# 换元代数代写|COMMUTATIVE ALGEBRA MATH247 University of Liverpool Assignment

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Commutative algebra is a branch of abstract algebra that studies commutative rings, which are algebraic structures that have two binary operations, usually denoted as addition and multiplication. Commutative algebra plays an important role in several areas of mathematics, including number theory, algebraic geometry, and linear algebra.

In commutative algebra, one studies commutative rings, which are rings in which the multiplication operation is commutative. Examples of commutative rings include the ring of integers, polynomial rings, and the field of real numbers. Commutative algebra also studies modules over commutative rings, which are generalizations of vector spaces over a field.

The theory of commutative algebra has many applications to other areas of mathematics. For example, it is used in algebraic geometry to study algebraic varieties, which are geometric objects defined by systems of polynomial equations. Commutative algebra is also used in number theory to study algebraic number fields, which are extensions of the rational numbers that are closed under addition, subtraction, multiplication, and division.

Finally, commutative algebra has applications to linear algebra, especially in the study of linear transformations of vector spaces over fields. For example, if one has a linear transformation of a vector space that is represented by a matrix, then commutative algebra can be used to study the properties of that matrix, such as its eigenvalues and eigenvectors.

Show that $k\left[\mathbb{A}^2 \backslash{0}\right]=k\left[\mathbb{A}^2\right]$. Conclude that $\mathbb{A}^2 \backslash{0}$ is not affine.
[Hint: Use the covering by two affine open subsets given by $x \neq 0$ and $y \neq 0$, where $x, y$ are coordinates on $\left.\mathbb{A}^2\right]$.

Let $U_1$ and $U_2$ be the affine open subsets of $\mathbb{A}^2$ defined by $x\neq0$ and $y\neq0$, respectively. We want to show that $k[U_1\cup U_2] = k[\mathbb{A}^2]$. Since $U_1$ and $U_2$ cover $\mathbb{A}^2\backslash{0}$, it is sufficient to show that $k[U_1\cup U_2] = k[\mathbb{A}^2\backslash{0}]$.

Note that $\mathbb{A}^2\backslash{0} = U_1\cup U_2 \cup (U_1\cap U_2)$. Therefore, we have the following chain of inclusions: \begin{align*} k[\mathbb{A}^2\backslash{0}] &\subseteq k[U_1\cup U_2 \cup (U_1\cap U_2)]\ &\subseteq k[U_1] \cap k[U_2] \cap k[U_1\cap U_2]\ &= k[\mathbb{A}^2]. \end{align*} The first inclusion follows from the fact that $k[\mathbb{A}^2\backslash{0}]$ is the subring of $k(\mathbb{A}^2)$ consisting of rational functions that are regular on $\mathbb{A}^2\backslash{0}$, and every such function is clearly regular on $U_1\cup U_2 \cup (U_1\cap U_2)$. The second inclusion follows from the fact that a rational function that is regular on $U_1\cup U_2 \cup (U_1\cap U_2)$ must be regular on each of these sets individually.

Since we have $k[U_1\cup U_2] = k[\mathbb{A}^2]$, it follows that $\mathbb{A}^2\backslash{0}$ cannot be affine, because an affine variety cannot be covered by two non-empty affine open subsets.

Let $C$ be a curve in $\mathbb{P}^2, x$ be a point in $C$ and $L$ a line passing through $x$. Let $m$ be the multiplicity of $C$ at $x$ and $M$ the multiplicity of intersection of $C$ and $L$ at $x$. Show that $m \leq M$ and that for given $C, x$ the equality $m=M$ holds for all but finitely many lines $L$ as above.

Let $f$ be a homogeneous polynomial of degree $d$ defining $C$. We may assume that $x = [1:0:0]$. Up to projective transformation we may also assume that $L$ is given by $y = 0$. Then $C \cap L$ is defined by the ideal $(f, y)$ in $k[x,y,z]$.

