Category Archives: 澳洲代写

高级物理学1A|PHYS1131 Higher Physics 1A代写 unsw

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This course provides an introduction to Physics. It is a calculus based
course. The course is examined at two levels, with Higher Physics 1A
being the higher of the two levels. While the same content is covered as
Physics 1A, Higher Physics 1A features more advanced assessment.

这是一份unsw新南威尔斯大学PHYS1131的成功案例

高级物理学1A|PHYS1131 Higher Physics 1A代写 unsw代写


\begin{prob}

As an oil well is drilled, each new section of drill pipe supports its own weight and that
of the pipe and drill bit beneath it. Calculate the stretch in a new 6.00 m length of steel
pipe that supports 3.00 km of pipe having a mass of 20.0 kg/m and a 100-kg drill bit.
The pipe is equivalent in strength to a solid cylinder 5.00 cm in diameter.

证明 .

Use the equation $\Delta L=\frac{1}{Y} \frac{F}{A} L_{0}$, where $L_{0}=6.00 \mathrm{~m}, Y=1.6 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}$. To calculate the mass supported by the pipe, we need to add the mass of the new pipe to the mass of the $3.00 \mathrm{~km}$ piece of pipe and the mass of the drill bit:
$$
\begin{aligned}
&m=m_{\mathrm{p}}+m_{3 \mathrm{~km}}+m_{\text {bit }} \
&=(6.00 \mathrm{~m})(20.0 \mathrm{~kg} / \mathrm{m})+\left(3.00 \times 10^{3} \mathrm{~m}\right)(20.0 \mathrm{~kg} / \mathrm{m})+100 \mathrm{~kg}=6.022 \times 10^{4} \mathrm{~kg}
\end{aligned}
$$
So that the force on the pipe is:
$$
F=w=m g=\left(6.022 \times 10^{4} \mathrm{~kg}\right)\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right)=5.902 \times 10^{5} \mathrm{~N}
$$
Finally the cross sectional area is given by: $A=\pi r^{2}=\pi\left(\frac{0.0500 \mathrm{~m}}{2}\right)^{2}=1.963 \times 10^{-3} \mathrm{~m}^{2}$





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PHYS1131 COURSE NOTES ：

Derive the equation for the vertical acceleration of a rocket

The force needed to give a small mass $\Delta m$ an acceleration $a_{\Delta m}$ is $F=\Delta m a_{\Delta m}$. To accelerate this mass in the small time interval $\Delta t$ at a speed $v_{\mathrm{e}}$ requires $v_{\mathrm{e}}=a_{\Delta m} \Delta t$, so $F=v_{e} \frac{\Delta m}{\Delta t}$. By Newton’s third law, this force is equal in magnitude to the thrust force acting on the rocket, so $F_{\text {thnust }}=v_{e} \frac{\Delta m}{\Delta t}$, where all quantities are positive. Applying Newton’s second law to the rocket gives $F_{\text {thrus }}-m g=m a \Rightarrow a=\frac{v_{e}}{m} \frac{\Delta m}{\Delta t}-g$, where $m$ is the mass of the rocket and unburnt fuel.




















物理学1A|PHYS1121 Physics 1A代写 unsw

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This course provides an introduction to Physics. It is a calculus based
course. The course is examined at two levels, with Physics 1A being the
lower of the two levels.
Mechanics: particle kinematics in one dimension, motion in two and
three dimensions, particle dynamics, work and energy, momentum and
collisions.
Thermal physics: temperature, kinetic theory and the ideal gas, heat and
the first law of thermodynamics.
Waves: oscillations, wave motion, sound waves.

这是一份unsw新南威尔斯大学PHYS1121的成功案例

物理学1A|PHYS1121 Physics 1A代写 unsw代写


\begin{prob}

(a) Calculate the momentum of a 2000-kg elephant charging a hunter at a speed of
7.50 m/s. (b) Compare the elephant’s momentum with the momentum of a 0.0400-kg
tranquilizer dart fired at a speed of 600 m/s. (c) What is the momentum of the 90.0-
kg hunter running at 7.40 m/s after missing the elephant?

证明 .

