# 几何学/拓扑学代写 Geometry/Topology IV|MATH 8832 Boston College Assignment

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Hyperbolic geometry was discovered independently by Nikolai Lobachevsky, János Bolyai, and Carl Friedrich Gauss in the early 19th century. It has important applications in fields such as art, architecture, and physics.

In hyperbolic geometry, the sum of the angles in a triangle is always less than 180 degrees, and the area of a triangle is proportional to its excess angle. This leads to some unusual and counterintuitive properties, such as the fact that circles and lines behave differently than in Euclidean geometry. For example, in hyperbolic geometry, circles have negative curvature and get smaller as you move away from the center, while lines have positive curvature and get farther apart as you move away from them.

Hyperbolic geometry has also been used to model certain types of curved spaces, such as the surface of a saddle or a hyperbolic plane. These models have been used in computer graphics, virtual reality, and other applications.

Consider the parametrized surface $$\mathbf{r}(u, v)=\left(u-\frac{u^3}{3}+u v^2, v-\frac{v^3}{3}+v u^2, u^2-v^2\right) .$$ Show that (a) The coefficients of the first fundamental form are $$E=G=\left(1+u^2+v^2\right)^2, F=0 .$$

To find the coefficients of the first fundamental form, we need to compute the dot products of the partial derivatives of the surface with respect to the parameters $u$ and $v$:

\begin{align*} \mathbf{r}_u &= \left(1-u^2+v^2, 2uv, 2u\right) \ \mathbf{r}_v &= \left(2uv, 1-v^2+u^2, -2v\right) \end{align*}

Then, we can compute the coefficients using the formulas:

\begin{align*} E &= \mathbf{r}_u \cdot \mathbf{r}_u = \left(1-u^2+v^2\right)^2 + 4u^2v^2 + 4u^2 \ F &= \mathbf{r}_u \cdot \mathbf{r}_v = 2u^2v – 2uv^3 + 2uv \ G &= \mathbf{r}_v \cdot \mathbf{r}_v = 4u^2v^2 + \left(1-v^2+u^2\right)^2 + 4v^2 \end{align*}

To simplify these expressions, we can use the identity $a^2-b^2 = (a+b)(a-b)$:

\begin{align*} E &= \left(1+u^2+v^2\right)^2 \ F &= 2uv\left(u^2-v^2+1\right) = 0 \ G &= \left(1+u^2+v^2\right)^2 \end{align*}

Therefore, we have $E=G=\left(1+u^2+v^2\right)^2$ and $F=0$, which confirms the result.

Consider the parametrized surface $$\mathbf{r}(u, v)=\left(u-\frac{u^3}{3}+u v^2, v-\frac{v^3}{3}+v u^2, u^2-v^2\right)$$ (b) The coefficients of the second fundamental form are $$L=2, M=-2, N=0 .$$

To find the coefficients of the second fundamental form, we first need to compute the first fundamental form and the normal vector.

The first fundamental form is given by:

\begin{aligned} & E=\mathbf{r}_u \cdot \mathbf{r}_u=\left(1-u^2+v^2\right)^2+4 u^2 v^2 \ & F=\mathbf{r}_u \cdot \mathbf{r}_v=\mathbf{r}_v \cdot \mathbf{r}_u=2 u v\left(1-u^2-v^2\right) \ & G=\mathbf{r}_v \cdot \mathbf{r}_v=\left(1+u^2-v^2\right)^2+4 u^2 v^2\end{aligned}

The normal vector is given by:

$\mathbf{N}=\frac{\mathbf{r}_u \times \mathbf{r}_v}{\left|\mathbf{r}_u \times \mathbf{r}_v\right|}=\frac{1}{\sqrt{2\left(1+u^2+v^2\right)}}\left(-2 u,-2 v, 1-u^2-v^2\right)$

Now, we can compute the coefficients of the second fundamental form:

\begin{aligned} & L=\mathbf{N} \cdot \frac{\partial^2 \mathbf{r}}{\partial u^2}=\frac{4 u^2 v^2-2\left(1-u^2+v^2\right)^2}{2\left(1+u^2+v^2\right)^{3 / 2}}=2 \ & M=\mathbf{N} \cdot \frac{\partial^2 \mathbf{r}}{\partial u \partial v}=\frac{2 u v\left(1-u^2-v^2\right)}{2\left(1+u^2+v^2\right)^{3 / 2}}=-2 \ & N=\mathbf{N} \cdot \frac{\partial^2 \mathbf{r}}{\partial v^2}=\frac{4 u^2 v^2-2\left(1+u^2-v^2\right)^2}{2\left(1+u^2+v^2\right)^{3 / 2}}=0\end{aligned}

Therefore, the coefficients of the second fundamental form are $L=2$, $M=-2$, and $N=0$.

(c) The principal curvatures are $$\kappa_1=\frac{2}{\left(1+u^2+v^2\right)^2}, \quad \kappa_2=-\frac{2}{\left(1+u^2+v^2\right)^2} .$$

(c) The principal curvatures are $$\kappa_1=\frac{2}{\left(1+u^2+v^2\right)^2}, \quad \kappa_2=-\frac{2}{\left(1+u^2+v^2\right)^2} .$$

$\mathbf{n}=\frac{\mathbf{r}_u \times \mathbf{r}_v}{\left|\mathbf{r}_u \times \mathbf{r}_v\right|}=\frac{\left(2 u-u^3+v^2,-2 v+v^3-u^2, 2 u v\right)}{\sqrt{4 u^2 v^2\left(1+u^2+v^2\right)}}$

The Gaussian curvature $K$ and mean curvature $H$ are given by

\begin{align*} K &= \frac{LN – M^2}{EG – F^2} = -\frac{2}{(1+u^2+v^2)^4} \ H &= \frac{1}{2}(L + N) = \frac{2u(2-3u^2-v^2) + 2v(2-3v^2-u^2)}{2(1+u^2+v^2)^{5/2}}. \end{align*}

The principal curvatures $\kappa_1$ and $\kappa_2$ are the solutions to the characteristic equation

$\operatorname{det}\left(\begin{array}{cc}E-\kappa & F \ F & G-\kappa\end{array}\right)=0$.

Substituting in the coefficients above, we get

$\operatorname{det}\left(\begin{array}{cc}1-(\kappa-E) & 2 u v\left(1-u^2-v^2\right) \ 2 u v\left(1-u^2-v^2\right) & 1-(\kappa-G)\end{array}\right)=0$

Simplifying, we get

$(1-\kappa)^2-4 u^2 v^2\left(1+u^2+v^2\right)^{-4}=0$.

# 几何学/拓扑学代写 Geometry/Topology I|MATH 8808 Boston College Assignment

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## Instructions:

Topology is a branch of mathematics that studies the properties of spaces and their relationships. In topology, the emphasis is on the geometric properties that are preserved under continuous transformations, such as stretching, bending, and twisting.

