# 代数代写|ALGEBRA MATHS4072 University of Glasgow Assignment

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Algebraic structures, such as groups, rings, and fields, are fundamental concepts in abstract algebra that have important applications in many areas of mathematics and beyond.

Group theory is the study of the algebraic structure called a group, which is a set of elements with a binary operation that satisfies certain axioms. Group theory has applications in many areas of mathematics, including number theory, geometry, and topology.

Ring theory is the study of rings, which are algebraic structures that generalize the properties of integers. Rings are important in algebraic geometry, number theory, and coding theory.

Field theory is the study of fields, which are algebraic structures that extend the properties of rational numbers. Fields are important in algebraic geometry, number theory, and physics.

The construction and application of quotient or factor groups and rings is an important aspect of algebraic structures. Homomorphisms between rings and groups are also essential in the study of algebraic structures.

Overall, this course will provide a solid foundation in the basic concepts and techniques of abstract algebra, which will be useful in many areas of mathematics and beyond.

Let $[n]$ denote the totally ordered set $\{0,1, \ldots, n\}$. Let $\phi:[m] \rightarrow[n]$ be an order preserving function (so that if $i \leq j$ then $\phi(i) \leq \phi(j)$ ). Identifying the elements of $[n]$ with the vertices of the standard simplex $\Delta^n, \phi$ extends to an affine $\operatorname{map} \Delta^m \rightarrow \Delta^n$ that we also denote by $\phi$. Give a formula for this map in terms of barycentric coordinates: If $\phi\left(s_0, \ldots, s_m\right)=\left(t_0, \ldots, t_n\right)$, what is $t_j$ as a function of $\left(s_0, \ldots, s_m\right) ?$

Let $\Delta^n$ be the standard $n$-simplex, that is, $\Delta^n = {(x_0, x_1, \ldots, x_n) \in \mathbb{R}^{n+1} : x_i \geq 0, \sum_{i=0}^n x_i = 1}$. We will use the barycentric coordinates for the points in $\Delta^n$, which are defined as follows: for $0 \leq i \leq n$, let $e_i \in \Delta^n$ be the point with $x_i = 1$ and $x_j = 0$ for $j \neq i$. Then any point $p \in \Delta^n$ can be written uniquely as a convex combination of the $e_i$’s, that is, $p = \sum_{i=0}^n \lambda_i e_i$, where $\lambda_i \geq 0$ and $\sum_{i=0}^n \lambda_i = 1$.

Now let $\phi: [m] \to [n]$ be an order-preserving function, and let $\phi:\Delta^m \to \Delta^n$ be the affine map that extends $\phi$ to the standard simplices. We want to express $\phi(p)$ in terms of the barycentric coordinates of $p = \sum_{i=0}^m \lambda_i e_i$.

Note that $\phi(e_i)$ is a point in $\Delta^n$, so we can write $\phi(e_i)$ as a convex combination of the $e_j$’s. Let $\phi(e_i) = \sum_{j=0}^n a_{ij} e_j$, where $a_{ij} \geq 0$ and $\sum_{j=0}^n a_{ij} = 1$. Then for any $p \in \Delta^m$, we have \begin{align*} \phi(p) &= \phi\left(\sum_{i=0}^m \lambda_i e_i\right) \ &= \sum_{i=0}^m \lambda_i \phi(e_i) \ &= \sum_{i=0}^m \lambda_i \sum_{j=0}^n a_{ij} e_j \ &= \sum_{j=0}^n \left(\sum_{i=0}^m \lambda_i a_{ij}\right) e_j. \end{align*}

Let $t_j = \sum_{i=0}^m \lambda_i a_{ij}$, so we have $\phi(p) = \sum_{j=0}^n t_j e_j$. Since $e_j$ is the point with $x_j = 1$ and $x_i = 0$ for $i \neq j$, we have $t_j = \phi(\lambda_j)$, where $\lambda_j$ is the $j$th barycentric coordinate of $p$. Therefore, we have

$t_j=\sum_{i=0}^m \lambda_i a_{i j}$,

where $a_{ij}$ is the $j$th barycentric coordinate of $\phi(e_i)$.

(a) Write $\pi_0(X)$ for the set of path-components of a space $X$. Construct an isomorphism $$\mathbb{Z} \pi_0(X) \rightarrow H_0(X)$$

Let $X$ be a topological space. We denote by $\pi_0(X)$ the set of path-components of $X$. That is, for any two points $x,y\in X$, we say that $x$ and $y$ are in the same path-component if there exists a path $\gamma:[0,1]\to X$ such that $\gamma(0)=x$ and $\gamma(1)=y$. The set $\pi_0(X)$ is a partition of $X$ into disjoint subsets, and each subset is a path-connected subspace of $X$.

Now, we construct an isomorphism between $\mathbb{Z} \pi_0(X)$ and $H_0(X)$. Let $\mathbb{Z}\pi_0(X)$ be the free abelian group generated by the set $\pi_0(X)$, and let $\phi:\mathbb{Z}\pi_0(X)\to H_0(X)$ be the homomorphism defined as follows. For each path-component $[x]\in \pi_0(X)$, choose a point $x\in [x]$, and define $\phi([x])=[x]$ to be the class in $H_0(X)$ represented by the point $x$. We extend $\phi$ to all of $\mathbb{Z}\pi_0(X)$ by linearity.

We claim that $\phi$ is an isomorphism. To see this, we construct its inverse. Let $\psi:H_0(X)\to \mathbb{Z}\pi_0(X)$ be the homomorphism defined by sending each class $[x]\in H_0(X)$ to the generator $[x]\in \mathbb{Z}\pi_0(X)$ corresponding to the path-component containing $x$. That is, if $[x]\in H_0(X)$, then there exists a chain $\sum_i n_i \sigma_i$ with $\sigma_i:\Delta^0\to X$ such that $\partial (\sum_i n_i \sigma_i)=[x]$, and we define $\psi([x])=[\sigma_i(0)]\in \mathbb{Z}\pi_0(X)$. Again, we extend $\psi$ to all of $H_0(X)$ by linearity.

It is easy to see that $\psi$ is well-defined and that $\phi$ and $\psi$ are inverses of each other, so $\phi$ is an isomorphism.

Let $A$ be a chain complex (of abelian groups). It is acyclic if $H(A)=0$, and contractible if it is chain-homotopy-equivalent to the trivial chain complex. Prove that a chain complex is contractible if and only if it is acyclic and for every $n$ the inclusion $Z_n A \hookrightarrow A_n$ is a split monomorphism of abelian groups.