Since $f$ vanishes at $x$, we may write $$f(x,y,z) = z^m g(x,y,z)$$ where $g(x,y,z)$ is a homogeneous polynomial of degree $d-m$ and $g(1,0,0) \neq 0$. Then $$(f, y) = (z^m g(x,y,z), y) = (z^m, y) + (g(x,y,z), y).$$ Since $y$ does not divide $z^m$, we have $(z^m, y) = (y)$, and so $$(f, y) = (y) + (g(x,y,z), y).$$ Thus the intersection multiplicity $M$ is equal to the order of vanishing of $g(x,y,z)$ at $[1:0:0]$, which is at least $m$.

To show that $m = M$ for all but finitely many lines $L$, it suffices to show that the intersection multiplicity $M$ is upper semicontinuous in $L$. In other words, we need to show that if $L_t$ is a family of lines converging to $L$ (in the Zariski topology) then $M_t \geq M$ for $t$ sufficiently close to $0$.

We may assume that the family $L_t$ is given by $y = t$. Let $g_t(x,z)$ be the polynomial of degree $d-m$ defining $C \cap L_t$ at $[1:0:0]$. Then $g_t(x,z) = f(x,t,z)$, and so the coefficient of $z^{d-m}$ in $g_t(x,z)$ is given by $$a_t = f_{0,d-m}(x,t,1).$$ Since $f_{0,d-m}(x,t,1)$ is a polynomial in $x$ and $t$, it is upper semicontinuous in $t$. Therefore, there exists a neighborhood $U$ of $0$ such that $a_t > 0$ for all $t \in U$.

Let $Z$ be an irreducible closed subset in an algebraic variety $X$. Show that if $\operatorname{dim}(Z)=\operatorname{dim}(X)$ then $Z$ is a component of $X$.

Suppose that $Z$ is not a component of $X$. Then there exists a proper closed subset $Y$ of $X$ such that $Z$ is contained in $Y$. Since $Y$ is a proper subset of $X$, we have $\operatorname{dim}(Y) < \operatorname{dim}(X)$. By the irreducibility of $Z$, we have $Z \not\subseteq Y$. Therefore, there exists a point $p\in Z$ such that $p\notin Y$.

Since $p$ is a point of $Z$, we have $\operatorname{dim}(\mathcal{O}{X,p}) \geq \operatorname{dim}(Z)$, where $\mathcal{O}{X,p}$ is the local ring of $X$ at $p$. On the other hand, since $p\notin Y$, we have $\mathcal{O}{X,p} \subseteq \mathcal{O}{Y,p}$. This implies $\operatorname{dim}(\mathcal{O}{X,p}) \leq \operatorname{dim}(\mathcal{O}{Y,p})$. By the definition of dimension of a local ring, we have $\operatorname{dim}(Y) = \operatorname{dim}(\mathcal{O}_{Y,p})$. Combining these inequalities, we obtain $\operatorname{dim}(Z) \leq \operatorname{dim}(Y)$. This contradicts the assumption that $\operatorname{dim}(Z) = \operatorname{dim}(X)$.

Therefore, our assumption that $Z$ is not a component of $X$ must be false. Thus, $Z$ is a component of $X$.

# 统计与概率代写|STATISTICS AND PROBABILITY II MATH254 University of Liverpool Assignment

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Assignment-daixieTM为您提供利物浦大学University of Liverpool STATISTICS AND PROBABILITY II MATH254统计与概率代写代考辅导服务！

## Instructions:

It’s good to hear that it covers both discrete and continuous random variables, as well as important distributions like the geometric, exponential, and normal distributions.

I also appreciate that the module highlights the importance of measure theory and probability as a foundation for this material. This should help students gain a deeper understanding of the underlying concepts and be better prepared to apply them in real-world situations.

Overall, it sounds like a valuable module for students interested in actuarial science, financial mathematics, statistics, and the physical sciences.

Corrupted by their power, the judges running the popular game show America’s Next Top Mathematician have been taking bribes from many of the contestants. Each episode, a given contestant is either allowed to stay on the show or is kicked off.

If the contestant has been bribing the judges she will be allowed to stay with probability 1. If the contestant has not been bribing the judges, she will be allowed to stay with probability $1 / 3$.

Suppose that $1 / 4$ of the contestants have been bribing the judges. The same contestants bribe the judges in both rounds, i.e., if a contestant bribes them in the first round, she bribes them in the second round too (and vice versa).
(a) If you pick a random contestant who was allowed to stay during the first episode, what is the probability that she was bribing the judges?

