Hamilton path is a simple path in $G$ that visits each vertex once (i.e., has $n$ vertices and $n-1$ edges), while a Hamilton cycle is a simple cycle in $G$ that visits each vertex once (i.e., has $n$ vertices and $n$ edges). Let HAMPATH and HAMCYCLE be the problems of deciding whether $G$ has a Hamilton path and Hamilton cycle respectively, given $G$ as input. (a) Show that if $G$ has a Hamilton cycle, then $G$ also has a Hamilton path.

证明 .

(a) If $G$ has a Hamilton cycle, then we can simply remove one edge from the cycle to obtain a Hamilton path. Since a Hamilton cycle has $n$ edges, removing one edge from it leaves a path with $n-1$ edges that visits every vertex exactly once. Therefore, if $G$ has a Hamilton cycle, it also has a Hamilton path.

问题 2.

(b) Give an example of a graph $G$ that has a Hamilton path but no Hamilton cycle.

证明 .

(b) Consider the graph $G$ consisting of three vertices $a,b,c$ and three edges $(a,b),(b,c),(c,a)$. This graph has a Hamilton path $a\rightarrow b\rightarrow c$, but no Hamilton cycle. To see this, note that any cycle in $G$ must contain all three vertices, and therefore must have length at least 3. But the only cycle in $G$ is of length 3, and it visits each vertex twice.

问题 3.

(c) Give a polynomial-time reduction from $H A M C Y C L E$ to $H A M P A T H$.

证明 .

(c) To reduce $HAMCYCLE$ to $HAMPATH$, we can simply remove one vertex $v$ from $G$ and ask whether the resulting graph $G’$ has a Hamilton path. If $G$ has a Hamilton cycle, then $G’$ has a Hamilton path by part (a). Conversely, if $G’$ has a Hamilton path, we can add the vertex $v$ back into the path to obtain a Hamilton cycle in $G$. This reduction takes polynomial time since we only need to remove one vertex and its adjacent edges, which can be done in constant time per vertex.

这是一份2023年的伦敦大学学院 MATH0013 Analysis 3: Complex Analysis复分析代写的成功案例

数学方法|MATH0010 Mathematical Methods 1代写2023

0
Posted on by

Assignment-daixieTM为您提供伦敦大学学院 London’s Global University MATH0010 Mathematical Methods 1数学方法代写代考和辅导服务！

Instructions:

Mathematical methods refer to the techniques and tools used in mathematics to solve problems and explore mathematical concepts. These methods can be used in various fields such as physics, engineering, economics, and computer science. Some of the key mathematical methods include:

  1. Calculus: This is a branch of mathematics that deals with functions, limits, derivatives, integrals, and infinite series. Calculus is widely used in physics, engineering, and economics.
  2. Linear algebra: This is a branch of mathematics that deals with linear equations, matrices, determinants, and vector spaces. Linear algebra is widely used in fields such as physics, engineering, computer science, and economics.
  3. Probability theory: This is a branch of mathematics that deals with the study of random events and the likelihood of their occurrence. Probability theory is widely used in statistics, finance, and engineering.
  4. Differential equations: This is a branch of mathematics that deals with equations that involve derivatives. Differential equations are widely used in physics, engineering, and economics.
  5. Fourier analysis: This is a branch of mathematics that deals with the study of periodic functions and their representation as a sum of simpler functions. Fourier analysis is widely used in signal processing, image processing, and data analysis.
  6. Numerical analysis: This is a branch of mathematics that deals with the development of algorithms and methods for solving mathematical problems using computers. Numerical analysis is widely used in scientific computing and engineering.

These methods are just a few examples of the many mathematical tools available to solve problems and explore concepts in various fields.

数学方法|MATH0010 Mathematical Methods 1代写2023

问题 1.

Assume that a velocity field, $u=u(x, t)$ exists. Show that the acceleration of an individual car is given by
$$
\frac{\partial u}{\partial t}+u \frac{\partial u}{\partial x}
$$

证明 .

To derive the acceleration of an individual car, we can start with the definition of acceleration, which is the rate of change of velocity with respect to time, i.e.,

$a=\frac{d u}{d t}$

Using the chain rule of differentiation, we can express this as

$a=\frac{\partial u}{\partial t}+\frac{\partial u}{\partial x} \frac{d x}{d t}$.

Here, the second term on the right-hand side represents the change in velocity due to a change in position, which is given by the product of the velocity gradient $\partial u / \partial x$ and the car’s velocity $dx/dt$. We can rewrite this in terms of the velocity field $u(x,t)$ by using the chain rule again:

$\frac{d x}{d t}=u(x, t)$

Substituting this into the previous equation gives

$a=\frac{\partial u}{\partial t}+u(x, t) \frac{\partial u}{\partial x}$.

This is the desired expression for the acceleration of an individual car in terms of the velocity field.

问题 2.

In how many ways can a positive integer $n$ be written as a sum of positive integers, taking order into account? For instance, 4 can be written as a sum in the eight ways $4=3+1=$ $1+3=2+2=2+1+1=1+2+1=1+1+2=1+1+1+1$

证明 .

The number of ways a positive integer $n$ can be written as a sum of positive integers, taking order into account, is given by the partition function $p(n)$. Unfortunately, there is no known closed-form expression for $p(n)$.

However, we can compute $p(n)$ recursively using the following recurrence relation:

$p(n)=\sum_{k=1}^n p(n-k)$

where the sum is taken over all positive integers $k$ such that $k \leq n$.

The base case for this recurrence relation is $p(0) = 1$, which represents the empty sum that adds up to $0$.