First, let’s assume that the chain complex $A$ is contractible. This means that there exists a chain map $s:A\rightarrow A$ (called a homotopy) such that $ds+sd=\mathrm{id}_A$, where $d$ is the differential of $A$. We can then define a chain map $r:A\rightarrow A$ by $r_n = \frac{1}{2}(1+s)$, which is well-defined because $s$ is a chain map. It’s easy to check that $r$ is also a chain map, and that $sr=\mathrm{id}_A$. This shows that $A$ is chain-homotopy-equivalent to the trivial chain complex.

Next, let’s assume that $A$ is acyclic and that the inclusion $Z_n A\hookrightarrow A_n$ is a split monomorphism for every $n$. Since $A$ is acyclic, we have $H_n(A)=0$ for all $n$. It remains to show that $A$ is chain-homotopy-equivalent to the trivial chain complex.

For each $n$, let $s_n:A_n\rightarrow A_{n+1}$ be a map such that $ds_n+s_{n-1}d=\mathrm{id}{A_n}$ (recall that we set $s{-1}=0$). Such a map exists because the inclusion $Z_n A\hookrightarrow A_n$ is a split monomorphism. In fact, we can take $s_n$ to be the composite of the projection $A_{n+1}\rightarrow Z_{n+1}A$ (which exists because of the split monomorphism) and the inclusion $Z_{n+1}A\hookrightarrow A_{n+1}$.

We claim that $s={s_n}$ is a chain homotopy between $A$ and the trivial chain complex. To see this, we need to show that $ds+sd=\mathrm{id}A$. We have: \begin{align*} (ds+sd)n &= d_ns{n-1}+s{n-1}d_n \ &= (s_{n-1}d-d_{n-1}s_n)+s_{n-1}d_n \ &= s_{n-1}(d_n+d_{n-1}) \ &= s_{n-1}^2 \ &= \frac{1}{4}(1+s_{n-1})^2 \ &= \frac{1}{4}(1+2s_{n-1}+s_{n-1}^2) \ &= \frac{1}{4}(1+2s_{n-1}+(1+s_{n-2})s_n) \ &= \frac{1}{4}(2s_n+s_{n-1}s_n+s_{n-2}s_n) \ &= \frac{1}{4}(2s_n+s_{n-1}d_n+s_{n-2}d_{n-1}) \ &= \frac{1}{4}(2s_n+d_{n-1}s_n+d_ns_{n-1}) \ &= \frac{1}{4}(2s_n+d_ns_{n-1}+s_{n-1}d_n) \ &= \frac{1}{4}(2s_n+\mathrm{id}{A_n}) \ &= r_n, \end{align*}

# 复分析代写|METHODS IN COMPLEX ANALYSIS MATHS4076 University of Glasgow Assignment

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It deals with functions that are differentiable in the complex sense, meaning that they satisfy the Cauchy-Riemann equations. These functions are also called holomorphic functions.

The theory of analytic functions has many important applications in various fields, including physics, engineering, and mathematics. Some of the key topics that you might encounter in this course include power series expansions, Laurent series, singularities, residues, conformal mappings, and the Cauchy integral formula.

Throughout the course, you will learn about the properties and behavior of analytic functions, and you will develop techniques for analyzing and manipulating these functions. You will also study the connections between analytic functions and other areas of mathematics, such as complex geometry and number theory.

Overall, the study of analytic functions is a fascinating and important topic in mathematics, and it is essential for anyone interested in pursuing a career in pure or applied mathematics.

Let $\sum_{n=1}^{\infty} z_n$ be a convergent series of complex numbers. (1) Let $\alpha<\pi / 2$. Show that if for all $n,\left|\operatorname{Arg} z_n\right|<\alpha$, then the series $\sum_{n=1}^{\infty} z_n$ converges absolutely.

(2) Does the conclusion of the previous part hold if $\alpha=\pi / 2$ ?

(2) The conclusion of part (1) does not hold if $\alpha = \frac{\pi}{2}$. For example, consider the series $\sum_{n=1}^\infty \frac{i^n}{n}$, where $i = \sqrt{-1}$. For all $n$, we have $|\operatorname{Arg} (i^n/n)| = \frac{\pi}{2}$. However, the series converges conditionally by the alternating series test.

(3) Assume that for all $n, \Re z_n \geq 0$ and that the series $\sum_{n=1}^{\infty} z_n^2$ also converges. Show that the series $\sum_{n=1}^{\infty} z_n^2$ converges absolutely.

(3) Since $\Re z_n \geq 0$ for all $n$, we have $|z_n| = \sqrt{z_n \overline{z_n}} = \sqrt{z_n^2} = z_n$. Therefore, we have $$|z_n^2| = |z_n|^2 = z_n^2$$ Since the series $\sum_{n=1}^\infty z_n^2$ converges, we have $$\sum_{n=1}^\infty |z_n^2| = \sum_{n=1}^\infty z_n^2$$ converges, and hence the series $\sum_{n=1}^\infty z_n$ converges absolutely.

# 度量空间和基本拓扑学代写|METRIC SPACE AND BASIC TOPOLOGY MATHS4077 University of Glasgow Assignment

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In a course on topology, you can expect to study the fundamental concepts of point-set topology, including sets, functions, continuity, convergence, compactness, and connectedness. You will learn how to use these concepts to prove theorems and solve problems in a variety of mathematical contexts.

The course will likely start with a review of basic set theory and logic, and then move on to the definition and properties of metric spaces. You will learn about various topological concepts and constructions, such as open sets, closed sets, neighborhoods, interior, closure, and boundary. You will study different types of convergence, such as pointwise, uniform, and Hausdorff convergence, and learn how to use these concepts to prove continuity and other results.

Later on, you may also study more advanced topics in topology, such as homotopy theory, algebraic topology, and differential topology. These areas of study involve the use of topological concepts and tools to solve more sophisticated problems in mathematics and other fields.

Overall, a course in topology will provide you with a deep understanding of the fundamental concepts and methods of topology, which are essential for further study in mathematics and other areas of science and engineering.

Consider the function $f: \mathbb{R} \rightarrow \mathbb{R}$ given by $f(x)=k x+b$ for $0<k<1$ and $b \in \mathbb{R}$. Show that $f$ is a contraction, find the fixed point of $f$, and directly show the fixed point is unique.

To show that $f$ is a contraction, we need to show that there exists a constant $0 \leqslant L < 1$ such that for all $x, y \in \mathbb{R}$, we have $|f(x) – f(y)| \leqslant L|x-y|$.

So let $x, y \in \mathbb{R}$ be arbitrary, and let $L = k$. Then we have

\begin{align*} |f(x) – f(y)| &= |kx + b – ky – b| \ &= |k(x-y)| \ &= k|x-y|. \end{align*}

Since $0<k<1$, we have $0\leqslant L<k<1$. Therefore, $f$ is a contraction.

To find the fixed point of $f$, we need to solve the equation $f(x) = x$. That is,

$$kx+b=x$$

which gives $x = \frac{b}{1-k}$. Therefore, the fixed point of $f$ is $\frac{b}{1-k}$.

