# 概率学和统计学代写|STATISTICS AND PROBABILITY I MATH253 University of Liverpool Assignment

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## Instructions:

It is true that data analysis has become an essential part of research in many fields, including medicine, pharmacology, and biology. Statistical methods are used to help researchers make sense of large amounts of data and to draw meaningful conclusions from their findings. In addition, data analysis is also important in many other fields, including finance, consultancy, and the public sector.

To perform statistical analysis on real data sets, researchers often use statistical software packages. These packages allow researchers to perform complex statistical analyses quickly and accurately, and to visualize their results in meaningful ways. Some popular statistical software packages include R, SAS, and SPSS.

When performing statistical analysis, it is important to have a strong understanding of the underlying statistical techniques being employed. This includes understanding concepts such as probability, hypothesis testing, and regression analysis. It is also important to understand how to apply these techniques appropriately to real-world data sets.

Interpreting the results of statistical analyses is also a crucial part of the process. Researchers must be able to communicate their findings effectively to others, including colleagues, stakeholders, and the general public. This requires a strong understanding of statistical concepts as well as effective communication skills.

Overall, statistical analysis is a critical tool for researchers and professionals in many fields. By applying statistical techniques appropriately and interpreting their results effectively, researchers can gain valuable insights into their data and make informed decisions based on their findings.

You roll a fair six sided die repeatedly until the sum of all numbers rolled is greater than 6 . Let $X$ be the number of times you roll the die. Let $F$ be the cumulative distribution function for $X$. Compute $F(1), F(2)$, and $F(7)$.

(15) $F(1)$ : Since you never get more than 6 on one roll we have $F(1)=0$.
\begin{aligned} & F(2)=P(X=1)+P(X=2): \ & P(X=1)=0 \ & P(X=2)=P(\text { total on } 2 \text { dice }=7,8,9,10,11,12)=\frac{21}{36}=\frac{7}{12} . \end{aligned}
$F(7)$ : The smallest total on 7 rolls is 7 , so $F(7)=1$.

(15) Suppose $X$ is a random variable with cdf
$$F(x)= \begin{cases}0 & \text { for } x<0 \\ x(2-x) & \text { for } 0 \leq x \leq 1 \\ 1 & \text { for } x>1\end{cases}$$
(a) Find $E(X)$.
(b) Find $P(X<0.4)$.

(a) $f(x)=F^{\prime}(x)=2-2 x$ on $[0,1]$. Therefore
\begin{aligned} E(X) & =\int_0^1 x f(x) d x \ & =\int_0^1 2 x-2 x^2 d x \ & =x^2-\left.\frac{2}{3} x^3\right|_0 ^1 \ & =\frac{1}{3} \end{aligned}
(b) $P(X \leq 0.4)=F(0.4)=0.4(2-0.4)=0.4(1.6)=0.64$.

A political poll is taken to determine the fraction $p$ of the population that would support a referendum requiring all citizens to be fluent in the language of probability and statistics.
(a) Assume $p=0.5$. Use the central limit theorem to estimate the probability that in a poll of 25 people, at least 14 people support the referendum.

Let $X \sim \operatorname{binomial}(25,0.5)=$ the number supporting the referendum.
We know that
$$E(X)=12.5, \quad \operatorname{Var}(X)=25 \cdot \frac{1}{4}=\frac{25}{4}, \quad \sigma_X=\frac{5}{2} .$$
Standardizing and using the CLT we have $Z=\frac{X-12.5}{5 / 2} \approx \mathrm{N}(0,1)$ Therefore,
$$P(X \geq 14)=P\left(\frac{X-12.5}{5 / 2} \geq \frac{14-12.5}{5 / 2}\right) \approx P(Z \geq 0.6)=\Phi(-0.6)=0.2743 .$$
where the last probability was looked up in the $Z$-table.

# 线性代数和几何代写|LINEAR ALGEBRA AND GEOMETRY MATH244 University of Liverpool Assignment

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## Instructions:

In abstract algebra, a vector space is defined as a collection of objects, called vectors, which can be added together and multiplied by scalars (usually real numbers or complex numbers). These operations satisfy certain axioms, such as closure, associativity, commutativity, distributivity, and existence of identity and inverses.

The key idea behind abstract vector spaces is that the specific nature of the vectors themselves is not important, only the operations that can be performed on them. In other words, abstract vector spaces are defined purely in terms of their algebraic properties, without any reference to their physical or geometric interpretation.

By working in this abstract setting, mathematicians can develop powerful tools for studying linear phenomena in a wide variety of contexts. For example, abstract vector spaces are used to study linear transformations, eigenvalues and eigenvectors, inner products, and many other topics in linear algebra and beyond.

For an algebraic variety $X$ over a field $k$ of characteristic $p$ the Frobenius twist $X^{\prime}$ of $X$ is defined as follows. $X^{\prime}=X$ as a topological space. A function $f$ on $U^{\prime} \subset X^{\prime}$ is regular iff $f(x)=\phi(x)^p$ where $\phi$ is a regular function on $U=U^{\prime} \subset X$. The identity map $X \rightarrow X^{\prime}$ defines a morphism $F r: X \rightarrow X^{\prime}$ called the Frobenius morphism. [Notice though that it does not define a morphism from $X^{\prime}$ to $X$.] (a) Check that if $X$ is a closed subvariety in $\mathbb{A}^n$ or $\mathbb{P}^n$ whose ideal is generated by polynomials with coefficients in $\mathbb{F}_p$, then $X^{\prime} \cong X$. Moreover, we have an isomorphism such that that composition $X \stackrel{F r}{\longrightarrow} X^{\prime} \cong X$ is given by $\left(x_i\right) \mapsto\left(x_i^p\right)$.

Let $X$ be a closed subvariety in $\mathbb{A}^n$ or $\mathbb{P}^n$ whose ideal is generated by polynomials with coefficients in $\mathbb{F}_p$. Then $X$ is defined by polynomials of the form $f(x_1,\ldots,x_n)^p – f(x_1,\ldots,x_n)$ where $f$ has coefficients in $\mathbb{F}_p$. Let $U \subseteq X$ be an open set and $\phi \in \mathcal{O}_X(U)$ be a regular function. Then $\phi = g^p$ for some $g \in \mathcal{O}_X(U)$, which implies that $\phi(x) = g(x)^p$ for all $x \in U^{\prime} = U$. Thus $\phi$ is a regular function on $U^{\prime}$ as well, and so $U^{\prime}$ and $U$ have the same regular functions. Hence $X^{\prime} \cong X$ as varieties.

Moreover, if we take the composition $X \xrightarrow{Fr} X^{\prime} \cong X$ and restrict it to an affine chart $U = \operatorname{Spec} A$, where $A = k[x_1,\ldots,x_n]/I$ and $I$ is generated by polynomials with coefficients in $\mathbb{F}_p$, then $Fr$ sends $U$ to $\operatorname{Spec} A^{\prime}$, where $A^{\prime} = k[x_1^{\prime},\ldots,x_n^{\prime}]/I$, with $x_i^{\prime p} = x_i^p$ for all $i$. It follows that the map $X \to X^{\prime}$ sends $\left(x_i\right) \mapsto\left(x_i^p\right)$.