(a) $p_{\mathrm{e}}=m_{\mathrm{e}} v_{e}=2000 \mathrm{~kg} \times 7.50 \mathrm{~m} / \mathrm{s}=1.50 \times 10^{4} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}$
(b) $p_{\mathrm{b}}=m_{\mathrm{b}} v_{\mathrm{b}}=0.0400 \mathrm{~kg} \times 600 \mathrm{~m} / \mathrm{s}=24.0 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}$, so
$$
\frac{p_{\mathrm{c}}}{p_{\mathrm{b}}}=\frac{1.50 \times 10^{4} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}}{24.0 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}}=625
$$
The momentum of the elephant is much larger because the mass of the elephant is much larger.
(c) $p_{\mathrm{b}}=m_{\mathrm{h}} v_{\mathrm{h}}=90.0 \mathrm{~kg} \times 7.40 \mathrm{~m} / \mathrm{s}=6.66 \times 10^{2} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}$
Again, the momentum is smaller than that of the elephant because the mass of the hunter is much smaller.





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PHYS1121 COURSE NOTES ：

A cruise ship with a mass of 1.00 10 kg 7 × strikes a pier at a speed of 0.750 m/s. It
comes to rest 6.00 m later, damaging the ship, the pier, and the tugboat captain’s
finances. Calculate the average force exerted on the pier using the concept of impulse.
(Hint: First calculate the time it took to bring the ship to rest.)

Given: $m=1.00 \times 10^{7} \mathrm{~kg}, v_{0}=0.75 \mathrm{~m} / \mathrm{s}, v=0 \mathrm{~m} / \mathrm{s}, \Delta x=6.00 \mathrm{~m}$. Find: net force on the pier. First, we need a way to express the time, $\Delta t$, in terms of known quantities.
Using the equations $\bar{v}=\frac{\Delta x}{\Delta t}$ and $\bar{v}=\frac{v_{0}+v}{2}$ gives:
$$
\Delta x=\bar{v} \Delta t=\frac{1}{2}\left(v+v_{0}\right) \Delta t \text { so that } \Delta t=\frac{2 \Delta x}{v+v_{0}}=\frac{2(6.00 \mathrm{~m})}{(0+0.750) \mathrm{m} / \mathrm{s}}=16.0 \mathrm{~s} .
$$
net $F=\frac{\Delta p}{\Delta t}=\frac{m\left(v-v_{0}\right)}{\Delta t}=\frac{\left(1.00 \times 10^{7} \mathrm{~kg}\right)(0-0750) \mathrm{m} / \mathrm{s}}{16.0 \mathrm{~s}}=-4.69 \times 10^{5} \mathrm{~N}$.
By Newton’s third law, the net force on the pier is $4.69 \times 10^{5} \mathrm{~N}$, in the original direction of the ship.




















统计学专题|MATH5805 Special Topics in Statistics代写 unsw代写

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This course will cover topics in the general area of Monte Carlo methods and their application
domains. The topics include Markov chain Monte Carlo and Sequential Monte Carlo methods,
Quantum and Diffusion Monte Carlo techniques, as well as branching and interacting particle
methodologies. The lectures cover discrete and continuous time stochastic models, starting from
traditional sampling techniques (perfect simulation, Metropolis-Hasting, and Gibbs-Glauber
models) to more refined methodologies such as gradient flows diffusions on constraint state space
and Riemannian manifolds, ending with the more recent and rapidly developing Branching and
mean field type Interacting Particle Systems techniques.

这是一份unsw新南威尔斯大学MATH5805的成功案例

统计学专题|MATH5805 Special Topics in Statistics代写 unsw代写


问题 1.

The mean and the standard deviation of the score at the exam of Statistics for the past 2 cohorts of students was:

  • cohort 2011-2012 (Italian; $n=171):$ mean $=25.5$, std $=3.959$
  • cohort 2012-2013 (English; $\mathrm{n}=16$ ): mean $=26.4$, std $=4.11$
    Compute the overall average and compare the variability.


证明 .

The overall mean is a weighted average:
$$
\text { Mean }=[(25.2 \cdot 171)+(26.4 \cdot 16)] /(171+16)=4782.9 / 187=25.58
$$
The variability is compared by looking at the coefficient of variation:
CV for cohort 2011-2012: $25.5 / 3.959=16 \%$
CV for cohort 2011-2012: $26.6 / 4.3=16 \%$