(a) We first compute $P\left(S_1\right)$ using the law of total probability.
$$P\left(S_1\right)=P\left(S_1 \mid B\right) P(B)+P\left(S_1 \mid H\right) P(H)=1 \cdot \frac{1}{4}+\frac{1}{3} \cdot \frac{3}{4}=\frac{1}{2} .$$
We therefore have (by Bayes’ rule) $P\left(B \mid S_1\right)=P\left(S_1 \mid B\right) \frac{P(B)}{P\left(S_1\right)}=1 \cdot \frac{1 / 4}{1 / 2}=\frac{1}{2}$.

(b) If you pick a random contestant, what is the probability that she is allowed to stay during both of the first two episodes?

(b) Using the tree we have the total probability of $S_2$ is
$$P\left(S_2\right)=\frac{1}{4}+\frac{3}{4} \cdot \frac{1}{3} \cdot \frac{1}{3}=\frac{1}{3}$$

(c) If you pick random contestant who was allowed to stay during the first episode, what is the probability that she gets kicked off during the second episode?

(c) We want to compute $P\left(L_2 \mid S_1\right)=\frac{P\left(L_2 \cap S_1\right)}{P\left(S_1\right)}$.
From the calculation we did in part (a), $P\left(S_1\right)=1 / 2$. For the numerator, we have (see the tree)
$$P\left(L_2 \cap S_1\right)=P\left(L_2 \cap S_1 \mid B\right) P(B)+P\left(L_2 \cap S_1 \mid H\right) P(H)=0 \cdot \frac{1}{3}+\frac{2}{9} \cdot \frac{3}{4}=\frac{1}{6}$$
Therefore $P\left(L_2 \mid S_1\right)=\frac{1 / 6}{1 / 2}=\frac{1}{3}$.

# 度量空间和微积分代写|METRIC SPACES AND CALCULUS MATH242 University of Liverpool Assignment

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Assignment-daixieTM为您提供利物浦大学University of Liverpool METRIC SPACES AND CALCULUS MATH242度量空间和微积分代写代考辅导服务！

## Instructions:

Metric spaces and calculus are two important areas of mathematics that are closely related to each other.

A metric space is a set of points, along with a metric or distance function that assigns a non-negative real number to every pair of points in the set. The metric satisfies certain properties such as the triangle inequality, which states that the distance between any two points in the set is always less than or equal to the sum of the distances between those points and a third point in the set. Examples of metric spaces include Euclidean space, the space of continuous functions, and the space of square integrable functions.

Calculus, on the other hand, is the branch of mathematics that deals with the study of rates of change and accumulation. It includes the study of limits, derivatives, integrals, and differential equations. Calculus is used in many areas of science and engineering, such as physics, economics, and engineering.

The study of calculus is closely related to metric spaces, as calculus often involves the analysis of functions defined on metric spaces. In particular, the concept of continuity, which is central to calculus, is closely related to the topology of a metric space. The notions of limit, derivative, and integral can all be defined in the context of metric spaces, and many of the results of calculus carry over to more general metric spaces.

Let $\gamma$ be the semi-circle connecting $(0,0)$ and $(2,0)$ that sits in the half plane where $y \geq 0$. Given $\mathbf{f}(x, y)=\left(2 x+\cos y,-x \sin y+y^7\right)$, calculate $\int \mathbf{f} \cdot d \gamma$. If your calculation requires justification from a theorem we proved in class, state the theorem you are using.

Notice that $D_1 f_2(x, y)=-\sin y=D_2 f_1(x, y)$. Since $f(x, y)$ is defined on all of $\mathbb{R}^2$, which is convex, we conclude $f(x, y)$ is a gradient field. Thus the integral of $f(x, y)$ from $(0,0)$ to $(2,0)$ is independent of the path. Let us integrate on a straight line $s:[0,2] \rightarrow \mathbb{R}^2$ defined by $s(t)=(t, 0)$ :
$$\int_C\left(2 x+\cos y,-x \sin y+y^7\right) d s=\int_0^2\left(2 t+\cos 0,-t \sin 0+0^7\right) \cdot(1,0) d t=\left[t^2+t\right]_0^2=6$$

Let $f(x, y, z)=x^2+y^2+z^2$. Prove $f$ is differentiable at $(1,1,1)$ with linear transformation $T(x, y, z)=2 x+2 y+2 z$.