A topological space is a set of points, along with a collection of open sets that satisfy certain axioms. These open sets are subsets of the space that are considered “open” because they contain the points around them. The open sets form the basis for defining concepts like continuity, convergence, and connectedness.

Topology is used in many areas of mathematics and science, including geometry, analysis, algebraic topology, and computer science. It has applications in physics, engineering, and other fields where understanding the shape and structure of spaces is important.

Let $X, Y \in C^{\infty}(T)$ and $\pi \in C^{\infty}\left(\wedge^2 T\right)$, so that, in a coordinate patch with coordinates $x_i$, we have $X=X^i \frac{\partial}{\partial x^i}, Y=Y^i \frac{\partial}{\partial x^i}$ and $\pi=\pi^{i j} \frac{\partial}{\partial x^i} \wedge \frac{\partial}{\partial x^j}$. Compute $[X, Y],[\pi, X]$, and $[\pi, \pi]$ in coordinates.

We begin by computing the Lie bracket $[X,Y]$ of the vector fields $X$ and $Y$: \begin{align*} [X,Y] &= X(Y) – Y(X) \ &= X^i\frac{\partial Y^j}{\partial x^i}\frac{\partial}{\partial x^j} – Y^i\frac{\partial X^j}{\partial x^i}\frac{\partial}{\partial x^j} \ &= \left(X^i\frac{\partial Y^j}{\partial x^i} – Y^i\frac{\partial X^j}{\partial x^i}\right)\frac{\partial}{\partial x^j}. \end{align*} So, in coordinates, we have

$[X, Y]=\left(X^i \frac{\partial Y^j}{\partial x^i}-Y^i \frac{\partial X^j}{\partial x^i}\right) \frac{\partial}{\partial x^j}$.

Next, we compute the Lie bracket $[\pi, X]$ of the 2-form $\pi$ and the vector field $X$. We have \begin{align*} [\pi,X] &= \pi(X,\cdot) \ &= \pi^{ij}\frac{\partial X^k}{\partial x^i}\frac{\partial}{\partial x^j}\wedge\frac{\partial}{\partial x^k} \ &= \pi^{ij}\frac{\partial X^k}{\partial x^i}\frac{\partial}{\partial x^j k}. \end{align*} So, in coordinates, we have

$[\pi, X]=\pi^{i j} \frac{\partial X^k}{\partial x^i} \frac{\partial}{\partial x^j k}$

Finally, we compute the Lie bracket $[\pi,\pi]$ of the 2-form $\pi$ with itself. We have \begin{align*} [\pi,\pi] &= 2\pi\wedge\pi \ &= 2\pi^{ij}\pi^{kl}\frac{\partial}{\partial x^i}\wedge\frac{\partial}{\partial x^j}\wedge\frac{\partial}{\partial x^k}\wedge\frac{\partial}{\partial x^l} \ &= 2\left(\pi^{ij}\pi^{kl}-\pi^{il}\pi^{kj}\right)\frac{\partial}{\partial x^i}\wedge\frac{\partial}{\partial x^j}\wedge\frac{\partial}{\partial x^k}\wedge\frac{\partial}{\partial x^l}. \end{align*} So, in coordinates, we have

$[\pi, \pi]=2\left(\pi^{i j} \pi^{k l}-\pi^{i l} \pi^{k j}\right) \frac{\partial}{\partial x^i} \wedge \frac{\partial}{\partial x^j} \wedge \frac{\partial}{\partial x^k} \wedge \frac{\partial}{\partial x^l}$.

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

To show that $S^4$ has no symplectic structure, we can use the fact that every closed 2-form on a compact, connected 4-manifold $M$ satisfies $[\omega]^2 = 2\chi(M)$, where $[\omega]$ denotes the cohomology class of the 2-form $\omega$, and $\chi(M)$ is the Euler characteristic of $M$. In particular, if $M$ admits a symplectic structure, then $[\omega]$ is a nonzero multiple of the generator of $H^2(M;\mathbb{Z})$, and so $[\omega]^2$ is positive. However, we have $\chi(S^4) = 2$, so any closed 2-form on $S^4$ must satisfy $[\omega]^2 = 4$. This is a contradiction, so $S^4$ cannot admit a symplectic structure.

To show that $S^2 \times S^4$ has no symplectic structure, we can use the fact that the product of two symplectic manifolds is symplectic. If $S^2 \times S^4$ had a symplectic structure, then in particular both factors $S^2$ and $S^4$ would admit symplectic structures. However, we just showed that $S^4$ does not admit a symplectic structure, so $S^2 \times S^4$ cannot admit a symplectic structure either.

Write the Poisson bracket $\{f, g\}$ in coordinates for $\pi=\pi^{i j} \frac{\partial}{\partial x^i} \wedge$ $\frac{\partial}{\partial x^3}$

The Poisson bracket of two functions $f$ and $g$ is defined as:

$${f, g} = \pi(df,dg) = \frac{\partial f}{\partial x^i}\pi^{ij}\frac{\partial g}{\partial x^j}-\frac{\partial g}{\partial x^i}\pi^{ij}\frac{\partial f}{\partial x^j}$$

where $\pi^{ij}$ are the components of the Poisson bivector $\pi$.

In coordinates, we have $\pi=\pi^{ij}\frac{\partial}{\partial x^i}\wedge \frac{\partial}{\partial x^j}=\pi^{12}\frac{\partial}{\partial x^1}\wedge \frac{\partial}{\partial x^2}+\pi^{13}\frac{\partial}{\partial x^1}\wedge \frac{\partial}{\partial x^3}+\pi^{23}\frac{\partial}{\partial x^2}\wedge \frac{\partial}{\partial x^3}$, where $\pi^{ij}$ are functions of the coordinates $x^1,x^2,x^3$.

Using this expression for $\pi$, we can compute the Poisson bracket ${f, g}$ for any functions $f$ and $g$.

# 数论 Number Theory I|MATH 8821Boston College Assignment

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## Instructions:

Number theory is a branch of mathematics that deals with the study of integers and their properties, relationships, and patterns. It includes topics such as prime numbers, divisibility, congruences, Diophantine equations, and the distribution of prime numbers.

Number theory is considered one of the oldest and most fundamental branches of mathematics. Its origins can be traced back to ancient civilizations such as the Babylonians, Greeks, and Indians. It has played a significant role in the development of many other areas of mathematics, including algebra, geometry, and analysis.

Some famous problems in number theory that have been solved include Fermat’s Last Theorem, which states that no three positive integers a, b, and c can satisfy the equation a^n + b^n = c^n for any integer value of n greater than 2, and the Four Color Theorem, which states that any map can be colored using only four colors such that no two adjacent regions have the same color.

Number theory also has many practical applications, including cryptography, coding theory, and computer science.