(b) Let $X$ be a normal irreducible variety of dimension $n$. Prove that $F r: X \rightarrow X^{\prime}$ is finite, find its degree and prove that every point is its ramification point.

(b) By the Noether normalization theorem, there exists a finite surjective morphism $\pi: X \rightarrow \mathbb{A}^n$. Since $\mathbb{A}^n$ is smooth and $X$ is normal, we have that $\pi$ is a finite morphism. Therefore, it suffices to prove that $F r: \mathbb{A}^n \rightarrow (\mathbb{A}^n)^{\prime}$ is finite, where $(\mathbb{A}^n)^{\prime}$ is the Frobenius twist of $\mathbb{A}^n$.

Let $x_1, \ldots, x_n$ be the coordinates on $\mathbb{A}^n$. Then the coordinate ring of $(\mathbb{A}^n)^{\prime}$ is given by $k[x_1^p, \ldots, x_n^p]$, and the Frobenius morphism is given by $F r(x_i) = x_i^p$. Thus, the map $F r$ is induced by the $p$-th power map on the coordinate ring, which is a finite morphism of rings. Therefore, $F r$ is a finite morphism.

To compute the degree of $F r$, note that $F r$ is a finite morphism between irreducible varieties of the same dimension, so its degree is given by the cardinality of the fiber over a generic point. Let $k’$ be the algebraic closure of $k$, and let $\overline{x}$ be a generic point of $\mathbb{A}^n_{k’}$. Then the fiber of $F r$ over $\overline{x}$ is given by the set of points $y$ in $(\mathbb{A}^n_{k’})^{\prime}$ such that $F r(y) = \overline{x}$. But $F r(y) = (\phi_1(y)^p, \ldots, \phi_n(y)^p)$, where $\phi_1, \ldots, \phi_n$ are regular functions on $y$. Therefore, $F r(y) = \overline{x}$ if and only if $\phi_1(y) = \cdots = \phi_n(y) = \overline{x}^{1/p}$. This is a system of $n$ equations in $n$ variables, and its solutions are the $p^n$ points $y$ in $(\mathbb{A}^n_{k’})^{\prime}$ with coordinates of the form $\alpha_1^{1/p}, \ldots, \alpha_n^{1/p}$, where $\alpha_1, \ldots, \alpha_n$ are the $p$-th power of elements in $k’$. Therefore, the degree of $F r$ is $p^n$.

(c) Describe the intersection points of the graph of Frobenius $F r: \mathbb{A}^1 \rightarrow$ $\mathbb{A}^1$ with the diagonal and check that each one has multiplicity one.

The Frobenius morphism for $\mathbb{A}^1$ is the map $F r: \mathbb{A}^1 \rightarrow \mathbb{A}^1$ given by $F r(x) = x^p$. The graph of $F r$ is the subset of $\mathbb{A}^1 \times \mathbb{A}^1$ defined by the equation $y = x^p$. The diagonal of $\mathbb{A}^1 \times \mathbb{A}^1$ is the subset defined by the equation $y = x$.

To find the intersection points of the graph of $F r$ with the diagonal, we need to solve the system of equations $y = x^p$ and $y = x$. Substituting $y=x$ into the first equation gives $x = x^p$, which implies that $x^{p-1} = 1$. Since we are working over a field of characteristic $p$, we have $p-1 \neq 0$, so $x^{p-1} = 1$ has exactly $p-1$ solutions in $\mathbb{A}^1$, namely $x = 0, 1, \omega, \omega^2, \dots, \omega^{p-2}$, where $\omega$ is a primitive $p$-th root of unity.

For each of these solutions $x$, we have $y=x^p$, so the intersection point is $(x, y) = (x, x^p)$. To check that each intersection point has multiplicity one, we need to compute the Jacobian matrix of the system of equations $y = x^p$ and $y = x$ at each intersection point. This matrix is $\begin{pmatrix} px^{p-1} & -1 \end{pmatrix}$, which has determinant $-p x^{p-1}$. Since $p$ is nonzero in the field we are working over, and $x^{p-1}$ is nonzero for each intersection point, the determinant is nonzero at each intersection point. Therefore, each intersection point has multiplicity one.

# 矢量微积分在流体力学中的应用代写|VECTOR CALCULUS WITH APPLICATIONS IN FLUID MECHANICS MATH225 University of Liverpool Assignment

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Assignment-daixieTM为您提供利物浦大学University of Liverpool DIFFERENTIAL EQUATIONS MATH221微分方程学代写代考辅导服务！

## Instructions:

Vector calculus is an important branch of mathematics that deals with the study of vector fields and their associated operations. The main vector operations studied in vector calculus are the gradient (grad), divergence (div), and curl. These operations are used to study various physical phenomena in fields like fluid mechanics and electromagnetism.

The gradient is a vector operator that describes the rate of change of a scalar field in a particular direction. Mathematically, the gradient of a scalar field is defined as the vector obtained by taking the partial derivative of the scalar field with respect to each of its variables. The gradient of a scalar field is a vector field, and it points in the direction of the maximum rate of increase of the scalar field.

The divergence is a scalar operator that measures the rate at which a vector field flows out of a given point in space. Mathematically, the divergence of a vector field is defined as the sum of the partial derivatives of its components with respect to each of the coordinates. The divergence of a vector field is zero if the field is “solenoidal” or “incompressible,” meaning that it does not have sources or sinks.

The curl is a vector operator that describes the rotational behavior of a vector field. Mathematically, the curl of a vector field is defined as the vector obtained by taking the cross product of the gradient of its components with respect to each of the coordinates. The curl of a vector field is zero if the field is “irrotational,” meaning that it does not have any swirl or vortices.

The motion of a point $P$ is given by the position vector $\vec{R}=3 \cos t \hat{\mathrm{i}}+3 \sin t \hat{\mathrm{j}}+t \hat{\mathrm{k}}$. Compute the velocity and the speed of $P$.

a) Find the area of the space triangle with vertices $P_0:(2,1,0), P_1:(1,0,1), P_2:(2,-1,1)$.

a) $\overrightarrow{P_0 P_1} \times \overrightarrow{P_0 P_2}=\left|\begin{array}{rrr}\hat{\imath} & \hat{\jmath} & \hat{k} \ -1 & -1 & 1 \ 0 & -2 & 1\end{array}\right|=\hat{\imath}+\hat{\jmath}+2 \hat{k} . \quad$ Area $=\frac{1}{2}\left|\overrightarrow{P_0 P_1} \times \overrightarrow{P_0 P_2}\right|=\frac{1}{2} \sqrt{6}$.

c) Find the intersection of this plane with the line parallel to the vector $\vec{V}=\langle 1,1,1\rangle$ and passing through the point $S:(-1,0,0)$.

c) Parametric equations for the line: $x=-1+t, y=t, z=t$.
Substituting: $-1+4 t=3, t=1$, intersection point $(0,1,1)$.