To show that the fixed point is unique, we need to show that if $x_1$ and $x_2$ are both fixed points of $f$, then $x_1 = x_2$.

So let $x_1$ and $x_2$ be fixed points of $f$. Then we have

\begin{align*} f(x_1) &= x_1 \ f(x_2) &= x_2 \end{align*}

Subtracting these equations, we get

\begin{align*} f(x_1) – f(x_2) &= x_1 – x_2 \ kx_1 + b – kx_2 – b &= x_1 – x_2 \ k(x_1 – x_2) &= x_1 – x_2 \ \end{align*}

Since $k < 1$, we can divide both sides by $k(x_1 – x_2)$ to obtain $1 < \frac{1}{k}$, which means $x_1 = x_2$. Therefore, the fixed point of $f$ is unique.

In class, we have defined a set $A \subset X$ to be closed if its complement is an open set in $X$. There is another useful definition of a closed set however. Show that $A \subset X$ is closed if and only if every convergent sequence in $A$ converges in $A$. In other words, if $\left\{x_n\right\}$ is a convergent sequence in $A$ such that $x_n \rightarrow x$, then $x \in A$.

To prove that $A \subset X$ is closed if and only if every convergent sequence in $A$ converges in $A$, we need to show two things:

$(\Rightarrow)$ If $A$ is closed, then every convergent sequence in $A$ converges in $A$. $(\Leftarrow)$ If every convergent sequence in $A$ converges in $A$, then $A$ is closed.

$(\Rightarrow)$ Suppose that $A$ is closed, and let $\left{x_n\right}$ be a convergent sequence in $A$ that converges to some $x \in X$. We want to show that $x \in A$. Since $\left{x_n\right}$ converges to $x$, we know that for any open set $U$ containing $x$, there exists an $N$ such that $x_n \in U$ for all $n \geq N$. In particular, for the open set $U = X \setminus A$, we have $x_n \notin U$ for all $n \geq N$, since $\left{x_n\right}$ is a sequence in $A$. Therefore, $x \notin U$, which implies that $x \in A$ (since $U = X \setminus A$ is closed). Thus, every convergent sequence in $A$ converges to a point in $A$, as required.

$(\Leftarrow)$ Suppose that every convergent sequence in $A$ converges in $A$, and let $U = X \setminus A$. We want to show that $U$ is open. Let $x \in U$ be arbitrary. Since $x \notin A$, there exists a sequence $\left{x_n\right}$ in $A$ that converges to $x$. But by assumption, this means that $x \in A$, which is a contradiction. Therefore, $U$ contains no points of $A$, and is therefore disjoint from $A$. Thus, $U$ is open, and $A$ is closed.

Therefore, we have shown that $A$ is closed if and only if every convergent sequence in $A$ converges in $A$.

Let $|\cdot|$ be a norm on a vector space $V$, and let $d(x, y)=|x-y|$ for all $x, y \in V$. Show the following three properties:
(a) $d(\lambda x, \lambda y)=|\lambda| d(x, y)$ for all $\lambda \in \mathbb{R}$, and for all $x, y \in V$.

To show that $d(\lambda x, \lambda y) = |\lambda| d(x, y)$ for all $\lambda \in \mathbb{R}$, and for all $x, y \in V$, we need to use the properties of the norm $|\cdot|$.

First, note that by the definition of $d$, we have:

$d(\lambda x, \lambda y)=|\lambda x-\lambda y|$

Using the properties of the norm, we can manipulate this expression as follows:

\begin{align*} |\lambda x – \lambda y| &= |\lambda (x – y)| \ &= |\lambda| |x – y| \ &= |\lambda| d(x, y). \end{align*}

Therefore, we have shown that $d(\lambda x, \lambda y) = |\lambda| d(x, y)$ for all $\lambda \in \mathbb{R}$, and for all $x, y \in V$. This property is known as homogeneity or scaling property of the norm.

# 分化和整合的分析代写|ANALYSIS OF DIFFERENTIATION AND INTEGRATION MATHS4073 University of Glasgow Assignment

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Differentiation and integration are two fundamental operations in calculus, which is a branch of mathematics that deals with continuous change and motion. Differentiation is the process of finding the derivative of a function, while integration is the process of finding the antiderivative of a function.

Differentiation:

The derivative of a function is the rate at which the function changes with respect to its input variable. It is defined as the limit of the difference quotient as the change in the input variable approaches zero. In simpler terms, it measures how much the output of the function changes when the input changes a little bit. The derivative is denoted by the symbol “dy/dx” or “f'(x)”.

Differentiation has a number of applications, such as determining the slope of a curve at a particular point, finding the velocity and acceleration of an object, and optimizing functions in optimization problems.

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

To solve this second-order differential equation, we first need to rewrite it as a first-order system. Let $y_1 = y$ and $y_2 = y’$, then we have:

\begin{align*} y_1′ &= y_2 \ y_2′ &= \frac{t}{y_2} \end{align*}

This gives us a first-order system that we can solve using standard techniques. To find the solution, we can use an integrating factor for the second equation. Multiplying both sides by $y_2$ gives:

y_2 y_2^{\prime}=t

We can then integrate both sides with respect to $t$ to get:

$\frac{1}{2} y_2^2=\frac{1}{2} t^2+C_1$

where $C_1$ is an arbitrary constant of integration. Solving for $y_2$, we get:

$y_2= \pm \sqrt{t^2+2 C_1}$

Using the initial condition $y'(1) = 1$, we have:

$y_2(1)= \pm \sqrt{1+2 C_1}=1$

Since $y_2$ must be positive, we can choose the positive sign and solve for $C_1$:

$\sqrt{1+2 C_1}=1 \quad \Rightarrow \quad C_1=0$

Thus, we have:

$y_2=\sqrt{t^2}=t$

Next, we can integrate the first equation:

$y_1=\int y_2 d t=\frac{1}{2} t^2+C_2$

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

$y_1(1)=\frac{1}{2}+C_2=2 \quad \Rightarrow \quad C_2=\frac{3}{2}$

Thus, the solution to the initial value problem is:

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

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

(a) To find the critical points, we need to solve the equation $y^{\prime}=0$, which is equivalent to $y(5-y)(y-4)^2=0$. Therefore, the critical points are $y=0, y=5, y=4$.

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

(b) To sketch the graph of $f(y)=y(5-y)(y-4)^2$, we need to analyze the behavior of the function around the critical points and at infinity.

First, we can determine the sign of $f(y)$ on different intervals using test points:

\begin{array}{c|c|c|c|c} y & (-\infty,0) & (0,4) & (4,5) & (5,\infty) \ \hline y-2 & – & – & + & + \ y-3 & – & – & – & + \ y-4 & – & 0 & + & + \ y-6 & – & – & – & – \end{array}

From this table, we can see that $f(y)$ is positive on $(0,4)$ and $(5,\infty)$, and negative on $(-\infty,0)$ and $(4,5)$. Also, $f(y)$ has a local maximum at $y=4$ and local minima at $y=0$ and $y=5$.