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MATH5805 COURSE NOTES ：

Exercise
There are two urns containing coloured balls. The first urn contains 50 red balls and 50 blue balls. The second urn contains 30 red balls and 70 blue balls. One of the two urns is randomly chosen (both urns have equal probability of being chosen) and then a ball is drawn at random from that urn.
(a) What is the probability to draw a red ball?
(b) If a red ball is drawn, what is the probability that it comes from the first urn?
Solution\begin{aligned}
&\mathrm{P}(\mathrm{U} 1)=\mathrm{P}(\mathrm{U} 2)=1 / 2 \
&\mathrm{P}(\text { Red } \mid \mathrm{U} 1)=1 / 2 \
&\mathrm{P}(\text { Red } \mid \mathrm{U} 2)=3 / 10
\end{aligned}
$$
a) $P($ Red $)=P($ Red|U1 $) \cdot P(U 1)+P($ Red| $U 2) \cdot P(U 2)=0.5 \cdot 0.5+0.3 \cdot 0.5=0.4$
b) $P(U 1 \mid$ Red $)=P(U 1 \&$ Red $) / P($ Red $)=P($ Red $\mid U 1) \cdot P(U 1) / P($ Red $)=0.5 \cdot 0.5 / 0.4=0.625$




















概率和随机过程|MATH3801 Probability and Stochastic Processes代写 unsw代写

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This course is an introduction to the theory of stochastic processes. Informally, a stochastic process is a random quantity that evolves over time, like a gambler’s net fortune and the price fluctuations of a stock on any stock exchange, for instance. The main aims of this course are: 1) to provide a thorough account of basic probability theory: 2) to introduce the ideas and tools of the theory of stochastic processes; and 3) to discuss in depth important classes of stochastic processes, including Markov Chains (both in discrete and continuous time), Poisson processes, the Brownian motion and Martingales. The course will also cover other important but less routine topies, like Markov decision processes and some elements of queucing theory.

这是一份unsw新南威尔斯大学MATH3801的成功案例

高等线性模型|MATH2931 Higher Linear Models代写 unsw代写


问题 1.

Suppose the claim is wrong and that
$$
\mathrm{P}\left[A_{\infty}=\infty, \sup {n}\left|X{n}\right|<\infty\right]>0 .
$$
Then,
$$
\mathrm{P}\left[T(c)=\infty ; A_{\infty}=\infty\right]>0
$$
where $T(c)$ is the stopping time
$$
T(c)=\inf \left{n|| X_{n} \mid>c\right} .
$$


证明 .

Now
$$
\mathrm{E}\left[X_{T(c) \wedge n}^{2}-A_{T(c) \wedge n}\right]=0
$$
and $X^{T(c)}$ is bounded by $c+K$. Thus
$$
\mathrm{E}\left[A_{T(c) \wedge n}\right] \leq(c+K)^{2}
$$
for all $n$. This is a contradiction to $\mathrm{P}\left[A_{\infty}=\infty, \sup {n}\left|X{n}\right|<\infty\right]>0$.







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MATH3801 COURSE NOTES ：

it is seen that the rate into the box around Node 0 is $\mu p_{1}$; the rate out of the box around Node 0 is $\lambda p_{0}$; thus, “rate in” = “rate out” yields
$$
\mu p_{1}=\lambda p_{0}
$$
The rate into the box around Node 1 is $\lambda p_{0}+\mu p_{2}$; the rate out of the box around Node 1 is $(\mu+\lambda) p_{1}$; thus
$$
\lambda p_{0}+\mu p_{2}=(\mu+\lambda) p_{1}
$$
Continuing in a similar fashion and rearranging, we obtain the system
$$
\begin{aligned}
p_{1} &=\frac{\lambda}{\mu} p_{0} \text { and } \
p_{n+1} &=\frac{\lambda+\mu}{\mu} p_{n}-\frac{\lambda}{\mu} p_{n-1} \text { for } n=1,2, \cdots .
\end{aligned}
$$




















随机分析入门|MATH5975 Introduction to Stochastic Analysis代写 unsw代写

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In this course, you will learn the basic concepts and techniques of Stochastic Analysis, such as: Brownian motion, martingales, Itˆo stochastic integral, Itˆo’s formula,
stochastic differential equations, equivalent change of a probability measure, integral representation of martingales with respect to a Brownian filtration, relations
to second order partial differential equations, the Feynman-Kac formula, and jump
processes.

这是一份unsw新南威尔斯大学MATH5975的成功案例

随机分析入门|MATH5975 Introduction to Stochastic Analysis代写 unsw代写


问题 1.