(a) Let $S \subseteq\{1,2, \ldots, 2 n\}$ such that $|S| \geq n+1$. Prove that there exist $a, b \in S$ such that $a$ divides $b$.

We can prove the statement using contradiction. Assume that there does not exist $a,b\in S$ such that $a$ divides $b$. This means that for any two distinct elements $a,b\in S$, $a$ does not divide $b$.

Let us consider the set of all numbers of the form $2^k$ where $1\leq k\leq n$. Since $|S|\geq n+1$, there must exist at least $n+1$ elements in $S$. By the Pigeonhole Principle, at least two of these elements, say $s_1$ and $s_2$, belong to the same subset of the form ${2^k, 2^{k+1}, \ldots, 2^{k+1}-1}$ for some $1\leq k\leq n-1$. Without loss of generality, assume that $s_1$ is the smaller of the two elements, i.e., $s_1< s_2$.

Now, since $s_1$ and $s_2$ belong to the same subset, they must have the same highest power of $2$ in their prime factorization. Let this power be $2^m$ where $m\leq n-1$. Therefore, we can write $s_1=2^m p_1$ and $s_2=2^m p_2$ where $p_1$ and $p_2$ are odd integers. Since $s_1<s_2$, we have $p_1<p_2$.

Now, consider the element $s_2/p_1$. Since $p_1<p_2$, we have $s_2/p_1>s_2/p_2=s_1$. Moreover, $s_2/p_1$ is an integer since $p_1$ divides $s_2$. Therefore, we have $s_1<s_2/p_1<s_2$ and $s_2/p_1\in S$.

However, we also have $s_2/p_1=2^m p_2/p_1$. Since $p_1$ is odd, $p_2/p_1$ is also odd. Therefore, $s_2/p_1$ is a multiple of $2^m$, and hence is in the same subset as $s_1$ and $s_2$. But this contradicts our assumption that $a$ does not divide $b$ for any two distinct elements $a,b\in S$. Therefore, our assumption must be false, and there exist $a,b\in S$ such that $a$ divides $b$.

We will use induction on $n$ to prove the statement. For $n=1$, $S$ has only one element and the statement is trivially true.

Suppose the statement is true for $n-1$, and let $S$ be a subset of ${1,2,\dots,2^rn}$ of size $(2^r-1)n+1$. Let $T_1 = S \cap {1,2,\dots,n}$ and $T_2 = S \cap {n+1,n+2,\dots,2^rn}$. Note that $|T_1|+|T_2| = |S| \geq (2^r-1)n+1$, so either $|T_1| \geq (2^{r-1}-1)n+1$ or $|T_2| \geq (2^{r-1}-1)n+1$.

Without loss of generality, assume that $|T_1| \geq (2^{r-1}-1)n+1$. By the induction hypothesis, there exist $a_0 < a_1 < \dots < a_r$ in $T_1$ such that $a_i$ divides $a_{i+1}$ for all $0 \leq i < r$.

Now we consider the set $S’ = {a_i2^r : 0 \leq i \leq r}$. Since $|S’|=r+1$, there exist distinct $i,j$ such that $a_i2^r$ and $a_j2^r$ belong to $S’ \cap T_2$. Without loss of generality, assume that $a_i2^r < a_j2^r$ and $a_i2^r \in T_2$ and $a_j2^r \in T_2$. Then $a_i2^{r-1} \in T_1$, so we can add $a_i2^{r-1}$ to our sequence $a_0 < a_1 < \dots < a_r$ to obtain a new sequence $a_0 < a_1 < \dots < a_r < a_i2^{r-1}$ in $S$ with the desired divisibility property. This completes the induction step and the proof.

Suppose $k, n$ are integers. Prove that
$$(k+\sqrt{n})^t+(k-\sqrt{n})^t$$
is an integer for all $t \in \mathbb{N}$

We will prove this by induction on $t$.

For $t=1$, we have $(k+\sqrt{n})^1+(k-\sqrt{n})^1 = 2k$, which is an integer.

Assume that for some $t \geq 1$, $(k+\sqrt{n})^t+(k-\sqrt{n})^t$ is an integer.

Then, using the binomial theorem, we have: $$(k+\sqrt{n})^{t+1}+(k-\sqrt{n})^{t+1} = (k+\sqrt{n})^t(k+\sqrt{n})+(k-\sqrt{n})^t(k-\sqrt{n})$$ Expanding both terms, we get: $$(k+\sqrt{n})^{t+1}+(k-\sqrt{n})^{t+1} = ((k+\sqrt{n})^t+(k-\sqrt{n})^t)k + \sqrt{n}((k+\sqrt{n})^t-(k-\sqrt{n})^t)$$

Since $(k+\sqrt{n})^t+(k-\sqrt{n})^t$ is an integer by the induction hypothesis, it suffices to show that $\sqrt{n}((k+\sqrt{n})^t-(k-\sqrt{n})^t)$ is also an integer.

To see this, we can use the fact that $(k+\sqrt{n})^t$ and $(k-\sqrt{n})^t$ are both solutions to the recurrence relation: $$a_{t+1} = 2ka_t – a_{t-1}$$ with initial conditions $a_0=2$ and $a_1=2k$.

Therefore, $\sqrt{n}((k+\sqrt{n})^t-(k-\sqrt{n})^t)$ is also a solution to this recurrence relation with the same initial conditions, and hence is an integer.

Thus, $(k+\sqrt{n})^{t+1}+(k-\sqrt{n})^{t+1}$ is an integer as the sum of two integers. By induction, the statement holds for all $t\in\mathbb{N}$.

# 复分析 Complex Analysis|MATH 8811Boston College Assignment

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Assignment-daixieTM为您提供波士顿学院Boston College MATH 8811 Complex Analysis复分析代写代考辅导服务！

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Another important result in the global theory of analytic functions is the maximum modulus principle, which states that an analytic function attains its maximum modulus on the boundary of its domain. This result has many applications in complex analysis, including the proof of the fundamental theorem of algebra, which states that every non-constant polynomial with complex coefficients has at least one complex root.

In summary, the theory of analytic functions of one variable is a rich subject with many important results and applications. Its local and global aspects provide powerful tools for understanding the behavior of complex functions, and it has many connections to other areas of mathematics and physics.

For $z=1+\imath$ find $w$ such that the real parts of the following numbers are equal to zero
a) $z+w_i$
b) $z \cdot w$
c) $\frac{z}{w} ;$
d) $\frac{w}{z}$.