Solution To prove $f$ is differentiable with total derivative $T$ as described we need to show
$$\lim {|\mathbf{v}| \rightarrow 0} \frac{f(\mathbf{v}+(1,1,1))-f(1,1,1)-T(\mathbf{v})}{|\mathbf{v}|}=0$$ Now observe that $$f(\mathbf{v}+(1,1,1))-f(1,1,1)-T(\mathbf{v})=\left(v_1+1\right)^2+\left(v_2+1\right)^2+\left(v_3+1\right)^2-3-2 v_1-2 v_2-2 v_3=v_1^2+v_2^2+v_3^2$$ Thus $$\lim {|\mathbf{v}| \rightarrow 0} \frac{f(\mathbf{v}+(1,1,1))-f(1,1,1)-T(\mathbf{v})}{|\mathbf{v}|}=\lim {|\mathbf{v}| \rightarrow 0} \frac{|\mathbf{v}|^2}{|\mathbf{v}|}=\lim {|\mathbf{v}| \rightarrow 0}|\mathbf{v}|=0 .$$
It follows that $f$ is differentiable at $(1,1,1)$ with the total derivative as described.

Consider $f(x, y)=(x y+y)^{10}$ on the square $Q=[0,1] \times[0,1]$. Evaluate $\iint_Q f d x d y$.

The function
$$f(x, y)=(x y+y)^{10}$$
is continous on $\mathbb{R}^2$. Thus, $\int_0^1 f(x, y) d x$ is integrable for all $y \in[0,1]$ so one can apply Fubini’s Theorem to get:
$$\iint_Q(x y+y)^{10} d x d y=\int_0^1 \int_0^1(x y+y)^{10} d x d y=\int_0^1\left[\frac{(x y+y)^{11}}{11 y}\right]_0^1 d y=\int_0^1 \frac{\left(2^{11}-1\right) y^{10}}{11} d y=\frac{2047}{121}$$

# 复数函数代写|COMPLEX FUNCTIONS MATH243 University of Liverpool Assignment

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Assignment-daixieTM为您提供利物浦大学University of Liverpool COMPLEX FUNCTIONS MATH243复数函数代写代考辅导服务！

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Harmonic functions are a class of functions that arise naturally in many areas of mathematics and physics, including the study of partial differential equations, potential theory, and complex analysis.

In essence, a function is said to be harmonic if it satisfies Laplace’s equation, which is a partial differential equation that arises frequently in the study of physical systems. Laplace’s equation has many interesting properties, and solutions to this equation often exhibit beautiful symmetries and patterns.

Harmonic functions have many practical applications as well. For example, they are used in electrical engineering to model the flow of electric current through conductive materials, and they are also important in fluid dynamics, where they are used to study the behavior of fluids in motion.

Overall, the theory of harmonic functions is a rich and fascinating subject, with many connections to other areas of mathematics and physics. It is definitely worth exploring further if you are interested in these topics.

Find all solutions $z$ to equation $z^3=-8 i$.

Evaluate the integral
$$\int_{|z-1|=\frac{1}{2}} \frac{d z}{(1-z)^3} .$$

\begin{aligned} \int_{|z-1|=\frac{1}{2}} \frac{d z}{(1-z)^3} & =\int_0^{2 \pi} \frac{i e^i t / 2}{\left(-e^{i t} / 2\right)^3} d t \ & =\frac{-i}{2} \cdot 8 \int_0^{2 \pi} e^{-i \cdot 2 t} d t \ & =-\left.4 i \frac{e^{-2 i t}}{-2 i}\right|_0 ^{2 \pi} \ & =0 .\end{aligned}

Evaluate the integral
$$\int_\gamma \frac{e^z+z}{z-2} d z$$
in the two cases: 1) $\gamma={z:|z|=1}$;
2) $\gamma={z:|z|=3}$.

1) $\int_{|z|=1} \frac{e^2+z}{z-2} d z=0$, since $2 \notin{z:|z|<1}$.
2) $\int_{|z|=3} \frac{e^z+z}{z-2} d z=2 \pi i\left(e^2+2\right)$, since $n(\gamma, 2)=1$.