Let $\left(X_{n}\right){n \in \mathbb{N}}$ be a submartingale w.r.t. the filtration $\left(\mathcal{F}{n}\right){n \in \mathbb{N}}$ such that $$ \forall n \in \mathbb{N}, \exists k \geq n: \mathbb{E}\left(X{k}\right) \leq \mathbb{E}\left(X_{n}\right)
$$
Show that $\left(X_{n}\right)_{n \in \mathbb{N}}$ is a martingale!


证明 .

We consider the Doob composition $X_{n}=M_{n}+A_{n}$ with $M_{n}$ a martingale and $A_{n} \geq 0$ an increasing predictable process. It follows
$$
\mathbb{E}\left[X_{n}\right]=\underbrace{\mathbb{E}\left[M_{n}\right]}{=\mathbb{E}\left[M{n+1}\right]}+\underbrace{\mathbb{E}\left[A_{n}\right]}{\geq \mathbb{E}\left[A{n-1}\right]} \geq \mathbb{E}\left[X_{n-1}\right]
$$
Hence, the sequence of $\left(\mathbb{E}\left[X_{n}\right]\right){n \in \mathbb{N}}$ is an increasing sequence of real numbers. On the other hand, by $$ \forall n \in \mathbb{N}, \exists k \geq n: \mathbb{E}\left(X{k}\right) \leq \mathbb{E}\left(X_{n}\right)
$$
we can a extract a decreasing a subsequence $\left(\mathbb{E}\left[X_{n_{k}}\right]\right){k \in \mathbb{N}}$. Hence, the whole sequence $\left(\mathbb{E}\left[X{n}\right]\right){n \in \mathbb{N}}$ must be constant. By (1), this is only possible if $\mathbb{E}\left[A{n}\right]=0$ for all $n$. Since $A_{n} \geq 0$, it follows $A_{n}=0$. Hence, $X_{n}=M_{n}$ is a martingale.







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MATH5975 COURSE NOTES ：

Exercise ：Let $\tau_{1}(\omega)$ and $\tau_{2}(\omega)$ be stopping times with respect to the filtration $\mathbb{F}=\left(F_{1}: t \in T\right)$ taking values in $T$. Here $T$ could be either $\mathbb{R}^{+}$or $\mathbb{N}$.

Use the definition of stopping time to show that $\sigma(\omega)=\min \left(\tau_{1}(\omega), \tau_{2}(\omega)\right)$ is a F-stopping time.
Solution ：We have
$$
{\sigma(\omega) \leq t}=\left{\omega: \tau_{1}(\omega) \leq t \text { or } \tau_{2}(\omega) \leq t\right}=\left{\tau_{1}(\omega) \leq t\right} \cup\left{\tau_{2}(\omega) \leq t\right} \in \mathcal{F}_{t},
$$
so $\sigma$ is a stopping time.




















高等线性模型|MATH2931 Higher Linear Models代写 unsw代写

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Statistics is about using probability models to make decisions from data in
the face of uncertainty. This course gives an introduction to the process of
building statistical models using an important class of models (linear models).
In a linear model we try to predict or explain variation in a response variable
in terms of related quantities (predictors). The relationship between the
expected response and predictors is assumed to be linear. Topics covered
in the course include how to estimate parameters in linear models, how to
compare models, how to select a good model or models when prediction of
the response is the goal. Concepts are illustrated with applications from the
natural and social sciences.

这是一份unsw新南威尔斯大学MATH2931的成功案例

高等线性模型|MATH2931 Higher Linear Models代写 unsw代写


问题 1.

The regression coefficients $\hat{\beta}{1}, \hat{\beta}{2}, \ldots, \hat{\boldsymbol{\beta}}{k}$ in $\hat{\boldsymbol{\beta}}{1}$ can be standardized so as to show the effect of standardized $x$ values (sometimes called $z$ scores). We illustrate this for $k=2$. The model in centered form
$$
\hat{y}{i}=\bar{y}+\hat{\beta}{1}\left(x_{i 1}-\bar{x}{1}\right)+\hat{\beta}{2}\left(x_{i 2}-\bar{x}_{2}\right)
$$


证明 .