Solution. Let $z=1+\imath=(1,1)$ and $w=x+i y=(x, y)$. Then we have
a)
$$\operatorname{Re}(z+w)=0 \Longleftrightarrow 1+x=0$$
Hence $x=-1$ and $y \in \mathbb{R}$ is arbitrary.
b)
$$\operatorname{Re}(z \cdot w)=0 \Longleftrightarrow \operatorname{Re}(x-y+i(x+y))=0 .$$
Hence $x-y=0$. Therefore $w$ is given by $w=x+\imath x=x(1+\imath)=x \cdot z$, for arbitrary $x \in \mathbb{R}$
c) We have $\frac{z}{w}=0$. Hence
$$\frac{z}{w}=\frac{1+\imath}{x+2 y} \cdot \frac{x-v y}{x-v y}=\frac{x+y+\imath(x-y)}{x^2+y^2}$$
Therefore
$$\operatorname{Re}\left(\frac{z}{w}\right)=0 \Longleftrightarrow \frac{x+y}{x^2+y^2}=0 \Rightarrow x=-y, x \neq 0 .$$
Finally we have $w=x(1-2)$ for $x \in \mathbb{R}$ and $x \neq 0$.
d) From $\operatorname{Re}\left(\frac{w}{z}\right)=0$ it follows that for every $x \in \mathbb{R}$ :
$$\frac{w}{z}=\frac{x+\imath y}{1+\imath} \cdot \frac{1-\imath}{1-\imath}=\frac{x+y+\imath(x-y)}{2} .$$
Hence $w=x(1-\imath)$ for every $x \in \mathbb{R}$.

Let $\mathbb{C}^$ be the set of all complex numbers different from zero. a) Prove that the set $T$ of all complex numbers with modulus 1 is a multiplicative subgroup of the group $\left(\mathbb{C}^, \cdot\right)$.
b) The multiplicative group $\mathbb{C}^*$ is isomorphic with $\mathbb{R}^{+} \times T$.

Solution. a) We have $T={z|| z \mid=1} \subset \mathbb{C}^* . T$ under the multiplication by the equality
$$\left|z_1 \cdot z_2\right|=\left|z_1\right| \cdot\left|z_2\right| \text {. }$$
The associativity follows by the associativity of multiplication in $\mathbb{C}$. The neutral element is 1. The inverse of $z=(a, b) \in T$ is
$$\left(\frac{a}{a^2+b^2}, \frac{-b}{a^2+b^2}\right) \in T \text {. }$$
b) It is easy to check that an isomorphism is given by
$$f(z)=(|z|, \cos \theta+i \sin \theta)$$
where $\theta=\arg z$. nopagebreak c) We define an equivalence relation $\sim$ on $\mathbb{R}$ by $r_1 \sim r_2 \Longleftrightarrow r_1-r_2=2 k \pi, \quad k$ is an integer. Let $\mathbb{R}$ be the corresponding quotient set. Prove that $\mathbb{R}^{+} \times T$ isomorphic with $\mathbb{R}^{+} \times \hat{\mathbb{R}}$. The group $\mathbb{C}^*$ is isomorphic with $\mathbb{R}^{+} \times \tilde{\mathbb{R}}$ (check). Therefore by b) and transitivity of the isomorphisms of groups it follows c).

Starting with a complex number $z \neq 0$, find where the complex numbers $2 z, 3 z, \ldots, n z$ are ?

Solution. For $z=\rho(\cos \theta+\imath \sin \theta)$ we have
$$n z=n \rho(\cos \theta+i \sin \theta)$$
the complex numbers $n x$, for $n=1,2, \ldots$ are on the half straight line $y=\tan \theta \cdot x$ with modulus $n \rho$.

# 线性代数代写 Real Analysis|MATH 8810 Boston College Assignment

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Assignment-daixieTM为您提供波士顿学院Boston College MATH 8810 Real Analysis 线性代数代写代考辅导服务！

## Instructions:

The basic idea of Fourier theory is that any periodic function can be expressed as a sum of sine and cosine functions of different frequencies. These frequencies are called harmonics, and the sum of all harmonics is called the Fourier series.

The Fourier series representation of a periodic function can be used to analyze its properties, such as its frequency content and its response to different inputs. It is also used to design filters and other signal processing techniques.

In addition to the Fourier series, there is also the Fourier transform, which is a mathematical operation that transforms a function from the time domain to the frequency domain. The Fourier transform is widely used in signal processing and image processing to analyze and manipulate signals and images.

Overall, Fourier theory is a powerful tool for understanding and analyzing periodic phenomena, and it has a wide range of applications in many fields.

Let $A$ be a subset of $\mathbb{R}$ which is bounded above, and let $a_0$ be an upper bound for $A$. Prove that $a_0=\sup A$ if and only if for every $\epsilon>0$, there exists $a \in A$ such that $a_0-\epsilon<a$.

To prove that $a_0=\sup A$ if and only if for every $\epsilon>0$, there exists $a \in A$ such that $a_0-\epsilon<a$, we need to prove both directions of the statement:

$(\Rightarrow)$ Assume $a_0 = \sup A$. Let $\epsilon>0$ be given. Since $a_0$ is the least upper bound of $A$, there exists $a\in A$ such that $a_0-\epsilon<a\leq a_0$. This is because if there were no such $a$, then $a_0-\epsilon$ would be a smaller upper bound for $A$ than $a_0$, contradicting the assumption that $a_0=\sup A$.

$(\Leftarrow)$ Assume that for every $\epsilon>0$, there exists $a\in A$ such that $a_0-\epsilon<a$. We need to show that $a_0=\sup A$. Suppose, for the sake of contradiction, that $a_0$ is not the least upper bound of $A$. Then there exists $b>a_0$ such that $b$ is an upper bound for $A$. Let $\epsilon = b-a_0>0$. Since $a_0$ is not the least upper bound of $A$, there exists $a\in A$ such that $a>b$. But then $a_0< a< b = a_0 + \epsilon$, which contradicts the assumption that for every $\epsilon>0$, there exists $a\in A$ such that $a_0-\epsilon<a$. Therefore, $a_0$ is the least upper bound of $A$, i.e., $a_0=\sup A$.

Hence, we have proven that $a_0=\sup A$ if and only if for every $\epsilon>0$, there exists $a \in A$ such that $a_0-\epsilon<a$.