Based on this information, we can sketch the graph of $f(y)$ as follows:

• The function is negative and decreasing on $(-\infty,0)$.
• The function has a local minimum at $y=0$ and increases from negative infinity to zero.
• The function is negative and decreasing on $(0,4)$.
• The function has a local maximum at $y=4$ and decreases from zero to the minimum value at $y=4$.
• The function is positive and increasing on $(4,5)$.
• The function has a local minimum at $y=5$ and increases from the minimum value at $y=5$ to positive infinity.
• The function is positive and increasing on $(5,\infty)$.

# 机械学代写|MECHANICS MATHS2034 University of Glasgow Assignment

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Mechanics is the branch of physics that deals with the study of the motion, forces, and energy of objects. It is concerned with how objects move and the forces that cause this motion. The study of mechanics involves the development of mathematical models and theories to describe and explain the behavior of physical systems, such as machines, structures, and particles. Some of the key concepts in mechanics include Newton’s laws of motion, conservation of energy, and momentum. Mechanics is essential in many fields of science and engineering, including aerospace engineering, mechanical engineering, and materials science.

Suppose the state of stress at a point in a $x, y, z$ coordinate system is given by
$$\left[\begin{array}{ccc} 100 & 1 & 180 \ 0 & 20 & 0 \ 180 & 0 & 20 \end{array}\right]$$
a. Calculate the three invariants of this stress tensor.

a) Invariants of stress tensor
Recall: these values do not change no matter the coordinate system selected
$$\begin{gathered} I_1=\sigma_x+\sigma_y+\sigma_z=-100+20+20 \ I_2=\sigma_x \sigma_y+\sigma_y \sigma_z+\sigma_z \sigma_x-\tau_{x y}{ }^2-\tau_{y z}{ }^2-\tau_{z x}{ }^2 \ =(-100 \times 20)+(-100 \times 20)+20^2-0-0-80^2 \ I_1=-60 \ I_2=-10000 \ =\sigma_x \sigma_y \sigma_z+2 \tau_{x y} \tau_{y z} \tau_{z x}-\sigma_x \tau_{y z}{ }^2-\sigma_y \tau_{z x}{ }^2-\sigma_z \tau_{x y}{ }^2 \ =(-100 \times 20 \times 20)+2 \times 0-0-20 \times(-80)^2-0 \ I_3=-168000 \end{gathered}$$

b. Determine the three principal stresses of this stress tensor.

The eigenvalue problem for a stress tensor

$$\left[\begin{array}{ccc} -100 & 0 & -80 \ 0 & 20 & 0 \ -80 & 0 & 20 \end{array}\right]$$
is given by
$$\operatorname{det}\left[\begin{array}{ccc} -100-\lambda & 0 & -80 \ 0 & 20-\lambda & 0 \ -80 & 0 & 20-\lambda \end{array}\right]=0$$
Solve for $\lambda$, we have three eigenvalues
\begin{aligned} & \lambda_1=-140 \ & \lambda_2=60 \ & \lambda_3=20 \end{aligned}
The principal stresses are the three eigenvalues of the stress tensor
\begin{aligned} \sigma_{x p} & =-140 \ \sigma_{y p} & =60 \ \sigma_{z p} & =20 \end{aligned}

Suppose the state of stress at a point relative to a $x, y, z$ coordinate system is given by:
$$\left[\begin{array}{cc} 15 & -10 \ -10 & -5 \end{array}\right]$$
Try to find a new coordinate system $\left(x^{\prime}, y^{\prime}\right)$ that corresponds to the principal directions of the stress tensor.
a. Find the principal stresses.

a) Determine principal stresses
The eigenvalue problem for a stress tensor
$$\left[\begin{array}{cc} 15 & -10 \ -10 & -5 \end{array}\right]$$
is given by
$$\operatorname{det}\left[\begin{array}{cc} 15-\lambda & -10 \ -10 & -5-\lambda \end{array}\right]=0$$
Solve for $\lambda$, we have two eigenvalues
\begin{aligned} & \lambda_1=19.14 \ & \lambda_2=-9.14 \end{aligned}
The principal stresses are the two eigenvalues of the stress tensor
\begin{aligned} & \sigma_{x p}=19.14 \ & \sigma_{y p}=-9.14 \end{aligned}

# 实分析代写|INTRODUCTION TO REAL ANALYSIS MATHS2032 University of Glasgow Assignment

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Assignment-daixieTM为您提供格拉斯哥大学University of Glasgow INTRODUCTION TO REAL ANALYSIS MATHS2032实分析代写代考辅导服务！

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Real analysis is a branch of mathematics that deals with the study of real numbers and their properties, such as continuity, convergence, differentiation, and integration. It provides a rigorous foundation for calculus and other mathematical topics, and it has many applications in science, engineering, economics, and other fields. The main focus of real analysis is on the concept of limits, which is used to study the behavior of functions and sequences. It also covers topics such as metric spaces, topology, and measure theory.

Let $S \subset \mathbb{R}$. A function $f: S \rightarrow \mathbb{R}$ is not continuous at $c \in S$ if

A function $f: S \rightarrow \mathbb{R}$ is not continuous at $c \in S$ if at least one of the following conditions hold:

1. $\lim_{x \rightarrow c} f(x)$ does not exist.
2. $\lim_{x \rightarrow c} f(x)$ exists, but it is not equal to $f(c)$.
3. $f(c)$ is undefined (i.e., $c$ is not in the domain of $f$).

In other words, a function $f$ is not continuous at $c$ if it has a discontinuity at $c$. A discontinuity is a point at which the function fails to be continuous, and it can occur for various reasons, such as a jump, a hole, or an asymptote in the graph of the function.

Let $S \subset \mathbb{R}$. A function $f: S \rightarrow \mathbb{R}$ is not uniformly continuous on $S$ if

There are different equivalent ways to define uniform continuity, but one common definition is:

A function $f: S \rightarrow \mathbb{R}$ is not uniformly continuous on $S$ if there exists $\epsilon > 0$ such that for all $\delta > 0$, there exist $x, y \in S$ such that $|x – y| < \delta$ but $|f(x) – f(y)| \geq \epsilon$.

In other words, there is a positive distance $\epsilon$ such that no matter how small we choose a distance $\delta$ around any point in $S$, we can always find two points in $S$ that are closer than $\delta$ but whose function values differ by at least $\epsilon$. This means that the function is “not uniformly continuous” because the amount of variation in the function values depends on the location in $S$ and cannot be controlled by a single choice of $\delta$ for all points in $S$.