This can be expressed in terms of standardized variables as
$$
\frac{\hat{y}{i}-\bar{y}}{s{y}}=\frac{s_{1}}{s_{y}} \hat{\beta}{1}\left(\frac{x{i 1}-\bar{x}{1}}{s{1}}\right)+\frac{s_{2}}{s_{y}} \hat{\beta}{2}\left(\frac{x{i 2}-\bar{x}{2}}{s{2}}\right),
$$
where $s_{j}=\sqrt{s_{j j}}$ is the standard deviation of $x_{j}$. We thus define the standardized coefficients as
$$
\hat{\beta}{j}^{*}=\frac{s{j}}{s_{y}} \hat{\beta}_{j}
$$







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MATH2931 COURSE NOTES ：

The hypothesis $H_{0}: \alpha_{1}=\alpha_{2}=\alpha_{3}$ can be expressed as $H_{0}: \alpha_{1}-\alpha_{2}=0$ and $\alpha_{1}-\alpha_{3}=0$. Thus $H_{0}$ is testable if $\alpha_{1}-\alpha_{2}$ and $\alpha_{1}-\alpha_{3}$ are estimable. To check $\alpha_{1}-\alpha_{2}$ for estimability, we write it as
$$
\alpha_{1}-\alpha_{2}=(0,1,-1,0,0,0) \boldsymbol{\beta}=\boldsymbol{\lambda}{1}^{\prime} \boldsymbol{\beta} $$ and then note that $\boldsymbol{\lambda}{1}^{\prime}$ can be obtained from $\mathbf{X}$ as
$$
(1,0,-1,0,0,0) \mathbf{X}=(0,1,-1,0,0,0)
$$
and from $\mathbf{X}^{\prime} \mathbf{X}$ as
$$
\left(0, \frac{1}{2},-\frac{1}{2}, 0,0,0\right) \mathbf{X}^{\prime} \mathbf{X}=(0,1,-1,0,0,0)
$$




















统计分析的数据管理|MATH2871 Data Management for Statistical Analysis代写 unsw代写

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The course covers the use of database and spreadsheet tools to organise and query statistical
data, programming in an advanced statistical package for file management, data manipulation and
cleaning; methods for data screening, cleaning, graphical displays and data analysis using a range
of statistical procedures; creation of data analysis reports using modern statistical and graphical
methods. The course is based around Microsoft Access and Excel as well as the SAS statistical
analysis system and programming tools. Knowledge and skills developed will be generic and
applicable to a range of modern statistical software tools.

这是一份unsw新南威尔斯大学MATH28711的成功案例

统计分析的数据管理|MATH2871 Data Management for Statistical Analysis代写 unsw代写


ods select datascoresmc;
proc npar1way data=ds;
class female;
var age;
exact scores=data / mc n=9999 alpha=.05;
run;
ods exclude none;
One-Sided Pr >= S
Estimate 0.1841
95% Lower Conf Limit 0.1765
95% Upper Conf Limit 0.1917
Two-Sided Pr >= |S - Mean|
Estimate 0.3615
95% Lower Conf Limit 0.3521
95% Upper Conf Limit 0.3710
Number of Samples 9999
Initial Seed 600843001





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MATH2871 COURSE NOTES ：

ods select datascoresmc;
proc npar1way data=ds;
class female;
var age;
exact scores=data / mc n=9999 alpha=.05;
run;
ods exclude none;
One-Sided Pr >= S
Estimate 0.1841
95% Lower Conf Limit 0.1765
95% Upper Conf Limit 0.1917
Two-Sided Pr >= |S - Mean|
Estimate 0.3615
95% Lower Conf Limit 0.3521
95% Upper Conf Limit 0.3710
Number of Samples 9999
Initial Seed 600843001




















数学|MATH1141/MATH1131 Mathematics代写 unsw代写

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MATH1131, Mathematics 1A, and MATH1141, Higher Mathematics 1A, are first year courses
taught by the School of Mathematics and Statistics in semester 1, and are each worth six units
of credit. MATH1131 is also taught in semester 2. Students who pass MATH1131 in semester
1 usually continue to study MATH1231, Mathematics 1B, in semester 2. Those students who
pass MATH1141 with a Credit usually continue to study MATH1241, Higher Mathematics 1B,
in semester 2. MATH1231 is also taught in Summer Session. MATH1131 and MATH1231 (or
MATH1141 and MATH1241) are generally specified in Engineering programs, as well as many
Science programs.

这是一份unsw新南威尔斯大学MATH1141/MATH1131的成功案例

数学|MATH1141/MATH1131 Mathematics代写 sydney代写


问题 1.