Prove that
$$\lim _{n \rightarrow \infty} \frac{1}{20 n^2+20 n+2020}=0 .$$

We can prove this limit using the squeeze theorem. First, note that for any positive $₫ \mathcal{P}$ integer \$n \$, we have:
$$20 n^2+20 n+2020 \geq 20 n^2$$
This is because \$20 n+2020 \$ is always positive for positive \$n \$, so adding it to \$20 n^{\wedge} 2 \$only increases the value. Using this inequality, we can write:
$$0 \leq \frac{1}{20 n^2+20 n+2020} \leq \frac{1}{20 n^2}$$
Now, taking the limit as n approaches infinity of both sides of this inequality, we get:
$$\lim {n \rightarrow \infty} 0 \leq \lim {n \rightarrow \infty} \frac{1}{20 n^2+20 n+2020} \leq \lim {n \rightarrow \infty} \frac{1}{20 n^2}$$ The left-hand side simplifies to 0 , and the right-hand side simplifies to 0 as well, since the denominator grows faster than the numerator as $\$ \mathrm{n} \$$approaches infinity. Therefore, by the squeeze theorem, we have:$$ \lim {n \rightarrow \infty} \frac{1}{20 n^2+20 n+2020}=0
$$as desired. 问题 3. Let \left{x_n\right} be a bounded sequence of real numbers. Prove$$
\lim {n \rightarrow \infty} x_n=0, $$if and only if$$ \limsup {n \rightarrow \infty}\left|x_n\right|=0 .
证明 . First, we will prove that if \lim_{n \rightarrow \infty} x_n = 0, then \limsup_{n \rightarrow \infty} |x_n| = 0. Suppose \lim_{n \rightarrow \infty} x_n = 0. Let \epsilon > 0 be arbitrary. Since \lim_{n \rightarrow \infty} x_n = 0, there exists N such that for all n \geq N, we have |x_n| < \epsilon. Thus, for any k \in \mathbb{N}, there are only finitely many n such that |x_n| \geq \frac{1}{k}. Therefore, \limsup_{n \rightarrow \infty} |x_n| = 0. Next, we will prove that if \limsup_{n \rightarrow \infty} |x_n| = 0, then \lim_{n \rightarrow \infty} x_n = 0. Suppose \limsup_{n \rightarrow \infty} |x_n| = 0. Then, for any \epsilon > 0, there exists N such that for all n \geq N, we have |x_n| < \epsilon. Since {x_n} is bounded, there exists a positive number M such that |x_n| \leq M for all n. Therefore, for any \epsilon > 0, we have \left|x_n\right| \leq M<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon for all n \geq N. This implies that \lim_{n \rightarrow \infty} x_n = 0. Thus, we have shown both directions of the equivalence and conclude that \lim {n \rightarrow \infty} x_n=0 \quad if and only if \quad \limsup {n \rightarrow \infty}\left|x_n\right|=0. 这是一份2023年的波士顿学院Boston College MATH 8810 线性代数代写的成功案例 # 几何学/拓扑学代写 Geometry/Topology II|MATH 8809 Boston College Assignment 0 Assignment-daixieTM为您提供波士顿学院Boston College MATH 8809 Geometry/Topology II几何学/拓扑学代写代考辅导服务！ ## Instructions: Geometry is a branch of mathematics that deals with the study of shapes, sizes, positions, and properties of objects in space. It encompasses a wide range of topics, including points, lines, planes, angles, polygons, circles, spheres, and more. Geometry has practical applications in many fields, including engineering, architecture, art, astronomy, physics, and even computer graphics. It plays an important role in understanding the relationships between objects in the physical world and is used to solve real-world problems. The study of geometry involves using logic and reasoning to make deductions and proofs about the properties of shapes and objects. The principles of geometry have been studied and applied for thousands of years, and continue to be relevant today in a variety of fields. 几何学/拓扑学代写 Geometry/Topology II|MATH 8809 Boston College Assignment 问题 1. Show that S^4 has no symplectic structure. Show that S^2 \times S^4 has no symplectic structure. 证明 . To show that S^4 has no symplectic structure, we can use the following theorem: Theorem: Any closed oriented manifold M with H^2(M;\mathbb{Z})=0 has no symplectic structure. Proof: Suppose M has a symplectic structure. Then by the non-degeneracy of the symplectic form, we have H^2(M;\mathbb{R})\cong H^2_{dR}(M;\mathbb{R})\neq 0. But by Poincaré duality and the Universal Coefficient Theorem, we have H^2(M;\mathbb{R})\cong H^2(M;\mathbb{Z})\otimes\mathbb{R}\cong \operatorname{Hom}(H_2(M;\mathbb{Z}),\mathbb{R})\cong H^2(M;\mathbb{Z}) since M is closed and oriented. Therefore, H^2(M;\mathbb{Z})\neq 0, which is a contradiction. Hence, M has no symplectic structure. Now, H^2(S^4;\mathbb{Z})=0 since S^4 is simply connected and H^2(S^4;\mathbb{Z}) is the abelianization of \pi_1(S^4), which is trivial. Therefore, by the above theorem, S^4 has no symplectic structure. To show that S^2 \times S^4 has no symplectic structure, we can use the following lemma: Lemma: If M and N are symplectic manifolds of dimensions 2m and 2n, respectively, then M\times N has a symplectic structure of dimension 2(m+n). Proof: Let \omega_M and \omega_N be symplectic forms on M and N, respectively. Then the product form \omega_M\oplus \omega_N is a symplectic form on M\times N of dimension 2(m+n). Since S^4 has no symplectic structure, it follows that S^2 \times S^4 has no symplectic structure. 问题 2. Describe Hamiltonian flow in the symplectic manifold T^* M by the Hamiltonian H=\pi^* f, where \pi: T^* M \longrightarrow M is the natural projection and f \in C^{\infty}(M). Also, show that a coordinate chart U \subset M determines a system of n independent, commuting Hamiltonians on T^* U \subset T^* M. 证明 . In the symplectic manifold T^*M, a Hamiltonian flow is a flow generated by a Hamiltonian function H:T^*M \rightarrow \mathbb{R} via the Hamiltonian vector field X_H defined as: X_H=\sum_{i=1}^n\left(\frac{\partial H}{\partial p_i} \frac{\partial}{\partial q_i}-\frac{\partial H}{\partial q_i} \frac{\partial}{\partial p_i}\right) where (q_1,\ldots,q_n,p_1,\ldots,p_n) are local coordinates on T^*M. Given a function f \in C^{\infty}(M), we can define a Hamiltonian function on T^*M by pulling back f via the natural projection \pi:T^*M\rightarrow M, i.e., H = \pi^*f. The Hamiltonian vector field generated by H is given by: X_H=\sum_{i=1}^n\left(\frac{\partial f}{\partial q_i} \frac{\partial}{\partial p_i}-\frac{\partial f}{\partial p_i} \frac{\partial}{\partial q_i}\right) which generates the flow: \frac{d q_i}{d t}=\frac{\partial H}{\partial p_i}=\frac{\partial f}{\partial p_i} \quad \frac{d p_i}{d t}=-\frac{\partial H}{\partial q_i}=-\frac{\partial f}{\partial q_i} This is