Intuitively, this means that the function can have arbitrarily rapid changes or oscillations, such that no matter how finely we try to sample or approximate it, there will always be some nearby points with very different function values. Alternatively, it may mean that the function has some kind of asymptotic behavior that prevents it from being “smoothly varying” or “locally approximable” at all points in $S$.

Let $S \subset \mathbb{R}$. A sequence of functions $f_n: S \rightarrow \mathbb{R}$ does not converge uniformly to $f: S \rightarrow \mathbb{R}$ if

There are a few equivalent ways to state the definition of uniform convergence, but one possible definition is:

A sequence of functions $f_n: S \rightarrow \mathbb{R}$ converges uniformly to $f: S \rightarrow \mathbb{R}$ if for every $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that for all $n \geq N$ and $x \in S$, we have $|f_n(x) – f(x)| < \epsilon$.

Therefore, a sequence of functions $f_n: S \rightarrow \mathbb{R}$ does not converge uniformly to $f: S \rightarrow \mathbb{R}$ if there exists an $\epsilon > 0$ such that for every $N \in \mathbb{N}$, there exists an $n \geq N$ and $x \in S$ such that $|f_n(x) – f(x)| \geq \epsilon$.

In other words, there is some fixed distance $\epsilon$ such that no matter how far out in the sequence you go, there will always be some point $x$ in $S$ and some function $f_n$ beyond that point that differ from $f(x)$ by at least $\epsilon$. This means that the sequence of functions does not converge uniformly to $f$.

# 数学方法和建模代写|MATHEMATICAL METHODS AND MODELLING MATHS2033 University of Glasgow Assignment

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Dynamical systems refer to the study of how complex systems change over time, and how these changes can be described mathematically. This can be applied to a wide range of fields, including physics, biology, economics, and engineering, among others.

Integral transforms involve transforming functions from one domain to another using an integral operation. This can be used to simplify calculations or solve differential equations, among other applications.

By learning about these topics, you will be able to build models that can help describe and predict real-world phenomena. This can be especially useful for honours students who are planning to pursue research in fields where mathematical modelling is an important tool.

Suppose that we have a hexagonal cavity filled with $\varepsilon=1$ and surrounded by perfectly conducting walls, as shown in fig. 1(i).
(a) List the symmetry operations and conjugacy classes, and give the character table of this symmetry group $\left(C_{6 \mathrm{~V}}\right)$.

The symmetry operations of the hexagonal cavity can be obtained by considering the transformations that leave the cavity invariant. These include:

1. Identity (E): No rotation or reflection is applied to the hexagonal cavity.
2. Rotation by $\frac{\pi}{3}$ (C$_{6}$): A $\frac{\pi}{3}$ counterclockwise rotation of the hexagonal cavity about its center leaves it invariant.
3. Rotation by $\frac{2\pi}{3}$ (C$_{6}^{2}$): A $\frac{2\pi}{3}$ counterclockwise rotation of the hexagonal cavity about its center leaves it invariant.
4. Rotation by $\pi$ (C$_{6}^{3}$): A $\pi$ rotation of the hexagonal cavity about its center leaves it invariant.
5. Rotation by $\frac{4\pi}{3}$ (C$_{6}^{4}$): A $\frac{4\pi}{3}$ counterclockwise rotation of the hexagonal cavity about its center leaves it invariant.
6. Rotation by $\frac{5\pi}{3}$ (C$_{6}^{5}$): A $\frac{5\pi}{3}$ counterclockwise rotation of the hexagonal cavity about its center leaves it invariant.
7. Reflection through a vertical plane ( $\sigma_{v}$): Reflection of the hexagonal cavity through a vertical plane containing two opposite vertices leaves it invariant.
8. Reflection through a horizontal plane ( $\sigma_{h}$): Reflection of the hexagonal cavity through a horizontal plane containing two opposite edges leaves it invariant.
9. Reflection through a diagonal plane ( $\sigma_{d}$): Reflection of the hexagonal cavity through a diagonal plane containing two opposite vertices and intersecting the opposite edges perpendicularly leaves it invariant.

Solve $u^{\prime}=-10 u$ by at least two of the second order methods. Stop at $t=1$, and check the accuracy (do the numbers support $(\Delta t)^2 ?$ ). How serious is the stability test on $-10 \Delta t$ ?

The differential equation $u’=-10u$ is a first-order ordinary differential equation (ODE), which can be solved using numerical methods. Here, we will use two second-order methods: the backward Euler method and the trapezoidal method.

1. Backward Euler method: The backward Euler method approximates $u(t)$ at time $t+\Delta t$ by the value $u_{n+1}$:

$u_{n+1}=u_n-10 \Delta t u_{n+1}$

Rearranging this equation gives:

$u_{n+1}=\frac{1}{1+10 \Delta t} u_n$

1. Trapezoidal method: The trapezoidal method approximates $u(t)$ at time $t+\Delta t$ by the value $u_{n+1}$:

$u_{n+1}=u_n+\frac{\Delta t}{2}\left(-10 u_n-10 u_{n+1}\right)$

Rearranging this equation gives:

$u_{n+1}=\frac{1+\frac{5}{2} \Delta t}{1-\frac{5}{2} \Delta t} u_n$

To check the accuracy of these methods, we can compare the numerical solution to the exact solution of the differential equation, which is $u(t)=e^{-10t}$.

Consider a semi-infinite highway $0 \leq x<\infty$ (with no entrances of exits other than at $x=0$ ). Show that the number of cars on the highway at time $t$ is: $$N_0+\int_0^t q(0, \tau) d \tau$$ where $N_0$ is the number of cars in the highway at time $t=0$. You may assume that $\rho(x, t) \rightarrow 0$ as $x \rightarrow \infty$. Justify the equation both: directly (by physical reasoning), as well as by using the equation $\rho_t+q_x=0$.

The equation $\rho_t + q_x = 0$ is a continuity equation, which states that the rate of change of the density of cars $\rho$ in a certain region is equal to the rate at which cars are entering or leaving that region, which is given by the flux $q$.

Now, let’s consider the region $[0, x]$ of the highway at time $t$. The number of cars in this region at time $t$ is given by the integral of the density over the region:

$N(x, t)=\int_0^x \rho(\xi, t) d \xi$

Taking the derivative with respect to time on both sides, we get:

$\frac{d N}{d t}=\int_0^x \frac{\partial \rho}{\partial t} d \xi=-\int_0^x \frac{\partial q}{\partial x} d \xi=-q(x, t)$

where we used the continuity equation. Therefore, the rate of change of the number of cars in the region $[0, x]$ is equal to the flux of cars leaving the region.