Using this recurrence relation, we can compute $p(n)$ for small values of $n$:

\begin{align*} p(1) &= 1 \ p(2) &= 2 \ p(3) &= 3 \ p(4) &= 5 \ p(5) &= 7 \ p(6) &= 11 \ p(7) &= 15 \ p(8) &= 22 \ p(9) &= 30 \ p(10) &= 42 \end{align*}

and so on.

For larger values of $n$, computing $p(n)$ using this recurrence relation becomes computationally expensive. However, there are more efficient algorithms for computing $p(n)$, such as the pentagonal number theorem and the Hardy-Ramanujan-Rademacher formula.

问题 3.

Let $p$ be a prime number and $1 \leq k \leq p-1$. How many $k$-element subsets $\left{a_1, \ldots, a_k\right}$ of ${1,2 \ldots, p}$ are there such that $a_1+\cdots+a_k \equiv 0(\bmod p)$ ?

证明 .

Let $S_k$ denote the number of $k$-element subsets of ${1,2,\ldots,p}$ that have a sum congruent to $0$ modulo $p$. We will derive a recursive formula for $S_k$.

Consider a fixed integer $a_1$. Then the remaining $k-1$ integers must sum to $-a_1$ modulo $p$. If $a_1=0$, then we have $S_{k-1}$ choices for the remaining $k-1$ integers. Otherwise, there are $S_{k-1}$ choices for the remaining $k-1$ integers when we consider the $k-1$-element subset of ${1,2,\ldots,p}\setminus{a_1}$ that sums to $-a_1$ modulo $p$. Thus, we have the recursion

$S_k=S_{k-1}+(p-1) S_{k-1}=p S_{k-1}$

where the factor of $p-1$ accounts for the fact that there are $p-1$ choices for $a_1$.

Since $S_1 = p$, we have $S_k = p^k$ for all positive integers $k$. Therefore, there are $p^k$ $k$-element subsets of ${1,2,\ldots,p}$ that have a sum congruent to $0$ modulo $p$.

这是一份2023年的剑伦敦大学学院伦敦大学学院 MATH0010 Mathematical Methods 1数学方法代写的成功案例

代数|MATH0006 Algebra 2代写2023

0
Posted on by

Assignment-daixieTM为您提供伦敦大学学院 London’s Global University MATH0006 Algebra 2代数代写代考和辅导服务！

Instructions:

  1. Check the syllabus: Make sure the book covers the topics that will be taught in the course. Look at the table of contents or the index to see if the book covers the topics you need.
  2. Consider your level of experience: Choose a book that matches your level of expertise in the subject. If you are new to the subject, choose a book that is more elementary and easy to understand. If you are more experienced, you may want a book that is more advanced and covers more material.
  3. Read reviews: Look for reviews of the book online or ask your classmates and professor for recommendations. Reviews can help you get a sense of how clear and helpful the book is.
  4. Consider your learning style: Everyone learns differently, so choose a book that matches your learning style. Some books have lots of examples and exercises, while others have more theory and proofs.
  5. Check the price: Textbooks can be expensive, so make sure you are comfortable with the price of the book you choose. You may want to consider buying a used copy or renting the book to save money.
代数|MATH0006 Algebra 2代写2023

问题 1.

Let $G$ be a group and let $H$ be a subgroup. Prove that the following are equivalent. (1) $H$ is normal in $G$.

证明 .

$(1) \Rightarrow (2)$: Assume that $H$ is a normal subgroup of $G$. We need to show that for every $g \in G, g H g^{-1}=H$. Let $g\in G$ and $h \in H$ be arbitrary. Since $H$ is normal in $G$, we have $ghg^{-1}\in H$. Hence, $gHg^{-1} \subseteq H$. To show the other inclusion, let $x\in H$. Then $g^{-1}xg \in H$ since $H$ is normal, and so $x=gg^{-1}xgg^{-1}=g(g^{-1}xg)g^{-1}\in gHg^{-1}$. Therefore, we have $H\subseteq gHg^{-1}$, which implies that $gHg^{-1}=H$.

问题 2.

(2) For every $g \in G, g H g^{-1}=H$.

证明 .

$(2) \Rightarrow (3)$: Assume that for every $g\in G, g H g^{-1}=H$. We need to show that for every $a \in G, a H=H a$. Let $a \in G$ be arbitrary. We want to show that $aH\subseteq Ha$. Let $h\in H$ be arbitrary. Then, we have $ah=g(g^{-1}ag)h(g^{-1}ag)^{-1}\in gHg^{-1}=H$, where we use that $g^{-1}ag$ is just some element of $G$ that we can conjugate with. Therefore, $ah\in Ha$ for all $h \in H$, which implies that $aH \subseteq Ha$. By a similar argument, we can show that $Ha \subseteq aH$, and so we have $aH=Ha$.

\begin{prob}

(3) For every $a \in G, a H=H a$.

\end{prob}

\begin{proof}

$(3) \Rightarrow (1)$: Assume that for every $a \in G, a H=H a$. We need to show that $H$ is normal in $G$. Let $g\in G$ and $h\in H$ be arbitrary. We want to show that $ghg^{-1}\in H$. To do this, note that $g^{-1}hg \in H$ since $H$ is closed under conjugation by elements of $G$. Therefore, $ghg^{-1}=g(g^{-1}hg)g^{-1} \in gHg^{-1}$, and so $H$ is normal in $G$.

这是一份2023年的伦敦大学学院 MATH0006 Algebra 2代数代写的成功案例