# 微分方程学代写|DIFFERENTIAL EQUATIONS MATH221 University of Liverpool Assignment

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Assignment-daixieTM为您提供利物浦大学University of Liverpool DIFFERENTIAL EQUATIONS MATH221微分方程学代写代考辅导服务！

## Instructions:

Differential equations are indeed fundamental to many areas of mathematics and science, including physics, engineering, economics, and biology, just to name a few. They allow us to model and analyze complex systems and phenomena that are too difficult or impossible to understand using algebraic or geometric techniques alone.

The five parts of the module you mentioned seem to cover a broad range of topics in differential equations, from basic first-order ODEs to more advanced PDEs. It’s great that the module emphasizes both theory and applications, as both are important for understanding and using differential equations effectively.

Solving differential equations can be a difficult and sometimes frustrating task, as there are often many different methods and techniques that can be used, and the solutions can be complex and difficult to interpret. However, with practice and patience, it’s possible to become proficient in solving and analyzing these equations.

Overall, it sounds like MATH201 will be a valuable and challenging module for students who are interested in pursuing further studies in mathematics, science, or engineering. Good luck to all who take it!

In (a)-(c) we consider the autonomous equation $\dot{x}=2 x-3 x^2+x^3$.
(a) Sketch the phase line of this equation.

(a) The equation is $\dot{x}=x(x-1)(x-2)$. The phase line has three equilibria $x=0,1,2$. For $x<0$, the arrow points down. For $02$, the arrow points up.

(b) Sketch the graphs of some solutions. Be sure to include at least one solution with values in each interval above, below, and between the critical points.

(b) The horizontal axis is $t$ and the vertical axis is $x$. There are three constant solutions $x(t) \equiv 0,1,2$. Their graphs are horizontal. Below $x=0$, all solutions are decreasisng and they tend to $-\infty$.

Between $x=0$ and $x=1$, all solutions are increasing and they approach $x=1$. Between $x=1$ and $x=2$, all solutions are decreasing and they approach $x=1$. Above $x=2$, all solutions are increasing and they tend to $+\infty$.

(c) Some solutions have points of inflection. What are the possible values of $x(a)$ if a nonconstant solution $x(t)$ has a point of inflection at $t=a$ ?

(c) A point of inflection $(a, x(a))$ is where $\ddot{x}$ changes sign. In particular, $\ddot{x}(a)$ must be zero. Differentiating the given equation with respect to $t$, we have
$$\ddot{x}=2 \dot{x}-6 x \dot{x}+3 x^2 \dot{x}=\dot{x}\left(2-6 x+3 x^2\right) \text {. }$$
If $x(t)$ is not a constant solution, $\dot{x}(a) \neq 0$ so that $x(a)$ must satisfy
$$2-6 x(a)+3 x(a)^2=0 \quad \Leftrightarrow \quad x(a)=1 \pm \frac{1}{\sqrt{3}} .$$

# 牛顿力学代写|NEWTONIAN MECHANICS MATH122 University of Liverpool Assignment

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Assignment-daixieTM为您提供利物浦大学University of Liverpool NEWTONIAN MECHANICS MATH122牛顿力学代写代考辅导服务！

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Newtonian mechanics, also known as classical mechanics, is a branch of physics that deals with the study of motion and its causes. It was founded by Sir Isaac Newton in the 17th century and is based on three laws known as Newton’s laws of motion.

Newton’s first law of motion states that an object at rest will remain at rest, and an object in motion will continue in motion with a constant velocity unless acted upon by an external force. This law is also known as the law of inertia.

Newton’s second law of motion states that the force acting on an object is equal to the mass of the object times its acceleration. Mathematically, F=ma, where F is the force, m is the mass, and a is the acceleration.

Newton’s third law of motion states that for every action, there is an equal and opposite reaction. This law implies that forces always occur in pairs.

In addition to these laws, Newtonian mechanics also includes the concepts of energy, work, and momentum. The laws of conservation of energy, work-energy theorem, and conservation of momentum are also important principles in this branch of physics.

Newtonian mechanics has been instrumental in the development of modern physics, and its principles are still widely used in many fields of science and engineering.

A rocket in zero gravitational field has a mass of $m_{r, i}=2.81 \times 10^7 \mathrm{~kg}$, which is the sum of the mass of the fuel $m_{f, i}=2.46 \times 10^7 \mathrm{~kg}$ and the dry mass of the rocket (empty of fuel) $m_{r, d} \equiv m_{r, i}-m_{f, i}=0.35 \times 10^7 \mathrm{~kg}$. The fuel is ejected at a speed $u=3000$ $\mathrm{m} / \mathrm{s}$ relative to the rocket. The total burn time is $510 \mathrm{~s}$ and the fuel is burned at a constant rate. (a) What is the final speed $v_f$ of the rocket in meters/second after all the fuel is burned assuming it starts from rest?

The total mass of the rocket decreases during the burn due to the ejection of fuel. The change in velocity of the rocket is given by the rocket equation:

$\Delta v=v_{f}-v_{i}=u \ln \left(\frac{m_{r, i}}{m_{r, f}}\right)$

where $v_i = 0$ is the initial velocity of the rocket, $v_f$ is the final velocity of the rocket, $m_{r,i}$ is the initial mass of the rocket, $m_{r,f}$ is the final mass of the rocket, and $u$ is the relative velocity of the ejected fuel. At the end of the burn, all the fuel has been ejected, so the final mass of the rocket is the dry mass $m_{r,d}$:

$m_{r, f}=m_{r, d}=0.35 \times 10^7 \mathrm{~kg}$

Thus,

$\Delta v=u \ln \left(\frac{m_{r, i}}{m_{r, d}}\right)=3000 \mathrm{~m} / \mathrm{s} \ln \left(\frac{2.81 \times 10^7 \mathrm{~kg}}{0.35 \times 10^7 \mathrm{~kg}}\right) \approx 7974.4 \mathrm{~m} / \mathrm{s}$

Therefore, the final speed of the rocket is $v_f = \Delta v = 7974.4$ m/s.

(b) Now suppose that the same rocket burns the fuel in two stages, expelling the fuel in each stage at the same relative speed $u=3000 \mathrm{~m} / \mathrm{s}$. In stage one, the available fuel to burn is $m_{f, 1, i}=2.03 \times 10^7 \mathrm{~kg}$ with burn time $150 \mathrm{~s}$. The total mass of the rocket after all the fuel in stage 1 is burned is $m_{r, 1, d}=m_{r, i}-m_{f, 1, i}=0.78 \times 10^7$ kg. What is the change in speed after stage one is complete?