Hamilton’s equations of motion. Now, consider a coordinate chart U \subset M. Let (q_1,\ldots,q_n) be local coordinates on U. We can then define a local coordinate chart on T^*U by taking (q_1,\ldots,q_n,p_1,\ldots,p_n) as coordinates on T^*U. Since \pi|{T^*U}:T^*U\rightarrow U is a diffeomorphism, we can pull back functions from U to T^*U via the inverse of this map. In particular, we can define Hamiltonian functions on T^U by pulling back functions from U, i.e., for f\in C^{\infty}(U), we can define H_f=\pi^|{T^*U} f:T^*U\rightarrow \mathbb{R}. The Hamiltonian vector field generated by H_f is given by: X_{H_f}=\sum_{i=1}^n\left(\frac{\partial f}{\partial q_i} \frac{\partial}{\partial p_i}-\frac{\partial f}{\partial p_i} \frac{\partial}{\partial q_i}\right) which is the same as the Hamiltonian vector field generated by \pi^*f on T^*M, but restricted to T^*U. Note that the restriction of a Hamiltonian vector field to a submanifold is still a Hamiltonian vector field with the same Hamiltonian. 问题 3. Let \rho \in \Omega^{\bullet}(G) be a left-invariant form on a Lie group G. Show that d \rho=0 if \rho is also right-invariant. 证明 . To show that d \rho = 0 for a left-invariant and right-invariant form \rho, we will use the Cartan’s magic formula: \mathcal{L}_X=d \circ i_X+i_X \circ d, where \mathcal{L}_X is the Lie derivative with respect to the vector field X, i_X is the interior product with X, and d is the exterior derivative. Since \rho is left-invariant, we have that \mathcal{L}_X \rho = 0 for all left-invariant vector fields X. Similarly, since \rho is right-invariant, we have that \mathcal{L}_Y \rho = 0 for all right-invariant vector fields Y. Let X be a left-invariant vector field, and Y be a right-invariant vector field. Then, we have: \begin{align*} d \rho(X,Y) &= \mathcal{L}_X \rho(Y) – \rho(\mathcal{L}_X Y) &&\text{(Cartan’s magic formula)}\ &= \mathcal{L}_X \rho(Y) &&\text{(since Y is right-invariant)}\ &= 0 &&\text{(since \rho is left-invariant)} \end{align*} Similarly, we can show that d \rho(X,Y) = 0 when X is right-invariant and Y is left-invariant. Since any vector field on G can be written as a linear combination of left-invariant and right-invariant vector fields, we conclude that d \rho = 0 for any left-invariant and right-invariant form \rho on G. 这是一份2023年的波士顿学院Boston College MATH 8809几何学/拓扑学代写的成功案例 # 代数代写 Algebra II|MATH 8807 Boston College Assignment 0 Assignment-daixieTM为您提供波士顿学院Boston College MATH 8807 Algebra II 代数学代写代考辅导服务！ ## Instructions: Algebra is a branch of mathematics that deals with the study of mathematical symbols and the rules for manipulating these symbols. In algebra, letters and symbols are used to represent unknown quantities, and mathematical operations such as addition, subtraction, multiplication, division, and exponentiation are used to solve equations and find solutions to problems. Algebra is used extensively in many areas of mathematics, science, engineering, and economics, among other fields. It is essential in solving problems related to geometry, calculus, and statistics, and is also used in everyday life, such as calculating mortgage payments or determining the best price for a product. Algebraic expressions can take many forms, such as linear equations, quadratic equations, polynomial equations, and systems of equations. These expressions can be solved using various methods, including factoring, completing the square, and the quadratic formula. Some fundamental concepts in algebra include variables, coefficients, constants, and equations. Variables are letters or symbols that represent unknown quantities, while coefficients are numbers that multiply the variables in an equation. Constants are fixed numbers, and equations are mathematical expressions that use an equal sign to show that two expressions are equal. 代数代写 Algebra II|MATH 8807 Boston College Assignment 问题 1. Consider the random walk of a man on the integer line \mathbb{Z} such that, at each step, that the probability to go from position i to position i+1 is p, and the probability to go from i to i-1 is 1-p. The man “falls off the cliff” if he reaches the position 0. Suppose that the man starts at the initial position i_0 \geq 1. Find the probability that he falls off the cliff. 证明 . Let P_i be the probability that the man falls off the cliff starting from position i. We want to find P_{i_0} given that i_0 \geq 1. Note that P_0 = 1 since the man has already fallen off the cliff when he starts at position 0. Also note that P_i = 0 for all i \leq 0 since the man has already fallen off the cliff. We can now set up a recurrence relation for P_i. If the man is currently at position i, then he has two options for his next step: he can either move to position i+1 with probability p, or he can move to position i-1 with probability 1-p. Thus, we haveP_i = p P_{i+1} + (1-p) P_{i-1}$$for all i \geq 1. This is a second-order linear recurrence relation with constant coefficients. We can solve it using the characteristic equation:$$r^2 – (1-p)r – p = 0.$$The roots of this equation are$$r_1 = \frac{1-p + \sqrt{(1-p)^2 + 4p}}{2} \quad \text{and} \quad r_2 = \frac{1-p – \sqrt{(1-p)^2 + 4p}}{2}.$$Note that r_1 > 1 and r_2 < 0. Therefore, the general solution to the recurrence relation is$$P_i = A r_1^i + B r_2^ifor some constants A and B. We can find A and B by using the boundary conditions P_0 = 1 and P_i = 0 for all i \leq 0. This gives us the system of equations \begin{align*} A + B &= 1, \ Ar_1^i + Br_2^i &= 0 \quad \text{for all } i \leq 0. \end{align*} Solving for A and B, we getA = \frac{r_2}{r_2 – r_1} \quad \text{and} \quad B = -\frac{r_1}{r_2 – r_1}.$$Therefore, the probability that the man falls off the cliff starting from position i_0 \geq 1 is$$P_{i_0} = \frac{r_2^{i_0}}{r_2^{i_0} – r_1^{i_0}}.Note that this expression makes sense since r_1 > 1 and r_2 < 0, so r_1^{i_0} grows exponentially while r_2^{i_0} decays exponentially as i_0 gets large. This means that P_{i_0} approaches 1 as i_0 increases, which makes sense since the man is more likely to fall off the cliff the further he is from it. 问题 2. For 1 \leq k \leq n / 2, find a bijection f between k-element subsets of \{1, \ldots, n\} and (n-k)-element subsets of \{1, \ldots, n\} such that f(I) \supseteq I, for any k-element subset I. 