Now, let’s consider the entire highway $[0,\infty)$ at time $t$. The number of cars on the highway at time $t$ is given by $N(\infty, t)$. We can write this as:

$N(\infty, t)=N(0, t)+\int_0^{\infty} \frac{d N}{d t} d x=N(0, t)-\int_0^{\infty} q(x, t) d x$

where we used the fact that the rate of change of the number of cars on the highway is equal to the flux of cars leaving the highway, and we assumed that $\rho(x,t) \rightarrow 0$ as $x \rightarrow \infty$.

We can rearrange this equation to get:

$N(\infty, t)=N(0, t)+\int_0^t \int_0^{\infty} \frac{\partial q}{\partial t} d x d \tau$

Now, let’s use the continuity equation to express $\partial q / \partial t$ in terms of $\partial \rho / \partial t$. We have:

$\frac{\partial q}{\partial t}=-\frac{\partial(\rho q)}{\partial x}+q \frac{\partial \rho}{\partial x}$

Integrating this equation over the entire highway, we get:

$\int_0^{\infty} \frac{\partial q}{\partial t} d x=-\int_0^{\infty} \frac{\partial(\rho q)}{\partial x} d x+\int_0^{\infty} q \frac{\partial \rho}{\partial x} d x$

The first term on the right-hand side can be integrated by parts:

$-\int_0^{\infty} \frac{\partial(\rho q)}{\partial x} d x=[\rho q]_0^{\infty}=-\rho(\infty, t) q(\infty, t)$

where we used the fact that $\rho(x,t) \rightarrow 0$ as $x \rightarrow \infty$. Therefore, we can rewrite the above equation as:

$\int_0^{\infty} \frac{\partial q}{\partial t} d x=-\rho(\infty$

# 线性代数代写|LINEAR ALGEBRA MATHS2004 University of Glasgow Assignment

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## Instructions:

Linear algebra is a branch of mathematics that deals with vector spaces and linear equations, and it has many practical applications in fields such as physics, computer science, economics, and more.

Throughout this course, you can expect to learn about the basics of linear algebra, such as vectors, matrices, and systems of linear equations. You will also learn about more advanced topics, such as eigenvalues and eigenvectors, linear transformations, and inner products. By the end of the course, you should be able to apply these concepts to solve a variety of problems in different areas of science and engineering.

It’s great that this course is emphasizing methods and applications, as it will help you develop the skills you need to use linear algebra in real-world situations. This course will be particularly important for students who plan to pursue honors-level studies in these fields, as it will provide them with a strong foundation in linear algebra that they can build upon in future courses.

(b) Find the coefficients $C$ and $D$ of the best curve $y=C+D 2^t$.

$$\begin{gathered} A^T A=\left[\begin{array}{lll} 1 & 1 & 1 \ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{ll} 1 & 1 \ 1 & 2 \ 1 & 4 \end{array}\right]=\left[\begin{array}{lc} 3 & 7 \ 7 & 21 \end{array}\right] \ A^T b=\left[\begin{array}{lll} 1 & 1 & 1 \ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{l} 6 \ 4 \ 0 \end{array}\right]=\left[\begin{array}{l} 10 \ 14 \end{array}\right] \end{gathered}$$
Solve $A^T A \hat{x}=A^T b$ :
$$\left[\begin{array}{cc} 3 & 7 \ 7 & 21 \end{array}\right]\left[\begin{array}{l} C \ D \end{array}\right]=\left[\begin{array}{l} 10 \ 14 \end{array}\right] \text { gives }\left[\begin{array}{l} C \ D \end{array}\right]=\frac{1}{14}\left[\begin{array}{rr} 21 & -7 \ -7 & 3 \end{array}\right]\left[\begin{array}{l} 10 \ 14 \end{array}\right]=\left[\begin{array}{r} 8 \ -2 \end{array}\right]$$

(a) Suppose $x_k$ is the fraction of MIT students who prefer calculus to linear algebra at year $k$. The remaining fraction $y_k=1-x_k$ prefers linear algebra.

At year $k+1,1 / 5$ of those who prefer calculus change their mind (possibly after taking 18.03). Also at year $k+1,1 / 10$ of those who prefer linear algebra change their mind (possibly because of this exam).

Create the matrix $A$ to give $\left[\begin{array}{l}x_{k+1} \ y_{k+1}\end{array}\right]=A\left[\begin{array}{l}x_k \ y_k\end{array}\right]$ and find the limit of $A^k\left[\begin{array}{l}1 \ 0\end{array}\right]$ as $k \rightarrow \infty$.

$$A=\left[\begin{array}{ll} .8 & .1 \ .2 & .9 \end{array}\right] .$$
The eigenvector with $\lambda=1$ is $\left[\begin{array}{l}1 / 3 \ 2 / 3\end{array}\right]$.
This is the steady state starting from $\left[\begin{array}{l}1 \ 0\end{array}\right]$.
$\frac{2}{3}$ of all students prefer linear algebra! I agree.

Solve these differential equations, starting from $x(0)=1, \quad y(0)=0$ :
$$\frac{d x}{d t}=3 x-4 y \quad \frac{d y}{d t}=2 x-3 y$$

$$A=\left[\begin{array}{ll} 3 & -4 \ 2 & -3 \end{array}\right]$$
has eigenvalues $\lambda_1=1$ and $\lambda_2=-1$ with eigenvectors $x_1=(2,1)$ and $x_2=(1,1)$. The initial vector $(x(0), y(0))=(1,0)$ is $x_1-x_2$.
So the solution is $(x(t), y(t))=e^t(2,1)+e^{-t}(1,1)$.

# 多变量微积分代写| MULTIVARIABLE CALCULUS MATHS2001 University of Glasgow Assignment

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Assignment-daixieTM为您提供格拉斯哥大学University of Glasgow MATHEMATICS 1 MATHS1017多变量微积分代写代考辅导服务！

## Instructions:

Multivariate calculus is a fundamental course that builds on the concepts learned in single-variable calculus and extends them to functions of multiple variables. This course is important for students who intend to pursue further studies in mathematics, engineering, physics, or any field that requires a solid understanding of calculus.

The course covers the concepts of partial derivatives, multiple integrals, line integrals, surface integrals, and vector calculus. These concepts are essential in various fields such as physics, engineering, economics, and computer science.

The emphasis in this course is on methods and applications. Students will learn how to use calculus to solve problems in real-world scenarios. They will also learn how to apply calculus concepts to different areas of study, including geometry, physics, and statistics.

Overall, multivariate calculus is a challenging but rewarding course that is essential for students who want to pursue advanced studies in mathematics, science, and engineering.

a) Find the equation in the form $A x+B y+C z=D$ of the plane $\mathcal{P}$ which contains the line $L$ given by $x=1-t, \quad y=1+2 t, \quad z=2-3 t$ and the point $(-1,1,2)$

a) Line $L$ has direction vector $\mathbf{v}=\langle-1,2,-3\rangle$ which lies in $\mathcal{P}$.