# 经典力学代写|CLASSICAL MECHANICS MATH228 University of Liverpool Assignment

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Assignment-daixieTM为您提供利物浦大学University of Liverpool CLASSICAL MECHANICS MATH228经典力学代写代考辅导服务！

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Classical mechanics is a branch of physics that deals with the study of the motion of objects under the influence of forces. It was first developed by Sir Isaac Newton in the 17th century and is based on three fundamental laws of motion:

1. The law of inertia: An object at rest will remain at rest and an object in motion will continue in a straight line at a constant speed unless acted upon by an external force.
2. The law of acceleration: The acceleration of an object is directly proportional to the force applied to it and inversely proportional to its mass. This is expressed mathematically as F = ma, where F is the force applied to the object, m is its mass, and a is its acceleration.
3. The law of action-reaction: For every action, there is an equal and opposite reaction.

Classical mechanics also includes the study of energy and momentum, and how they are conserved in various physical processes. It has numerous applications in engineering, physics, and astronomy, and is the foundation for the development of modern physics.

A driven oscillator is described by
$$\ddot{x}+\omega_o^2 x=\frac{F}{m} \cos (\gamma t+\alpha) .$$
We found that the solution off resonance is
$$x(t)=B \cos \left(\omega_o t+\beta\right)+\frac{F / m}{\omega_o^2-\gamma^2} \cos (\gamma t+\alpha) .$$
which we can rearrange to
$$x(t)=C \cos \left(\omega_o t+\kappa\right)+\frac{F / m}{\omega_o^2-\gamma^2}\left(\cos (\gamma t+\alpha)-\cos \left(\omega_o t+\alpha\right)\right) .$$
with new constants $C$ and $\kappa$.
a) If the oscillator is driven close to the natural frequency $\omega_o$, we can write $\omega_o=\gamma+\epsilon$ with $\epsilon \ll \omega_o$. Keeping terms only linear in $\epsilon$ (i.e. set any $\epsilon$ with higher power to zero), show that we can write
$$x(t)=C \cos \left(\omega_o t+\kappa\right)+\frac{F / m}{2 \omega_o \epsilon}\left(\cos \left(\omega_o t+\alpha-\epsilon t\right)-\cos \left(\omega_o t+\alpha\right)\right)$$

a) Starting from the expression for $x(t)$ given in the problem, we substitute $\omega_o=\gamma+\epsilon$ and keep only terms linear in $\epsilon$. Using the identity $\cos(A+B)=\cos A\cos B-\sin A\sin B$, we have

\begin{align} x(t) &= C \cos(\gamma t + \epsilon t + \kappa) + \frac{F/m}{(\gamma+\epsilon)^2 – \gamma^2}\left[\cos(\gamma t + \alpha) – \cos(\gamma t + \epsilon t + \alpha)\right]\ &= C \cos(\gamma t + \kappa)\cos(\epsilon t) – C \sin(\gamma t + \kappa)\sin(\epsilon t)\ &\quad + \frac{F/m}{2\gamma\epsilon + \epsilon^2}\left[\cos(\gamma t + \alpha – \epsilon t) – \cos(\gamma t + \alpha)\right] + O(\epsilon^2)\ &= C \cos(\gamma t + \kappa)\cos(\epsilon t) + \frac{F/m}{2\gamma\epsilon}\sin(\gamma t + \kappa)\epsilon t\ &\quad + \frac{F/m}{2\gamma\epsilon + \epsilon^2}\left[\cos(\gamma t + \alpha)\cos(\epsilon t) + \sin(\gamma t + \alpha)\sin(\epsilon t) – \cos(\gamma t + \alpha)\right] + O(\epsilon^2) \end{align}