In stage one, the initial mass of the rocket is $m_{r,1,i}=m_{r,i}=2.81 \times 10^7 \mathrm{~kg}$ and the initial mass of fuel is $m_{f,1,i}=2.03 \times 10^7 \mathrm{~kg}$. The mass of the rocket after all the fuel in stage 1 is burned is $m_{r,1,f}=m_{r,1,i}-m_{f,1,i}=0.78 \times 10^7$ kg. By the principle of conservation of momentum, the change in velocity of the rocket after stage one is complete is given by:

$\Delta v_1=u\ln \left(\frac{m_{r,1,i}}{m_{r,1,f}}\right)$

where $u=3000 \mathrm{~m}/\mathrm{s}$ is the relative speed of the ejected fuel and $m_{r,1,i}$ and $m_{r,1,f}$ are the initial and final masses of the rocket in stage one, respectively. Substituting the values:

$\Delta v_1=3000 \mathrm{~m}/\mathrm{s} \ln \left(\frac{2.81 \times 10^7 \mathrm{~kg}}{0.78 \times 10^7 \mathrm{~kg}}\right) \approx 3066 \mathrm{~m}/\mathrm{s}$

Therefore, the change in speed after stage one is complete is approximately $3066 \mathrm{~m}/\mathrm{s}$.

(c) Next, the empty fuel tank and accessories from stage one are disconnected from the rest of the rocket. These disconnected parts had a mass $m=1.4 \times 10^6 \mathrm{~kg}$, hence the remaining dry mass of the rocket is $m_{r, 2, d}=2.1 \times 10^6 \mathrm{~kg}$. All the remaining fuel with mass $m_{f, 2, i}=4.3 \times 10^6 \mathrm{~kg}$ is burned during stage 2 with burn time of $360 \mathrm{~s}$. What is the change in speed in meters/second after stage two is complete?

To calculate the change in speed after stage two, we can use the rocket equation, which relates the change in velocity of a rocket to the mass ratio of the rocket before and after the burn, and the effective exhaust velocity of the propellant.

The mass ratio of the rocket before and after stage two is:

$$\frac{m_{i}}{m_{f, 2}}=\frac{m_{r, i}+m_{f, i}+m}{m_{r, 2, d}}=\frac{2.81 \times 10^7 \mathrm{~kg}+2.46 \times 10^7 \mathrm{~kg}+1.4 \times 10^6 \mathrm{~kg}}{2.1 \times 10^6 \mathrm{~kg}}=22.95$$

where $m_{i}$ is the initial mass of the rocket (including all the fuel and the dry mass) and $m_{f,2}$ is the final mass of the rocket after stage two (including the dry mass and all the fuel burned in stage two).

The effective exhaust velocity $v_{e}$ is given by:

$$v_{e}=u \ln\frac{m_{i}}{m_{f,2}}=3000 \mathrm{~m/s} \ln\frac{2.81 \times 10^7 \mathrm{~kg}}{2.1 \times 10^6 \mathrm{~kg}+4.3 \times 10^6 \mathrm{~kg}}=2445.5 \mathrm{~m/s}$$

where $u$ is the speed of the ejected fuel relative to the rocket.

Now we can use the rocket equation:

$$\Delta v=v_{e} \ln\frac{m_{i}}{m_{f,2}}=2445.5 \mathrm{~m/s} \ln\frac{2.81 \times 10^7 \mathrm{~kg}}{2.1 \times 10^6 \mathrm{~kg}+4.3 \times 10^6 \mathrm{~kg}}=2434.4 \mathrm{~m/s}$$

Therefore, the change in speed after stage two is $\Delta v = 2434.4$ $\mathrm{m/s}$.

# 计算物理学代写|INTRODUCTION TO COMPUTATIONAL PHYSICS PHYS105 University of Liverpool Assignment

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Assignment-daixieTM为您提供利物浦大学University of Liverpool INTRODUCTION TO COMPUTATIONAL PHYSICS PHYS105计算物理学代写代考辅导服务！

## Instructions:

Computational physics is a subfield of physics that involves the use of computers to solve problems in physics that are too complex to be solved analytically. The use of computational techniques has become increasingly important in physics research and has led to many breakthroughs in our understanding of the physical world.

To get started with computational physics, it’s important to have a strong foundation in mathematics and programming. In terms of programming languages, Python is a popular choice for computational physics because of its ease of use and large user community.

There are many resources available online to help you learn computational physics. Some good places to start include:

• The Open Source Physics Project: This project provides a collection of open-source software tools and simulations for teaching and researching physics. They have a range of resources and materials available for learning computational physics, including tutorials and sample code.
• Coursera: Coursera offers a range of online courses in computational physics, including courses on Python programming and numerical methods for physics.
• GitHub: GitHub is a platform for collaborative software development, and there are many open-source repositories available on GitHub that contain code for solving physics problems. You can use these repositories to learn from existing code and contribute to the development of new code.
• Textbooks: There are many textbooks available on computational physics, including “Computational Physics” by Mark Newman and “An Introduction to Computational Physics” by Tao Pang.

I hope this information is helpful! Let me know if you have any other questions.

(a) Write a procedure that solves quadratic equations using the quadratic formula: It should take as arguments three numbers a, b, and c. It should print error messages if a is zero, or if the roots are complex. Otherwise it should print the two roots.

# a.) code for roots function
def roots(a, b, c):
#form of equation is a*x**2 + b*x + c = 0
#quadratic formula is   x = (-b + sqrt(b**2 - 4 * a * c))/(2 * a)
#                   or  x = (-b - sqrt(b**2 - 4 * a * c))/(2 * a)
#roots are complex when the discriminant (b**2 - 4*a*c) is negative
discriminant = b**2 - (4 * a * c)
if discriminant < 0:
return "Roots are complex"
return "x = "+str((-b + math.sqrt(discriminant)) / (2 * a))+" or x = "+\
str((-b - math.sqrt(discriminant)) / (2 * a))

# Test Cases (assertions are just to automate)
##print '\nTesting roots'
##
##print roots(1, 2, 1) #(x+1)^2: double root at x = -1
##assert roots(1, 2, 1) == 'x = -1.0 or x = -1.0'
##
##print roots(1, -2, -3) #(x+1)(x-3): roots at x = 3 or x = -1
##assert roots(1, -2, -3) == 'x = 3.0 or x = -1.0'
##
##print roots(2, 2, 2) #2x^2 + 2x + 2: complex roots
##assert roots(2, 2, 2) == 'Roots are complex'

(b) Modify your procedure to handle the case of complex roots.