To get a point $P_0$ on $L$ take $t=0 \Rightarrow P_0=(1,1,2)$.
$\Rightarrow \overrightarrow{\mathbf{P}{\mathbf{0}} \mathbf{Q}}=\langle-1,1,2\rangle-\langle 1,1,2\rangle=\langle-2,0,0\rangle$ also lies in $\mathcal{P}$. $\Rightarrow$ A normal to $\mathcal{P}$ is $$\mathbf{n}=\mathbf{v} \times \overrightarrow{\mathbf{P}{\mathbf{0}} \mathbf{Q}}=\left|\begin{array}{rrr} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -1 & 2 & 3 \ -2 & 0 & 0 \end{array}\right|=\mathbf{i}(0)-\mathbf{j}(-6)+\mathbf{k}(4)=\langle 0,6,4\rangle .$$
So, the equation of $\mathcal{P}$ is
$$0(x-1)+6(y-1)+4(z-2)=0 \quad \text { or } \quad 6 y=4 z=14 \quad \text { or } \quad 3 y+2 z=7 .$$
b) $\mathbf{n}_Q=\langle 2,1,1\rangle \Rightarrow \widehat{\mathbf{n}}=\frac{1}{\sqrt{6}}\langle 2,1,1\rangle, \quad \mathbf{v}=\langle-1,2,-3\rangle \Rightarrow \widehat{\mathbf{v}}=\frac{1}{\sqrt{14}}\langle-1,2,-3\rangle$ Component of $\widehat{\mathbf{n}}$ on $\widehat{\mathbf{v}}$ is
$$\widehat{\mathbf{n}} \cdot \widehat{\mathbf{v}}=\frac{1}{\sqrt{6 \cdot 14}}(2+2-3)=-\frac{3}{\sqrt{84}}$$

Let $\mathbf{r}(t)=\left\langle\cos \left(e^t\right), \sin \left(e^t\right), e^t\right\rangle$.
a) (5 pts.) Compute and simplify the unit tangent vector $\mathbf{T}(t)=\frac{\mathbf{r}^{\prime}(t)}{\left|\mathbf{r}^{\prime}(t)\right|}$.
b) (5 pts.) Compute $\mathbf{T}^{\prime}(t)$

\begin{aligned} & \text { a) } \mathbf{r}^{\prime}(t)=\left\langle-\sin \left(\mathrm{e}^t\right) \mathrm{e}^t, \cos \left(\mathrm{e}^t\right) \mathrm{e}^t, \mathrm{e}^t\right\rangle \Rightarrow\left|\mathbf{r}^{\prime}(t)\right|=\mathrm{e}^t \sqrt{1+1}=\mathrm{e}^t \sqrt{2} \ & \Rightarrow \mathbf{T}(t)=\frac{\mathbf{r}^{\prime}(t)}{\left|\mathbf{r}^{\prime}(t)\right|}=\frac{1}{\sqrt{2}}\left\langle-\sin \left(\mathrm{e}^t\right), \cos \left(\mathrm{e}^t\right), 1\right\rangle \end{aligned}
b)
$$\mathbf{T}^{\prime}(t)=\frac{1}{\sqrt{2}}\left\langle-\cos \left(\mathrm{e}^t\right),-\sin \left(\mathrm{e}^t\right), 0\right\rangle=-\frac{\mathrm{e}^t}{\sqrt{2}}\left\langle\cos \left(\mathrm{e}^t\right), \sin \left(\mathrm{e}^t\right), 0\right\rangle .$$

Let $f(x, y)=x+4 y+\frac{2}{x y}$.
a) Find the critical point(s) of $f(x, y)$
b) Use the second-derivative test to test the critical point(s) found in part(a).

\begin{aligned} & \left.\left.\begin{array}{l} f_x=1-2 /\left(x^2 y\right)=0 \ f_y=4-2 /\left(x y^2\right)=0 \end{array}\right} \Rightarrow \begin{array}{l} x^2 y=2 \ x y^2=1 / 2 \end{array}\right} \Rightarrow x=4 y \ & \Rightarrow 4 y^3=\frac{1}{2} \Rightarrow y^3=\frac{1}{8} \Rightarrow y=\frac{1}{2} \Rightarrow x=2 . \ & \end{aligned}
There is one critical point at $(x, y)=(2,1 / 2)$.
b) $f_{x x}=4 /\left(x^2 y\right), \quad f_{y y}=4 /\left(x y^3\right), \quad f_{x y}=f_{y x}=2 /\left(x^2 y^2\right)$
$A=f_{x x}(2,1 / 2)=1, \quad C=f_{y y}(2,1 / 2)=16, \quad B=f_{x y}(2,1 / 2)=2$
$\Rightarrow A C-B^2=12>0, A>0 \Rightarrow f$ has a relative minimum at $(2,1 / 2)$.

# 数学代写|MATHEMATICS 1 MATHS1017 University of Glasgow Assignment

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That sounds like a great course! As an AI language model, I can provide some general information about the topics that might be covered in an advanced mathematics course:

1. Calculus: This includes topics such as differentiation, integration, limits, and series.
2. Linear Algebra: This includes topics such as matrix operations, determinants, eigenvalues and eigenvectors, and systems of linear equations.
3. Differential Equations: This includes topics such as first-order equations, higher-order equations, and systems of equations.
4. Real Analysis: This includes topics such as sequences, continuity, convergence, and theorems related to limits and functions.
5. Abstract Algebra: This includes topics such as group theory, ring theory, and field theory.
6. Probability and Statistics: This includes topics such as probability distributions, statistical inference, hypothesis testing, and regression analysis.

Overall, this course would likely cover a broad range of mathematical topics, with a focus on developing problem-solving skills and effective communication of mathematical ideas.

Prove that if the potential for a moving particle is $U(q)=-B(q, q)$, where $B$ is a nonnegative definite symmetric bilinear form on a Euclidean space $V$, then for any $q_1, q_2 \in V$ and $t_1<t_2 \in \mathbb{R}$ there exists a unique solution of the Newton equation with $q\left(t_1\right)=q_1$ and $q\left(t_2\right)=q_2$. Show that it provides not only an extremum but also a minimum for the action with these boundary conditions. What happens if $B$ is not nonnegative?