The term proportional to $\sin(\gamma t + \kappa)\epsilon t$ comes from expanding the sine term to first order in $\epsilon$ and keeping only the linear term. We can simplify the expression by using the trigonometric identity $\sin^2\theta + \cos^2\theta = 1$ and discarding terms of order $\epsilon^2$ or higher. This gives

\begin{align} x(t) &= C \cos(\gamma t + \kappa)\cos(\epsilon t) + \frac{F/m}{2\gamma\epsilon}\sin(\gamma t + \kappa)\epsilon t\ &\quad + \frac{F/m}{2\gamma\epsilon}\sin(\alpha)\sin(\epsilon t) + O(\epsilon^2)\ &= C \cos(\omega_o t + \kappa) + \frac{F/m}{2\gamma\epsilon}\left[\cos(\omega_o t + \alpha – \epsilon t) – \cos(\omega_o t + \alpha)\right] + O(\epsilon^2) \end{align}

where we have used the definition of $\omega_o$ and the fact that $\sin(\gamma t + \kappa) = \sin(\omega_o t + \alpha)$ and $\cos(\gamma t + \kappa) = \cos(\omega_o t + \alpha)$.

b) Show that this evolves to the on resonance solution (LL 22.5) for $\epsilon \rightarrow 0$. Note: you may carry out the calculation using trigonometric identities or complex notation. Note: to compare with LL 22.5 , convert the above as follows: $$C \rightarrow a, \quad F \rightarrow f, \omega_0 \rightarrow \omega, \quad \kappa \rightarrow \alpha, \alpha \rightarrow \beta$$

To show that the off-resonance solution evolves to the on-resonance solution for $\epsilon \rightarrow 0$, we need to take the limit of the off-resonance solution as $\epsilon \rightarrow 0$ and show that it matches the on-resonance solution given by LL 22.5.

The on-resonance solution given by LL 22.5 is:

$x(t)=a \cos (\omega t+\alpha)+\frac{f}{2 m \omega} t \sin (\omega t+\alpha)$

where $\omega = \omega_0$ and $\alpha = \beta$.

Substituting the constants given in the note, we have:

$x(t)=a \cos \left(\omega_0 t+\beta\right)+\frac{f}{2 m \omega_0} t \sin \left(\omega_0 t+\beta\right)$

Now, let’s take the limit of the off-resonance solution as $\epsilon \rightarrow 0$: \begin{align*} x(t) &= C \cos(\omega_0 t + \kappa) + \frac{F/m}{\omega_0^2 – \gamma^2}(\cos(\gamma t + \alpha) – \cos(\omega_0 t + \alpha)) \ &= C \cos(\omega_0 t + \kappa) + \frac{F/m}{\omega_0^2 – \gamma^2}\cos(\gamma t + \alpha) – \frac{F/m}{\omega_0^2 – \gamma^2}\cos(\omega_0 t + \alpha) \end{align*}

We can use $a, b, c$ and $d$ from the previous problem to make the matrix $M$ such that $\vec{x}(t+\Delta t)=M \vec{x}(t)$. Find the eigenvalues of $M$. Take $\Delta t=4 \pi / \omega_o$ and find the eigenvectors.

Recall that the system of differential equations is given by:

\begin{aligned} \dot{x}_1 & =x_2 \ \dot{x}_2 & =-\frac{k}{m} x_1-\frac{c}{m} x_2 \ \dot{x}_3 & =x_4 \ \dot{x}_4 & =-\frac{k}{m} x_3-\frac{c}{m} x_4+\frac{F_0}{m} \cos \left(\omega_d t\right)\end{aligned}

We can rewrite this system of differential equations in matrix form as:

$\frac{d}{d t}\left(\begin{array}{l}x_1 \ x_2 \ x_3 \ x_4\end{array}\right)=\left(\begin{array}{cccc}0 & 1 & 0 & 0 \ -\frac{k}{m} & -\frac{c}{m} & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & -\frac{k}{m} & -\frac{c}{m}\end{array}\right)\left(\begin{array}{l}x_1 \ x_2 \ x_3 \ x_4\end{array}\right)+\left(\begin{array}{c}0 \ 0 \ 0 \ \frac{F_0}{m} \cos \left(\omega_d t\right)\end{array}\right)$

Let’s define the matrix $M$ as:

$M=\left(\begin{array}{cccc}0 & 1 & 0 & 0 \ -\frac{k}{m} & -\frac{c}{m} & 0 & 0 \ 0 & 0 & 0 & 1 \ 0 & 0 & -\frac{k}{m} & -\frac{c}{m}\end{array}\right)$

# 应用数学的数值方法代写|NUMERICAL METHODS FOR APPLIED MATHEMATICS MATH226 University of Liverpool Assignment

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Assignment-daixieTM为您提供利物浦大学University of Liverpool NUMERICAL METHODS FOR APPLIED MATHEMATICS MATH226应用数学的数值方法代写代考辅导服务！

## Instructions:

Applied mathematics often deals with complex problems that cannot be solved exactly using analytical methods. In such cases, numerical methods provide an alternative approach to solve these problems using computers. These methods involve approximating solutions by breaking down the problem into smaller parts and performing iterative calculations until a desired level of accuracy is achieved.