# b.) same code, modified to handle complex roots
def roots(a, b, c):
discriminant = b**2 - (4 * a * c)
if discriminant < 0:
discriminant = discriminant + 0j
return "x = "+str((-b + discriminant**0.5) / (2 * a))+ " or x = " +\
str((-b - discriminant**0.5) / (2 * a)) # math.sqrt does not work

# Test Cases (assertions are just to automate)
##print '\nTesting roots'
##
##print roots(1, 2, 1) #(x+1)^2: double root at x = -1
##assert roots(1, 2, 1) == 'x = -1.0 or x = -1.0'
##
##print roots(1, -2, -3) #(x+1)(x-3): roots at x = 3 or x = -1
##assert roots(1, -2, -3) == 'x = 3.0 or x = -1.0'
##
##print roots(2, 2, 2) #2x^2 + 2x + 2: complex roots
##assert roots(2, 2, 2) == 'x = (-0.5+0.866025403784j) or x = (-0.5-0.866025403784j)'

###########################################
## 2.) procedure for evaluating polynomials
###########################################
def eval_poly(x, coeffs):
total=0 #will keep a running total of the sum
coeffs.reverse() # to put the low order coeffs first
for i in range(len(coeffs)):
total+=coeffs[i]*(x**i) #add the curent term to the total

# Test cases
##print '/nTesting Polynomial'
##
##print eval_poly(1,[1,2,3])
##assert eval_poly(1,[1,2,3]) == 6
##
##print eval_poly(2,[1,2,3,4])
##assert eval_poly(2,[1,2,3,4]) == 26


Write a procedure that evaluates polynomials. It should take two arguments. The first is a number $x$. The second is a list of of coefficients ordered from highest to lowest:
$$a_n, a_{n-1}, \ldots, a_2, a_1, a_0$$
Your procedure should return the value of the polynomial evaluated at $x$ :
$$a_n x^n+a_{n-1} x^{n-1}+\ldots+a_2 x+a_1 x+a_0$$

) procedure for evaluating polynomials
###########################################
def eval_poly(x, coeffs):
total=0 #will keep a running total of the sum
coeffs.reverse() # to put the low order coeffs first
for i in range(len(coeffs)):
total+=coeffs[i]*(x**i) #add the curent term to the total

# Test cases
##print '/nTesting Polynomial'
##
##print eval_poly(1,[1,2,3])
##assert eval_poly(1,[1,2,3]) == 6
##
##print eval_poly(2,[1,2,3,4])
##assert eval_poly(2,[1,2,3,4]) == 26

###########################################
## 3.) procedure to make change
###########################################
def make_change(cost, paid):
change=paid-cost #calculate the change due
bills={20:0,10:0,5:0,2:0,1:0} #dictionary that maps each bill to the amount of it
bill_list=bills.keys() #bill_list is a list of the available bills
bill_list.sort() #sort it in ascending order
bill_list.reverse() #now reverse it to be in descending order
#this is to make sure you get the smallest number of bills
for bill in bill_list:
while change >=bill:
bills[bill]+=1 #increment the amount of that bill
change-=bill #decrease the change by that bill

print "Change is:"
for bill in bill_list:
if bills[bill] == 1:
print bills[bill], bill, "dollar bill"
elif bills[bill] > 1:
print bills[bill], bill, "dollar bills"

#return bills #return the dictionary with amounts of each bill needed

make_change(1, 6)
make_change(4, 109)

# 量子物理学基础代写|FOUNDATIONS OF QUANTUM PHYSICS PHYS104 University of Liverpool Assignment

0

Assignment-daixieTM为您提供利物浦大学University of Liverpool FOUNDATIONS OF QUANTUM PHYSICS PHYS104微积分代写代考辅导服务！

## Instructions:

The experiments in question, such as the double-slit experiment and the photoelectric effect, challenged classical physics and led to the development of quantum mechanics, which has since become a fundamental theory in physics. It explains the behavior of matter and energy at the atomic and subatomic level, where classical mechanics fails. Quantum mechanics has led to the development of many technologies, including lasers, transistors, and quantum computing.

Consider a two-state system with Hamiltonian $$H(t)=\left(\begin{array}{cc} +E & v(t) \\ v(t) & -E \end{array}\right)$$ where $v(t)$ is real and where $v \rightarrow 0$ for $t \rightarrow \pm \infty$. (a) Suppose that at $t=-\infty$ the system is in the state $|1\rangle$. Use time dependent perturbation theory to determine the probability that at $t=+\infty$ the system is in the state $|2\rangle$, to lowest order in $v$.

$P_{i \rightarrow f}=\frac{1}{\hbar^2}\left|\int_{-\infty}^{+\infty}\left\langle\psi_f(t)\left|H^{\prime}(t)\right| \psi_i(t)\right\rangle d t\right|^2$

where $H'(t)$ is the time-dependent perturbation, and $\langle \psi_f(t)|$ and $|\psi_i(t) \rangle$ are the time-dependent states of the system. In this case, the initial state is $|1\rangle$ and the final state is $|2\rangle$.

To lowest order in $v$, the perturbation to the Hamiltonian is given by $H'(t) = v(t) \left(\begin{smallmatrix} 0 & 1 \ 1 & 0 \end{smallmatrix}\right)$. Therefore, we have: \begin{align*} P_{1 \rightarrow 2} &= \frac{1}{\hbar^2} \left| \int_{-\infty}^{+\infty} \langle 2 | H'(t) | 1 \rangle e^{i(E_2-E_1)t/\hbar} , dt \right|^2 \ &= \frac{1}{\hbar^2} \left| \int_{-\infty}^{+\infty} v(t) e^{i2Et/\hbar} , dt \right|^2 \ &= \frac{4}{\hbar^2} \left| \int_{0}^{+\infty} v(t) \cos(2Et/\hbar) , dt \right|^2 \end{align*} where we used the fact that $E_2-E_1=2E$ and $\langle 2 | 1 \rangle = 0$.

Since $v \rightarrow 0$ for $t \rightarrow \pm \infty$, we can assume that $v(t)$ is non-zero only for a finite interval of time $T$ around $t=0$. Therefore, we have: \begin{align*} P_{1 \rightarrow 2} &= \frac{4}{\hbar^2} \left| \int_{-T/2}^{T/2} v(t) \cos(2Et/\hbar) , dt \right|^2 \ &\approx \frac{16}{\hbar^4} \left| \int_{0}^{T/2} v(t) \cos(2Et/\hbar) , dt \right|^2 \end{align*} where we used the fact that $v(t)$ is an even function and that the integral over $(-T/2,0)$ gives a contribution of order $v^2$, which we neglect to lowest order in $v$.

Therefore, the lowest-order probability of transition from $|1\rangle$ to $|2\rangle$ is proportional to the squared integral of $v(t)$ over a finite interval of time around $t=0$.

(b) If $E=0$, the eigenstates of $H(t)$ do not depend on $t$. Use this fact to calculate the probability of a transition from $|1\rangle$ to $|2\rangle$ exactly, in this case. What is the result obtained from time-dependent perturbation theory in this case? What is the condition that the perturbative result is a good approximation to the exact result?

When $E=0$, the Hamiltonian becomes

$H(t)=\left(\begin{array}{cc}0 & v(t) \ v(t) & 0\end{array}\right)$

The eigenstates of this Hamiltonian are given by

$| \pm\rangle=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1 \ \pm 1\end{array}\right)$,

with eigenvalues $\pm |v(t)|$. Note that these eigenstates do not depend on time.

Suppose the system starts out in the state $|1\rangle$. The probability of a transition from $|1\rangle$ to $|2\rangle$ is given by

$P_{1 \rightarrow 2}=\left|\int_{-\infty}^{\infty} d t\langle 2|v(t)| 1\rangle e^{i \omega_{21} t}\right|^2$

where $\omega_{21}$ is the energy difference between states $|2\rangle$ and $|1\rangle$. In this case, we have $\omega_{21}=2|v(t)|$.

