# 热力学代写 Thermodynamics代考2023

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## 热力学代写Thermodynamics

### 统计力学Statistical mechanics代写

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

## 热力学的历史

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

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.

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
$$d N=\frac{(V-B N) d P+\left(P-\left(A N^2 / V^2\right)+\left(2 A B N^3 / V^3\right)\right) d V-R N d T}{B P+R T+\left(3 A B N^2 / V^2\right)-(2 A N / V)}$$

# 实分析|MATH0051 Analysis 4: Real Analysis代写2023

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

## Instructions:

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 $$\lim _{x \rightarrow \infty} \frac{x^2}{x^2+1}=1$$

(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 $\lim_{x \rightarrow \infty} \frac{x^2}{x^2+1} = 1$.

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

(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.

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 $S$.

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 $S$.

# 数学方法|MATH0016 Mathematical Methods 3代写2023

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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.

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

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

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

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

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 $4$.

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

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.

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

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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.

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 $A$.

(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.

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

(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
$$c_1\left[\begin{array}{l} 1 \ 3 \ 0 \end{array}\right]+c_2\left[\begin{array}{l} 2 \ 8 \ 4 \end{array}\right]$$

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

(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
$$B=\left[\begin{array}{rr} \operatorname{rref}(A) & \operatorname{rref}(A) \ 0 & 0 \end{array}\right], \operatorname{rref}(A)=\left[\begin{array}{llll} 0 & 1 & 0 & 1 \ 0 & 0 & 1 & \frac{1}{2} \ 0 & 0 & 0 & 0 \end{array}\right]$$

# 流体力学|MATH0015 Fluid Mechanics代写2023

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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.

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: $\mathrm{x}=o, \mathrm{y}=o ; \mathrm{x}=0, \mathrm{y}=\frac{-2 \pi}{k}$

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:

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 $\mathrm{t}=\mathrm{o}$

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 $t=\frac{2\pi}{\omega}$

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?

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.

# 光学代写 optics代考2023

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## 光学代写optics

### 量子光学Quantum optics代写

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

## 光学的历史

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

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)

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:
$$R=\lambda \frac{3.274}{(2 \pi)^2\left(\frac{\Delta \theta}{360}\right)}=500 \mathrm{~nm} \times\left(\frac{3.274}{(2 \pi)^2 \frac{2}{360}}\right)=7463 \mathrm{~nm}=7.4 \mu \mathrm{m}$$

# 电磁学代写 Electromagnetism代考2023

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## 电磁学代写Electromagnetism

### 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。

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

## 电磁学的历史

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

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

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.

# 复分析|MATH0013 Analysis 3: Complex Analysis代写2023

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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.

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.

(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.

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

(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.

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

(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.

# 数学方法|MATH0010 Mathematical Methods 1代写2023

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## 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.

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

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.

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 $4=3+1=$ $1+3=2+2=2+1+1=1+2+1=1+1+2=1+1+1+1$

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.

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 $a_1+\cdots+a_k \equiv 0(\bmod p)$ ?

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 $p$.

# 代数|MATH0006 Algebra 2代写2023

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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.
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.

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

$(1) \Rightarrow (2)$: 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 $gHg^{-1}=H$.

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

$(2) \Rightarrow (3)$: 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 $G$.