To prove that there exists a unique solution to the Newton equation with the given potential, we can use the fact that the equation of motion for a particle with potential energy $U(q)$ is given by:

$m \frac{d^2 q}{d t^2}=-\nabla U(q)$

where $m$ is the mass of the particle. Substituting $U(q)=-B(q,q)$ into this equation, we get:

$m \frac{d^2 q}{d t^2}=B^{\prime}(q) \dot{q}$

where $B'(q)$ is the derivative of $B(q,q)$ with respect to $q$. Since $B$ is nonnegative definite and symmetric, we have $B'(q)$ is a symmetric matrix that is also nonnegative definite. This means that $B'(q)$ can be diagonalized by an orthogonal matrix $O$, such that $B'(q)=O^T\Lambda O$, where $\Lambda$ is a diagonal matrix with nonnegative entries. Thus, we can write the equation of motion as:

$m \frac{d^2}{d t^2}(O q)=\left(O^T \Lambda O\right)(O \dot{q})=\Lambda\left(O^T \dot{q}\right)$,

where we have used the fact that $O$ is orthogonal, so $O^TO=I$. Defining $p=O^T\dot{q}$, we obtain the system of first-order differential equations:

$\frac{d p}{d t}=\frac{1}{m} \Lambda p, \quad \frac{d q}{d t}=O p$

with initial conditions $q(t_1)=q_1$ and $q(t_2)=q_2$. Since $\Lambda$ is diagonal with nonnegative entries, the solutions to the first equation are of the form $p_i(t)=c_ie^{\omega_i t}$, where $\omega_i=\sqrt{\lambda_i/m}$ and $c_i$ are constants determined by the initial conditions. The solutions to the second equation are of the form $q_i(t)=d_ie^{\omega_i t}$, where $d_i$ are constants determined by the initial conditions. Thus, we have a unique solution for $p(t)$ and $q(t)$.

To show that this solution provides a minimum for the action, we need to compute the action:

$S=\int_{t_1}^{t_2}\left[\frac{1}{2} m \dot{q}^2+U(q)\right] d t=\int_{t_1}^{t_2}\left[\frac{1}{2} p^T \Lambda^{-1} p-B(q, q)\right] d t$

Using the solutions for $p(t)$ and $q(t)$, we can evaluate the action as:

$S=\sum_{i=1}^n\left[\frac{m \omega_i^2}{2} \int_{t_1}^{t_2} p_i^2 d t-B(q, q)\right]=-\sum_{i=1}^n \frac{m \omega_i^2}{2} c_i^2-B\left(q_1, q_2\right)$

Since $B$ is nonnegative definite, we have $B(q_1,q_2)\leq 0$. Also, since $\omega_i^2\geq 0$ and $c_i$ are constants, we have $-\sum_{i=1}^n\frac{m\omega_i^2}{2}c_i^2\geq 0$.

Find the generating function $\sum a_n z^n / n$ ! for the numbers $a_n$ of labeled $\mathrm{n}-$ vertex trees with 1-valent and 4 -valent vertices. You may express the answer via inverse functions to polynomials.

Let $f(z)$ be the generating function for labeled trees with 1-valent and 4-valent vertices. Then, we can decompose a 4-valent vertex into two 1-valent vertices and two 2-valent vertices, which gives us the following recursion: $$f(z) = z + z\left(f(z)\right)^2$$ The first term, $z$, accounts for the 1-valent vertex, while the second term accounts for the 4-valent vertex. The factor of $z$ in the second term accounts for the fact that the 4-valent vertex can appear anywhere in the tree.

We want to find the generating function for labeled trees with 1-valent and 4-valent vertices, but with the additional restriction that the total degree of the vertices is equal to $n$. Let $g_n$ be the number of such trees. Then, we have: $$\frac{1}{n}\sum_{k=1}^{n-1} g_k g_{n-k} = [z^n]\left(f(z)\right)^2$$ The factor of $1/n$ in the sum is to account for the fact that we’re dividing by the total number of edges in the tree. The coefficient $[z^n]\left(f(z)\right)^2$ counts the number of ways to split a tree of size $n$ into two smaller trees.

Let $g(z) = \sum_{n=0}^\infty g_n z^n$. Then, we have: $$g(z) = z + z\left(g(z)\right)^2$$ Multiplying both sides by $g(z)$ and differentiating with respect to $z$, we get: $$g(z)\left(1 + 2z g'(z)\right) = 1 + g(z)^2$$ Solving for $g(z)$, we get: $$g(z) = \frac{1 – \sqrt{1 – 4z^2}}{2z}$$ Therefore, the generating function for the labeled trees with 1-valent and 4-valent vertices, weighted by $1/n$, is given by: $$\frac{1}{z}\left(\frac{1 – \sqrt{1 – 4z^2}}{2}\right)$$

Prove Mumford’s theorem (see the notes, Th. 5.2) for $g=1$.

Mumford’s theorem for $g=1$ states that any algebraic curve of genus $1$ over an algebraically closed field is isomorphic to an elliptic curve in Weierstrass form.

To prove this, we start with an algebraic curve $C$ of genus $1$ defined over an algebraically closed field $K$. By definition of genus, $C$ is a smooth projective curve of degree $2$.

Since $C$ has genus $1$, it has exactly one nontrivial invertible sheaf $\mathcal{L}$ of degree $1$. By the Riemann-Roch theorem, we have $\dim_K H^0(C,\mathcal{L})=\dim_K H^0(C,K)-\deg(\mathcal{L})+1=2-1+1=2$. Thus, there exist two linearly independent global sections $f$ and $g$ of $\mathcal{L}$.

We can use these sections to define a morphism $\phi:C\to\mathbb{P}^2_K$ as follows. Let $P\in C$ be a point, and let $(f(P):g(P):1)$ be the corresponding point in $\mathbb{P}^2_K$. Then define $\phi(P)=(f(P)^2:g(P)^2:f(P)g(P))$. It can be shown that $\phi$ is a morphism of algebraic curves, and that its image is contained in the zero set of the Weierstrass equation $y^2=x^3+ax+b$, where $a$ and $b$ are constants in $K$.

To complete the proof, we need to show that $\phi$ is an isomorphism. Since $C$ and the zero set of the Weierstrass equation are both smooth and projective, it suffices to show that $\phi$ is bijective and that its differential is everywhere nonzero.

To show that $\phi$ is bijective, we note that the inverse image of a point $(x:y:1)$ in $\mathbb{P}^2_K$ under $\phi$ consists of two points if $x$ and $y$ are nonzero, and of one point if either $x$ or $y$ is zero. It can be shown that these points are distinct, so $\phi$ is injective. To show that $\phi$ is surjective, we note that any point $(x:y:z)$ in the zero set of the Weierstrass equation can be mapped to by the point $(\frac{y}{z},\frac{x}{z})$ on $C$, so $\phi$ is surjective.

Finally, we need to show that the differential $d\phi$ is everywhere nonzero. It can be shown that $d\phi$ is nonzero at the point $(0,0)$ of $C$, which corresponds to the point at infinity in $\mathbb{P}^2_K$. Since $C$ and the zero set of the Weierstrass equation are smooth, $d\phi$ is nonzero everywhere.

Therefore, $\phi$ is an isomorphism of algebraic curves, so $C$ is isomorphic to an elliptic curve in Weierstrass form.