Since $|1\rangle$ and $|2\rangle$ are not the eigenstates of the Hamiltonian, we need to use time-dependent perturbation theory to calculate the transition probability. The first-order perturbation theory gives

$P_{1 \rightarrow 2}^{(1)}=\frac{|\langle 2|v(t)| 1\rangle|^2}{\omega_{21}^2}$.

Substituting $|1\rangle$ and $|2\rangle$ and using the fact that they are orthonormal, we get

$P_{1 \rightarrow 2}^{(1)}=\frac{1}{2 \omega_{21}^2}\left|\left\langle 2\left|\sigma_x\right| 1\right\rangle\right|^2$,

where $\sigma_x$ is the Pauli matrix.

Using the fact that $|\pm\rangle$ are eigenstates of $\sigma_x$, we can write

$P_{1 \rightarrow 2}^{(1)}=\frac{1}{4 \omega_{21}^2}$

To compare the perturbative result with the exact result, we need to consider the condition for which the perturbation is small. The perturbation is small when $|v(t)|$ is small compared to $E$. This is because the unperturbed energy gap is $E$, and if the perturbation is much smaller than this, then the perturbative result will be a good approximation to the exact result. Therefore, the condition for the perturbative result to be a good approximation is

$|v(t)| \ll E$.

Consider $s$-wave scattering for a particle of mass $m$ off a potential $V(r)$ which vanishes at the origin, rises steadily as $r$ increases from zero, reaches a maximum at $r=c$, and then goes quickly to zero as $r$ increases further.

For $\ell=0$, the radial wave function $u(r)$ satisfies the same Schrödinger equation as that for a particle in one dimension with potential $V$, subject to the boundary condition $u(0)=0$.

Consider scattering with energy $E$ where $0 \ll E \ll V(c)$. The classical turning points are at $r=a$ and $r=b$ with $a<c<b$.
(a) What is the semiclassical approximation to the wave function in the classically allowed region, $0 \leq r<a$ ?

In the classically allowed region, the semiclassical approximation to the radial wave function is given by $$u(r) \approx \frac{1}{\sqrt{p(r)}} \sin \left( \int_{0}^{r} p(r’) dr’ + \frac{\ell \pi}{2} \right)$$ where $p(r) = \sqrt{2m(E – V(r))}$ is the classical momentum.

In this case, we have $\ell = 0$, so the phase term reduces to $\frac{\ell \pi}{2} = 0$.

Since $V(r)$ vanishes at the origin, we have $p(0) = \sqrt{2mE}$. Near $r=0$, the potential $V(r)$ rises steadily, so we can approximate $V(r)$ by its first-order Taylor expansion around $r=0$: $$V(r) \approx V(0) + rV'(0) = 0$$ since $V(0) = 0$ and $V'(0)$ is finite. Therefore, we have $$p(r) \approx \sqrt{2mE} – \frac{1}{2}\sqrt{2m}V”(0)r + \mathcal{O}(r^2).$$

Integrating $p(r)$ from $0$ to $r$, we obtain $$\int_{0}^{r} p(r’) dr’ \approx \sqrt{2mE} r – \frac{1}{6}\sqrt{2m}V”(0)r^3.$$

Substituting this expression back into the semiclassical approximation for $u(r)$, we get $$u(r) \approx \frac{1}{\sqrt{\sqrt{2mE} – \frac{1}{2}\sqrt{2m}V”(0)r}} \sin \left( \sqrt{2mE} r – \frac{1}{6}\sqrt{2m}V”(0)r^3 \right)$$ in the classically allowed region $0 \leq r < a$.

# 微积分代写|CALCULUS II MATH102 University of Liverpool Assignment

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Assignment-daixieTM为您提供利物浦大学University of Liverpool CALCULUS II MATH102微积分代写代考辅导服务！

## Instructions:

Part I of the module focuses on power series and their applications. Power series are infinite series that can represent functions as a sum of powers of a variable. You will learn about the properties of power series and their convergence, as well as the relationship between power series and functions, including how to use Taylor and Maclaurin series to approximate functions.

Part II of the module introduces functions of several variables and partial differentiation. You will learn how to find partial derivatives, including the chain rule, total differential, directional derivative, and tangent planes. Additionally, you will learn how to find extrema of functions of several variables and how to use Taylor expansions to approximate functions.

Finally, part III of the module focuses on double integrals and their applications. You will learn how to evaluate double integrals using different methods, including iterated integrals and change of variables. You will also learn how to use double integrals to find areas, volumes, and centroids of two-dimensional and three-dimensional objects.

Overall, this module, together with MATH101 and MATH103, provides a strong foundation in calculus and linear algebra, which is essential for further studies in mathematics and many other fields, such as physics, engineering, and economics.

$\vec{R} \cdot \vec{V}=0$, so $\frac{d}{d t}(\vec{R} \cdot \vec{V})=\vec{V} \cdot \vec{V}+\vec{R} \cdot \vec{A}=0$. Therefore $\vec{R} \cdot \vec{A}=-|\vec{V}|^2$.

We can start by differentiating the expression $\vec{R} \cdot \vec{V}$ with respect to time $t$ using the product rule:

$$\frac{d}{dt}(\vec{R} \cdot \vec{V}) = \frac{d \vec{R}}{dt} \cdot \vec{V} + \vec{R} \cdot \frac{d \vec{V}}{dt}$$

Since $\vec{R} \cdot \vec{V}$ is a scalar quantity, its derivative must be a scalar as well. We can simplify the above expression using the fact that $\vec{R}$ and $\vec{V}$ are perpendicular (i.e. $\vec{R} \cdot \vec{V} = 0$), which implies that $\frac{d \vec{R}}{dt} \cdot \vec{V} = 0$. Therefore:

$$\frac{d}{dt}(\vec{R} \cdot \vec{V}) = \vec{R} \cdot \frac{d \vec{V}}{dt}$$

Now, we can substitute the expression for the acceleration $\vec{A} = \frac{d \vec{V}}{dt}$:

$$\frac{d}{dt}(\vec{R} \cdot \vec{V}) = \vec{R} \cdot \vec{A}$$

Using the chain rule, we can rewrite the left-hand side as:

$$\frac{d}{dt}(\vec{R} \cdot \vec{V}) = \frac{d\vec{R}}{dt} \cdot \vec{V} + \vec{R} \cdot \frac{d\vec{V}}{dt} = \vec{V} \cdot \vec{V} + \vec{R} \cdot \vec{A}$$

Substituting this into the previous expression, we get:

$$\vec{V} \cdot \vec{V} + \vec{R} \cdot \vec{A} = 0$$

Solving for $\vec{R} \cdot \vec{A}$, we get:

$$\vec{R} \cdot \vec{A} = -|\vec{V}|^2$$

Therefore, we have shown that $\vec{R} \cdot \vec{A} = -|\vec{V}|^2$ given the initial condition $\vec{R} \cdot \vec{V} = 0$.

a) $P=(1,0,0), Q=(0,2,0)$ and $R=(0,0,3)$. Therefore $\overrightarrow{Q P}=\hat{\boldsymbol{\imath}}-2 \hat{\boldsymbol{\jmath}}$ and $\overrightarrow{Q R}=-2 \hat{\boldsymbol{\jmath}}+3 \hat{\boldsymbol{k}}$.