Next, if $\alpha>\omega$ is a limit ordinal, then according to what we have said before, $\sum_{\xi<\omega^{\alpha}} \xi=\sup {\beta<\alpha} \sum{\xi<\omega^{\sigma}} \xi$, and here in the supremum we can take the supremum for successor ordinals $\beta$ smaller than $\alpha$. Thus, according to what we have just proved, in this case
$$
\sum_{\xi<\omega^{\alpha}} \xi=\sup {\beta<\alpha} \omega^{\beta \cdot 2}=\omega^{\sigma}, $$ where $\sigma=\sup {\beta<\alpha} \beta \cdot 2$. Here if $\alpha$ equals one of the powers of $\omega$, say $\alpha=\omega^{\tau}$, then
$$
\omega^{\tau}=\sup {\beta<\alpha} \beta \leq \sup {\beta<\alpha} \beta \cdot 2 \leq \sup _{\beta<\alpha} \beta \cdot \omega=\omega^{\tau},
$$


证明 .

$$
\beta=\omega^{\xi_{n}} \cdot a_{n}+\cdots+\omega^{\xi_{0}} \cdot\left(a_{0}-1\right)+\delta,
$$
where $\delta<\omega^{\xi_{0}}$, and here
$$
\beta \cdot 2=\omega^{\varepsilon_{n}} \cdot\left(2 a_{n}\right)+\omega^{\xi_{n-1}} \cdot a_{n-1}+\cdots+\omega^{\varepsilon_{0}} \cdot\left(a_{0}-1\right)+\delta,
$$
thus
$$
\sigma=\sup {\beta<\alpha} \beta \cdot 2=\omega^{\xi{n}} \cdot\left(2 a_{n}\right)+\cdots+\omega^{\xi_{0}} \cdot a_{0}=\alpha \cdot 2 .
$$
In summary, the sum in question is equal to $\omega^{2 \alpha-1}$ if $\alpha$ is finite, it equals $\omega^{\alpha}$ if $\alpha$ is a power of $\omega$, and in all other cases it equals $\omega^{\alpha \cdot 2}$.







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MATH1141/MATH1131 COURSE NOTES ：

Actually, the method we used for the existence can be easily extended to yield both the existence and unicity of the representation. In fact, if the normal form of $\alpha$ is $\alpha=\omega^{\xi_{n}} \cdot a_{n}+\cdots+\omega^{\varepsilon_{0}} \cdot a_{0}$ and set $\delta_{0}=\xi_{0}$ and choose $\delta_{i}$, $1 \leq i \leq n$, so that $\xi_{i}=\xi_{i-1}+\delta_{i}$, then $\delta_{0} \geq 0$ and $\delta_{i}>0$ for $1 \leq i \leq n$. Now
$$
\alpha=\omega^{\delta_{0}} \cdot a_{0} \cdot\left(\omega^{\delta_{1}}+1\right) \cdot a_{1} \cdots a_{n-1} \cdot\left(\omega^{\delta_{n}}+1\right) \cdot a_{n}
$$
and if $\delta_{0}=\omega^{\gamma_{m}}+\cdots+\omega^{\gamma_{0}}$ with $\gamma_{m} \geq \cdots \geq \gamma_{0}$, then
$$
\alpha=\omega^{\omega^{\gamma m}} \cdots \omega^{\omega v 0} \cdot a_{0} \cdot\left(\omega^{\delta_{1}}+1\right) \cdot a_{1} \cdots a_{n-1} \cdot\left(\omega^{\delta_{n}}+1\right) \cdot a_{n}
$$




















量子、统计和复合物理|PHYS3034/PHYS3935 Quantum, Statistical and Comp Phys代写 sydney代写

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You will learn how to use computer-aided design (CAD) software to design 3D printable molecular models that convey intermolecular interactions like hydrogen-bonding. The models will incorporate movable (i.e., rotation, flexion) and magnetic design elements to represent the formation of dynamic/weak bonds, which will serve as a useful visualisation and communication tool for complex molecular structures. The designs will be printed remotely and students will be able to keep their printed designs.

这是一份sydney悉尼大学PHYS3034/PHYS3935的成功案例

量子、统计和复合物理|PHYS3034/PHYS3935 Quantum, Statistical and Comp Phys代写 sydney代写


问题 1.

is also a nonnegative-definite Hermitian matrix, and we obtain the matrix inequality
$$
\sum_{j=1}^{m}\left(x_{j}^{2}+y_{j}^{2}\right) M_{j} \geq X^{2}+Y^{2} \pm i[X, Y]
$$
Since $S=S^{*}>0$ from the assumption, the above formula results in
$$
\sum_{j=1}^{m}\left(x_{j}^{2}+y_{j}^{2}\right) S^{\frac{1}{2}} M_{j} S^{\frac{1}{2}} \geq S^{\frac{1}{2}}\left(X^{2}+Y^{2}\right) S^{\frac{1}{2}} \pm i S^{\frac{1}{2}}[X, Y] S^{\frac{1}{2}}
$$


证明 .