Numerical methods are commonly used in a variety of fields, such as physics, engineering, finance, and biology, to solve problems that involve real-world data. For example, in finance, numerical methods are used to calculate the value of complex financial derivatives, while in biology, these methods are used to simulate the behavior of large-scale biochemical networks.

One of the key advantages of numerical methods is their ability to handle complex problems that would be impractical or impossible to solve by hand. These methods allow for the efficient solution of problems that involve large amounts of data, complex equations, or nonlinear relationships.

However, numerical methods are not without their drawbacks. One major issue is the propagation of errors, where small errors in the initial data or calculations can lead to significant errors in the final results. To mitigate this issue, it is essential to carefully analyze the sources of error and design algorithms that minimize their impact.

In this module, you will learn about various numerical methods, including methods for finding roots, approximating integrals, and interpolating data. You will examine the advantages and disadvantages of different approaches in terms of accuracy and efficiency, and you will learn how to write computer programs to handle calculations automatically. By the end of the module, you will have a solid understanding of numerical methods and be able to apply them to solve complex problems in various fields.

Create a real $4 \times 4$ matrix with rank 2 . Propose a vector $\mathbf{b} \in \mathbb{R}^4$ such that the system of equations $\mathbf{A x}=\mathbf{b}$ has a family solutions. What is that family of solutions? Propose a vector $\mathbf{b} \in \mathbb{R}^4$ such that the system of equations $\mathbf{A x}=\mathbf{b}$ has no solutions. Explain why this is the case.

There are many possible answers. A diagonal matrix with rank 2 would be:
$$\left(\begin{array}{llll} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \end{array}\right)$$
A value of $\mathbf{b}=(1,0,0,0)$ would give a family of solutions $\mathbf{x}=(1,0, a, b)$ where $a$ and $b$ are arbitrary. A value of $\mathbf{b}=(0,0,1,0)$ would give no solution since this vector is not in the range space of the matrix.

Write a MATLAB ${ }^6$ function that takes advantage of this sparsity pattern to compute the product of $\mathbf{S}$ with a vector. Your function should take as an input the fluxes associated with each reaction in the cycle and return the rate of change for the concentration of each species in the cycle. Be sure that your function does not compute the stoichiometry matrix explicitly.

function $y=$ sparse_mult $(x)$
$$y(1)=-x(1)+x(\text { end })$$
for $i=2:$ length $(x)$
$$y(i)=x(i-1)-x(i) ;$$
end;

Develop an expression for the characteristic polynomial of the $\mathrm{N}$ component stoichiometry matrix, S. The roots represent the eigenvalues of $\mathbf{S}$.

$$\operatorname{det}(\mathbf{S}-\lambda \mathbf{I})=(-1-\lambda) M_{11}+(-1)^{N-1} M_{1 N}=0$$
Because the minors are determinants of diagonal/upper triangular matrices, these are simple to calculate: $M_{11}=(-1-\lambda)^{N-1}, M_{1 N}=1$. Therefore the secular polynomial satisfies $(-1-\lambda)^N+(-1)^{N-1}=0$.

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

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Assignment-daixieTM为您提供利物浦大学University of Liverpool STATISTICS AND PROBABILITY I MATH253概率学和统计学代写代考辅导服务！

## 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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Assignment-daixieTM为您提供利物浦大学University of Liverpool LINEAR ALGEBRA AND GEOMETRY MATH244线性代数和几何代写代考辅导服务！

## 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
$$2-6 x(a)+3 x(a)^2=0 \quad \Leftrightarrow \quad x(a)=1 \pm \frac{1}{\sqrt{3}} .$$