To calculate the vector $\overrightarrow{QP}$, we subtract the coordinates of $P$ from the coordinates of $Q$:

$$\overrightarrow{QP} = \begin{pmatrix}0-1 \ 2-0 \ 0-0\end{pmatrix} = \begin{pmatrix}-1 \ 2 \ 0\end{pmatrix} = -1\hat{\boldsymbol{\imath}} + 2\hat{\boldsymbol{\jmath}}$$

To calculate the vector $\overrightarrow{QR}$, we subtract the coordinates of $R$ from the coordinates of $Q$:

$$\overrightarrow{QR} = \begin{pmatrix}0-0 \ 2-0 \ 3-0\end{pmatrix} = \begin{pmatrix}0 \ 2 \ 3\end{pmatrix} = 2\hat{\boldsymbol{\jmath}} + 3\hat{\boldsymbol{k}}$$

So, as you said, $\overrightarrow{QP} = \hat{\boldsymbol{\imath}} – 2 \hat{\boldsymbol{\jmath}}$ and $\overrightarrow{QR} = -2\hat{\boldsymbol{\jmath}} + 3\hat{\boldsymbol{k}}$.

$\vec{N} \cdot \vec{r}(t)=6$, where $\vec{N}=\langle 4,-3,-2\rangle$.

The equation $\vec{N} \cdot \vec{r}(t)=6$ represents a plane in three-dimensional space, where $\vec{N}$ is the normal vector to the plane and $\vec{r}(t)$ is a vector pointing to any point on the plane at time $t$.

Given that $\vec{N}=\langle 4,-3,-2\rangle$, we can write the equation of the plane as:

$$4x – 3y – 2z = 6$$

where $x$, $y$, and $z$ are the coordinates of any point on the plane.

Alternatively, we can express the equation in vector form as:

$$\langle 4,-3,-2\rangle \cdot \langle x,y,z\rangle = 6$$

or

$$\langle 4,-3,-2\rangle \cdot \vec{r}(t) = 6$$

where $\vec{r}(t)=\langle x(t),y(t),z(t)\rangle$ is a vector function that describes the position of any point on the plane at time $t$.

Note that there are infinitely many points on the plane described by this equation, since the equation has three variables but only one constraint. Any point that satisfies the equation $4x – 3y – 2z = 6$ lies on the plane, and there are infinitely many such points.

# 微积分代写|CALCULUS I MATH101 University of Liverpool Assignment

0

Assignment-daixieTM为您提供利物浦大学University of Liverpool CALCULUS I MATH101微积分代写代考辅导服务！

## Instructions:

To add more details, calculus is a branch of mathematics that deals with the study of rates of change and accumulation. It is divided into two main branches: differential calculus and integral calculus.

Differential calculus deals with the study of rates of change, such as slopes of tangent lines, velocity, and acceleration. It involves concepts such as derivatives, limits, and continuity. Derivatives are used to calculate the instantaneous rate of change of a function at a specific point.

Integral calculus deals with the study of accumulation, such as finding the area under a curve or the volume of a solid. It involves concepts such as integrals, limits, and the fundamental theorem of calculus. Integrals are used to calculate the total amount of a quantity that accumulates over an interval of time or a region of space.

Calculus has numerous practical applications in various fields such as physics, engineering, economics, and biology. It provides a powerful tool for solving real-world problems by modeling them mathematically and then applying calculus techniques to obtain solutions.

$\mathrm{Q}=$ top of the ladder: $\overrightarrow{\mathrm{OQ}}=\langle 0, L \sin \theta\rangle ; \quad \mathrm{R}=$ bottom of the ladder: $\overrightarrow{\mathrm{OR}}=\langle-L \cos \theta, 0\rangle$.
Midpoint: $\overrightarrow{\mathrm{OP}}=\frac{1}{2}(\overrightarrow{\mathrm{OQ}}+\overrightarrow{\mathrm{OR}})=\left\langle-\frac{L}{2} \cos \theta, \frac{L}{2} \sin \theta\right\rangle$.
Parametric equations: $x=-\frac{L}{2} \cos \theta, y=\frac{L}{2} \sin \theta$.

It looks like you have provided the vector representations and parametric equations for three points in a coordinate plane: point Q at the top of a ladder, point R at the bottom of the ladder, and the midpoint P between points Q and R.

The vector representation of point Q is $\overrightarrow{\mathrm{OQ}}=\langle 0, L \sin \theta\rangle$, which means that point Q is located at the origin of the coordinate plane (since the x-coordinate is 0) and has a y-coordinate of $L \sin \theta$.

The vector representation of point R is $\overrightarrow{\mathrm{OR}}=\langle-L \cos \theta, 0\rangle$, which means that point R is located on the x-axis (since the y-coordinate is 0) and has an x-coordinate of $-L \cos \theta$.

The vector representation of point P is found by taking the average of the vectors for points Q and R: $\overrightarrow{\mathrm{OP}}=\frac{1}{2}(\overrightarrow{\mathrm{OQ}}+\overrightarrow{\mathrm{OR}})=\left\langle-\frac{L}{2} \cos \theta, \frac{L}{2} \sin \theta\right\rangle$. This means that point P is located at $x=-\frac{L}{2} \cos \theta$ and $y=\frac{L}{2} \sin \theta$.

The parametric equations for these points are simply the x and y coordinates expressed as functions of the parameter $\theta$: $x=-\frac{L}{2} \cos \theta, y=\frac{L}{2} \sin \theta$. These equations allow you to plot the points on the coordinate plane as $\theta$ varies.

a) $\frac{d}{d t}(\vec{R} \cdot \vec{R})=\vec{V} \cdot \vec{R}+\vec{R} \cdot \vec{V}=2 \vec{R} \cdot \vec{V}$.

We can use the product rule of differentiation and the fact that the derivative of a constant vector is zero to find the derivative of $\vec{R} \cdot \vec{R}$ with respect to time:

$$\frac{d}{dt}(\vec{R} \cdot \vec{R}) = \frac{d}{dt}(\sum_{i=1}^{n} R_i^2) = \sum_{i=1}^{n} \frac{d}{dt}(R_i^2)$$

Using the chain rule, we have:

$$\frac{d}{dt}(R_i^2) = 2R_i \frac{dR_i}{dt}$$

Therefore:

$$\frac{d}{dt}(\vec{R} \cdot \vec{R}) = \sum_{i=1}^{n} 2R_i \frac{dR_i}{dt} = 2\sum_{i=1}^{n} R_i \frac{dR_i}{dt}$$

Using the product rule of differentiation, we have:

$$\frac{d}{dt}(\vec{R} \cdot \vec{R}) = 2\sum_{i=1}^{n} R_i \frac{dR_i}{dt} = 2\sum_{i=1}^{n} R_i \frac{dR_i}{dt} + 2\sum_{i=1}^{n} R_i \frac{dR_i}{dt} = 2\sum_{i=1}^{n} R_i \frac{dR_i}{dt} + 2\sum_{i=1}^{n} \frac{dR_i}{dt} R_i$$

Using the dot product rule, we have:

$$\frac{d}{dt}(\vec{R} \cdot \vec{R}) = 2\sum_{i=1}^{n} R_i \frac{dR_i}{dt} + 2\sum_{i=1}^{n} \frac{dR_i}{dt} R_i = 2\vec{R} \cdot \vec{V}$$

Therefore, we have:

$$\frac{d}{d t}(\vec{R} \cdot \vec{R})=\vec{V} \cdot \vec{R}+\vec{R} \cdot \vec{V}=2 \vec{R} \cdot \vec{V}.$$

b) Assume $|\vec{R}|$ is constant: then $\frac{d}{d t}(\vec{R} \cdot \vec{R})=2 \vec{R} \cdot \vec{V}=0$, i.e. $\vec{R} \perp \vec{V}$.