If $[X, Y]=0$, then $\Gamma(X, Y ; S)=\operatorname{Tr}\left[S\left(X^{2}+Y^{2}\right)\right]$ holds. In this case, if we define $M=\left{M_{j} ; x_{j} ; y_{j}\right}$ by (4)-(6), then (1) and (2) obviously hold and
$$
\sum_{j=1}^{m} x_{j}^{2} M_{j}=X^{2}, \quad \sum_{j=1}^{m} y_{j}^{2} M_{j}=Y^{2}
$$
hold, so that we have $\Delta(M ; S)=\operatorname{Tr}\left[S\left(X^{2}+Y^{2}\right)\right]$. Therefore, it gives the minimum of $\Delta$ and $\Delta_{*}(X, Y ; S)=\Gamma(X, Y ; S)$ holds.







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PHYS3034/PHYS3935 COURSE NOTES ：

$$
\frac{1}{2 \pi \mu} \int(x+i y) E_{x, y} d x d y=X+i Y,
$$
which can be also written as
$$
\frac{1}{2 \pi \mu} \int x E_{x, y} d x d y=X, \quad \frac{1}{2 \pi \mu} \int y E_{x, y} d x d y=Y .
$$
Moreover, if we multiply $X-i Y$ from the right in both sides, from $[X, Y]=i \mu I$, we have
$$
\frac{1}{2 \pi \mu} \int\left(x^{2}+y^{2}\right) E_{x, y} d x d y=X^{2}+Y^{2}+\mu I .
$$




















电动力学和光学|PHYS3035/PHYS3935 Electrodynamics and Optics代写 sydney代写

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这是一份sydney悉尼大学PHYS3035/PHYS3935 的成功案例

电动力学和光学|PHYS3035/PHYS3935 Electrodynamics and Optics代写 sydney代写


问题 1.

$$
\varrho_{-}^{\prime}=\frac{\varrho}{\sqrt{1-\left(u-u_{\mathrm{e}}\right)^{2} / c^{2}}}=\varrho\left(1+\frac{1}{2 c^{2}}\left(u^{2}-2 u u_{\mathrm{e}}\right)\right),
$$
and for the lower section of windings, just above (1),
$$
\varrho_{-}^{\prime}=\frac{\varrho}{\sqrt{1-\left(u+u_{e}\right)^{2} / c^{2}}}=\varrho\left(1+\frac{1}{2 c^{2}}\left(u^{2}+2 u u_{e}\right)\right) \text {. }
$$


证明 .


For the upper section of the coil containing $N$ wires , combiningand yields an excess of positive charges:
$$
\Delta Q_{+}^{\prime}=N \Delta q_{+}^{\prime}=\frac{N q u_{\mathrm{e}} u}{c^{2}} .
$$
For the lower section of windings (1), the combination of yields an excess of negative charges:
$$
\Delta Q_{-}^{\prime}=N \Delta q_{-}^{\prime}=-\frac{N q u_{\mathrm{e}} u}{c^{2}}
$$







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PHYS3035/PHYS3935 COURSE NOTES ：

Derivation: Each end of the bar (or coil) produces a flux density $B_{t}=$ $\frac{\Phi}{4 \pi R^{2}}$ at the point of observation according. Only the difference of the two values is important, so that in the first principal orientation
$$
B=\frac{\phi}{4 \pi}\left(\frac{1}{(R-l / 2)^{2}}-\frac{1}{(R+1 / 2)^{2}}\right)
$$
When the distance $R$ is sufficiently large compared to the length $I$ of the bar or coil, we can neglect $P^{2}$ relative to $R^{2}$, and for the magnitude of $B$, we then obtain
$$
B=\frac{1}{2 \pi} \frac{\Phi l}{R^{3}}=\frac{\mu_{0}}{2 \pi} \frac{m}{R^{3}}
$$
Correspondingly, for the second principal orientation, we find
$$
B=\frac{\mu_{0}}{4 \pi} \frac{m^{3}}{R^{3}}
$$