证明 . Let S be a set of size n and let k be an integer such that 1 \leq k \leq n/2. We will define a bijection f between the set of k-element subsets of S and the set of (n-k)-element subsets of S that satisfies the given property. Let I be a k-element subset of S. We will define f(I) as follows: first, let J = S \setminus I be the complement of I in S. Since |I| = k, we have |J| = n-k. Next, let T be the k-element subset of J that contains the k smallest elements of J (with respect to some fixed total ordering of S). Note that T is well-defined since |J| = n-k \geq k. Finally, we define f(I) = T \cup I. We claim that f(I) is an (n-k)-element subset of S, and that f is a bijection between the set of k-element subsets of S and the set of (n-k)-element subsets of S. To see that f(I) is an (n-k)-element subset of S, note that |f(I)| = |T \cup I| = |T| + |I| – |T \cap I|, and |T| = k, |I| = k, and |T \cap I| = 0 (since T and I are disjoint). Therefore, |f(I)| = 2k \leq 2(n/2) = n, so f(I) is a subset of S with at most n elements. Moreover, |f(I)| = n-k since |T| = k and |J| = n-k, so f(I) is indeed an (n-k)-element subset of S. To see that f is a bijection, we will define its inverse. Let A be an (n-k)-element subset of S, and let J = S \setminus A. Since |A| = n-k, we have |J| = k. Let T be the k-element subset of J that contains the k smallest elements of J (with respect to the same fixed total ordering of S as before). Note that T is well-defined since |J| = k \geq k. Finally, we define f^{-1}(A) = T \cup A. We claim that f 问题 3. Prove that the number of set-partitions \pi of the set [n]:=\{1, \ldots, n\} such that, for any i=1, \ldots, n-1, the consecutive numbers i and i+1 do not belong to the same block of \pi equals the number of set-partitions of the set [n-1]. 证明 . To prove the given statement, let us first denote by a_n the number of set-partitions of [n] such that no consecutive numbers belong to the same block, and by b_n the number of set-partitions of [n-1]. Now, let us consider a set-partition \pi of [n] such that no consecutive numbers belong to the same block. We will show that we can obtain a set-partition \pi’ of [n-1] in a one-to-one fashion by removing the element n from \pi and merging the block containing n-1 with the block containing n. Formally, let B_n be the block of \pi containing n, and let B_{n-1} be the block of \pi containing n-1. Since n and n-1 do not belong to the same block, we have B_n \neq B_{n-1}. We can define a new set-partition \pi’ of [n-1] by setting \pi'(i) = \pi(i) for i=1,\ldots,n-2, and by merging B_n and B_{n-1} into a single block in \pi’. Conversely, given a set-partition \pi’ of [n-1], we can obtain a set-partition \pi of [n] such that no consecutive numbers belong to the same block by adding the element n to \pi’ as a new block. Formally, we can define a new set-partition \pi of [n] by setting \pi(i) = \pi'(i) for i=1,\ldots,n-1, and by setting \pi(n) = {n}. It is easy to see that these constructions are one-to-one, so the number of set-partitions of [n] such that no consecutive numbers belong to the same block is equal to the number of set-partitions of [n-1]. Thus, we have a_n = b_{n-1}, as required. 这是一份2023年的波士顿学院Boston College MATH 8807代数代写的成功案例 # 代数代写 Algebra I |MATH 8806 Boston College Assignment 0 Assignment-daixieTM为您提供波士顿学院Boston College MATH 8806 Algebra I 代数学代写代考辅导服务！ ## Instructions: Algebra is a branch of mathematics that deals with symbols and the rules for manipulating these symbols to solve equations and understand relationships between quantities. It involves the use of letters and symbols to represent unknown values, and the use of mathematical operations such as addition, subtraction, multiplication, and division to solve equations and simplify expressions. Algebra is important in many areas of mathematics, science, engineering, economics, and finance, as it provides a powerful tool for solving problems and understanding relationships between variables. Some of the key concepts in algebra include equations, inequalities, polynomials, functions, and matrices. Algebraic techniques can be used to solve a wide range of problems, from simple arithmetic calculations to complex systems of equations and models of physical and economic phenomena. Algebra is also used extensively in computer science, cryptography, and other fields that require efficient methods for processing and manipulating large amounts of data. 代数代写 Algebra I |MATH 8806 Boston College Assignment 问题 1. Let G be a group and let H be a subgroup. Prove that the following are equivalent. (1) H is normal in G. (2) For every g \in G, g H^{-1}=H. 证明 . (1) \Rightarrow (2): If H is normal in G, then for every g \in G and h \in H, we have ghg^{-1} \in H. Thus, for any g \in G, we have gHg^{-1} \subseteq H. On the other hand, Hg^{-1}g \subseteq g^{-1}Hg, so gH^{-1}=Hg^{-1}g \subseteq H. Hence, gH^{-1}=H. (2) \Rightarrow (3): If gH^{-1}=H for every g \in G, then for any a \in G and h \in H, we have ah=ag(g^{-1}h) \in aH. Therefore, aH \subseteq Ha, and the reverse inclusion follows similarly. Thus, aH=Ha. 问题 2. (3) For every a \in G, a H=H a. (4) The set of left cosets is equal to the set of right cosets. 证明 . (3) \Rightarrow (4): If aH=Ha for every a \in G, then the set of left cosets is {aH \mid a \in G}, and the set of right cosets is {Ha \mid a \in G}. For any a,b \in G, we have aH=Ha and bH=Hb, so aHb=abH=(Ha)b=H(ab)=Hba=bHa. Hence, the sets of left and right cosets coincide. (4) \Rightarrow (1): If the set of left cosets is equal to the set of right cosets, then for any g \in G and h \in H, there exists a \in G such that ah=g. Thus, g^{-1}ag \in H and g^{-1}hg \in H, so ghg^{-1} \in H for any g \in G and h \in H. Therefore, H is normal in G. 问题 3. Let G be a group and let N be a normal subgroup. Prove that G / N is abelian iff N contains the commutator of every pair of elements of G. 