Starting with $\vec{R}\cdot\vec{R}=|\vec{R}|^2$, we can take the derivative with respect to time $t$ using the chain rule:

$$\frac{d}{d t}(\vec{R} \cdot \vec{R})=\frac{d}{d t}(|\vec{R}|^2)=2|\vec{R}|\frac{d|\vec{R}|}{dt}.$$

Since we are assuming $|\vec{R}|$ is constant, we have $\frac{d|\vec{R}|}{dt}=0$, so

$$\frac{d}{d t}(\vec{R} \cdot \vec{R})=2|\vec{R}|\frac{d|\vec{R}|}{dt}=0.$$

Therefore, $\vec{R}\cdot\vec{V}=0$ because the dot product of any two vectors is zero if and only if the vectors are perpendicular. Thus, we have $\vec{R}\perp\vec{V}$ as required.

# 动态系统代写|DYNAMICAL SYSTEMS MATHS4074 University of Glasgow Assignment

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Assignment-daixieTM为您提供格拉斯哥大学University of Glasgow DYNAMICAL SYSTEMS MATHS4074动态系统代写代考辅导服务！

## Instructions:

Stability analysis involves studying the behavior of solutions to differential equations or maps as time goes on. An equilibrium solution is a solution that doesn’t change over time, so it’s important to determine whether these solutions are stable (perturbations from the equilibrium decay over time) or unstable (perturbations grow over time). Depending on the system and the equilibrium solution, there are different methods for determining stability, including linearization, Lyapunov functions, and phase portraits.

Bifurcation theory is concerned with how the behavior of a dynamic system changes as parameters are varied. When a system undergoes a bifurcation, its behavior can change drastically, for example, an equilibrium solution may suddenly appear or disappear, or the system may exhibit periodic or chaotic behavior. Understanding the bifurcations that a system undergoes can provide insight into the underlying mechanisms that govern its behavior.

Overall, the topics covered in this course should provide you with a solid foundation for understanding the behavior of nonlinear systems and the methods used to analyze them.

Let $A \in \mathbb{C}^{n \times n}$, and $B \in \mathbb{C}^{m \times m}$. Show that $X(t)=e^{A t} X(0) e^{B t}$ is the solution to $\dot{X}=$ $A X+X B$

Recalling the definition of matrix exponential, $e^{A t}=\sum_{i=0}^{\infty} \frac{1}{i !}(A t)^i$, it is clear that, for any matrix $A, e^{A t}=I$ for $t=0$, and $d e^{A t} / d t=A e^{A t}=e^{A t} A$.
Hence,
\begin{aligned} \frac{d}{d t}\left(e^{A t} X(0) e^{B t}\right)=\left(\frac{d}{d t} e^{A t}\right) X(0) e^{B t}+e^{A t} & \left(\frac{d}{d t} X(0)\right) e^{B t}+e^{A t} X(0)\left(\frac{d}{d t} e^{B t}\right) \ & =A\left(e^{A t} X(0) e^{B t}\right)+0+\left(e^{A t} X(0) e^{B t}\right) B \end{aligned}
Furthermore, for $t=0$,
$$\left.\left(e^{A t} X(0) e^{B t}\right)\right|_{t=0}=X(0)$$
Hence we can conclude that the proposed function is in fact the solution to the initialvalue problem under consideration.

Use the projection theorem to solve the problem:
$$\min _{x \in \mathbb{R}^n}\left{x^T Q x: A x=b\right}$$
where $Q$ is a positive-definite $n \times n$ matrix, $A$ is a $m \times n$ real matrix, with rank $m<n$, and $b$ is a real $m$-dimensional vector. Is the solution unique?

(Note that $Q$ being positive-definite implies it is self-adjoint, i.e., Hermitian.) Let $x_0$ be such that $A x_0=b$, and consider the change of variables $z=x-x_0$. In the inner product space $\mathbb{R}^n$, with inner product $\langle u, v\rangle=u^T Q v$, it is desired to minimize $|x|^2=x^T Q x=\left|z+x_0\right|^2$, subject to the constraint that $z$ lies in the subspace $M:=\left{z \in \mathbb{R}^n: A z=0\right}$. Using the projection theorem, we know that an optimal solution $\hat{z}=\hat{x}-x_0$ must be such that $\left(\hat{z}+x_0=\hat{x}\right) \perp M$, i.e., $\langle\hat{x}, y\rangle=\hat{x}^T Q y=0$, for all $y \in M$. Summarizing, we know that
\begin{aligned} \hat{x}^T Q y & =0, \quad \forall A y=0 \ A x & =b . \end{aligned}
In order to satisfy the first equation for all $y$ such that $A y=0, \hat{x}$ must be of the form $\hat{x}=Q^{-1} A^T v$, for some $v \in \mathbb{R}^m$. The vector $v$ can be found using the constraint $A x=b$, i.e.,
$$A \hat{x}=A Q^{-1} A^T v=b,$$
and hence
$$v=\left(A Q^{-1} A^T\right)^{-1} b .$$
Concluding,
$$\hat{x}=Q^{-1} A^T\left(A Q^{-1} A^T\right)^{-1} b$$

Let $|A|<1$. Show that $\left|(I-A)^{-1}\right| \geq \frac{1}{1+|A|}$.

First of all, for any vector $x_0$, with $\left|x_0\right|=1$,
$$\left|(I-A) x_0\right| \geq\left|x_0\right|-\left|A x_0\right| \geq 1-|A|>0$$

which shows that the matrix $I-A$ is invertible, i.e., there is no vector $x_0$, with $\left|x_0\right|=1$, such that $(I-A) x_0=0$.
Furthermore, the following chain of inequalities holds:
\begin{aligned} 1=|I|=\left|(I-A)(I-A)^{-1}\right| & \leq|I-A| \cdot\left|(I-A)^{-1}\right| \ & \leq(|I|+|A|) \cdot\left|(I-A)^{-1}\right|=(1+|A|) \cdot\left|(I-A)^{-1}\right|, \end{aligned}
and the result follows. The definition of induced norm implies that $|I|=1$. The first inequality is due to the submultiplicative property of induced norms. The second inequality can be derived from the triangle inequality.