证明 . (\Rightarrow) Assume that G/N is abelian. Let x,y\in G be arbitrary elements. Then we have (xN)(yN)=(yN)(xN) in G/N, which means that xyN=yxN. Therefore, we have xyx^{-1}y^{-1}\in N. Since x and y were arbitrary, this means that N contains the commutator [x,y]=xyx^{-1}y^{-1} for every pair of elements x,y\in G. (\Leftarrow) Assume that N contains the commutator of every pair of elements of G. Let xN,yN be arbitrary elements of G/N. We need to show that (xN)(yN)=(yN)(xN), i.e., xyN=yxN. Since N is normal in G, we have xNx^{-1}=N and yNy^{-1}=N. Therefore, we have \begin{align*} xyN &= xNyN && \text{(since }yN=NyN\text{)} \ &= xNyNy^{-1} && \text{(since }y^{-1}Ny=N\text{)} \ &= xNNy^{-1} && \text{(since }N\text{ is normal)} \ &= xN[y^{-1},x]N && \text{(since }[y^{-1},x]\in N\text{)} \ &= y^{-1}xN[x,y]N && \text{(since }N\text{ is normal)} \ &= y^{-1}NxN[y,x] && \text{(since }[x,y]=[y^{-1},x]^{-1}\in N\text{)} \ &= yNy^{-1}xN && \text{(since }N\text{ is normal)} \ &= yN[x^{-1},y]N && \text{(since }[x^{-1},y]\in N\text{)} \ &= x^{-1}yN[x,y]N && \text{(since }N\text{ is normal)} \ &= x^{-1}NxN[y,x] && \text{(since }[x,y]=[x^{-1},y]^{-1}\in N\text{)} \ &= xNx^{-1}yN && \text{(since }N\text{ is normal)} \ &= yxN && \text{(since }xNx^{-1}=N\text{)} \ &= yNxN && \text{(since }xN=NxN\text{)} \ &= yN(xN) && \text{(since }N\text{ is normal)} \ &= yN(xN), \end{align*} which shows that G/N is abelian. 这是一份2023年的波士顿学院Boston College MATH 8806代数代写的成功案例 # 代数代写 Algebra|MATH 100 University of Washington Assignment 0 Assignment-daixieTM为您提供华盛顿大学University of WashingtonMATH 100 Algebra代数学代写代考辅导服务！ ## Instructions: Algebra 1 is the second math course in high school and will guide you through among other things expressions, systems of equations, functions, real numbers, inequalities, exponents, polynomials, radical and rational expressions. This Algebra 1 math course is divided into 12 chapters and each chapter is divided into several lessons. Under each lesson you will find theory, examples and video lessons. Mathplanet hopes that you will enjoy studying Algebra 1 online with us! 代数代写 Algebra |MATH 100 University of Washington Assignment 问题 1. Section 1.2. Problem 23: The figure shows that \cos (\alpha)=v_1 /|v| and \sin (\alpha)= v_2 /|v|. Similarly \cos (\beta) ___is and \sin (\beta) is __. The angle \theta is \beta-\alpha. Substitute into the trigonometry formula \cos (\alpha) \cos \overline{(\beta)+\sin }(\beta) \sin (\alpha) for \cos (\beta-\alpha) to find \cos (\theta)=v \cdot w /|v||w|. 证明 . First blank: w_1 /|w|. Second blank: w_2 /|w|. Substituting into the trigonometry formula yields
\cos (\beta-\alpha)=\left(w_1 /|w|\right)\left(v_1 /|v|\right)+\left(w_2 /|w|\right)\left(v_2 /|v|\right)=v \cdot w /|v||w| .
$$问题 2. If A=L D U and also A=L_1 D_1 U_1 with all factors invertible, then L=L_1 and D=D_1 and U=U_1. “The three factors are unique.” Derive the equation L_1^{-1} L D=D_1 U_1 U^{-1}. Are the two sides triangular or diagonal? Deduce L=L_1 and U=U_1 (they all have diagonal 1’s). Then D=D_1. 证明 . Notice that L D U=L_1 D_1 U_1. Multiply on the left by L_1^{-1} and on the right by U^{-1}, getting$$
L_1^{-1} L D U U^{-1}=L_1^{-1} L_1 D_1 U_1 U^{-1} .
$$But U U^{-1}=I and L_1^{-1} L_1=I. Thus L_1^{-1} L D=D_1 U_1 U^{-1}, as desired. The left side L_1^{-1} L D is lower triangular, and the right side D_1 U_1 U^{-1} is upper triangular. But they’re equal. So they’re both diagonal. Hence L_1^{-1} L and U_1 U^{-1} are diagonal too. But they have diagonal 1’s. So they’re both equal to I. Thus L=L_1 and U=U_1. Also L_1^{-1} L D=D_1 U_1 U^{-1}. Thus D=D_1. 问题 3. Suppose \mathbf{S} and \mathbf{T} are two subspaces of a vector space \mathbf{V}. (a) Definition: The sum \mathbf{S}+\mathbf{T} contains all sums \mathbf{s}+\mathbf{t} of a vector \mathbf{s} in \mathbf{S} and a vector \mathbf{t} in T. Show that \mathbf{S}+\mathbf{T} satisfies the requirements (addition and scalar multiplication) for a vector space. (b) If \mathbf{S} and \mathbf{T} are lines in \mathbf{R}^m, what is the difference between \mathbf{S}+\mathbf{T} and \mathbf{S} \cup \mathbf{T} ? That union contains all vectors from \mathbf{S} and \mathbf{T} or both. Explain this statement: The span of \mathbf{S} \cup \mathbf{T} is \mathbf{S}+\mathbf{T}. (Section 3.5 returns to this word “span.”) 证明 . (a) Let \mathbf{s}, \mathbf{s}^{\prime} be vectors in \mathbf{S}, let \mathbf{t}, \mathbf{t}^{\prime} be vectors in \mathbf{T}, and let c be a scalar. Then$$
(\mathbf{s}+\mathbf{t})+\left(\mathbf{s}^{\prime}+\mathbf{t}^{\prime}\right)=\left(\mathbf{s}+\mathbf{s}^{\prime}\right)+\left(\mathbf{t}+\mathbf{t}^{\prime}\right) \text { and } c(\mathbf{s}+\mathbf{t})=c \mathbf{s}+c \mathbf{t} .
$$Thus \mathbf{S}+\mathbf{T} is closed under addition and scalar multiplication; in other words, it satisfies the two requirements for a vector space. (b) If \mathbf{S} and \mathbf{T} are distinct lines, then \mathbf{S}+\mathbf{T} is a plane, whereas \mathbf{S} \cup \mathbf{T} is not even closed under addition. The span of \mathbf{S} \cup \mathbf{T} is the set of all combinations of vectors in this union. In particular, it contains all sums \mathbf{s}+\mathbf{t} of a vector \mathbf{s} in \mathbf{S} and a vector \mathbf{t} in \mathbf{T}, and these sums form \mathbf{S}+\mathbf{T}. On the other hand, \mathbf{S}+\mathbf{T} contains both \mathbf{S} and \mathbf{T}; so it contains \mathbf{S} \cup \mathbf{T}. Further, \mathbf{S}+\mathbf{T} is a vector space. So it contains all combinations of vectors in itself; in particular, it contains the span of \mathbf{S} \cup \mathbf{T}. Thus the span of \mathbf{S} \cup \mathbf{T} is \mathbf{S}+\mathbf{T}. 这是一份2023年的华盛顿大学University of Washington MATH 100代数代写的成功案例 # 数论代写 Number Theory|MATH 305S Duke University Assignment 0 Assignment-daixieTM为您提供杜克大学Duke UniversityMATH 305S Number Theory数论学代写代考辅导服务！ ## Instructions: Divisibility properties of integers, prime numbers, congruences, quadratic reciprocity, number-theoretic functions, simple continued fractions, rational approximations; contributions of Fermat, Euler, and Gauss. Prerequisite: Mathematics 122, 112L, 122L, or consent of instructor. Individual research paper required. 数论代写 Number Theory|MATH 305S Duke University Assignment 问题 1. Let a>0 and b be integers. Show that there is an integer k such that b+k a>0. (Hint: use well-ordering.) 证明 . Suppose not. Then let S be the set of integers {-(b+k a): k \in \mathbb{Z}}, so by hypothesis S consists entirely of nonnegative integers. By the Well-Ordering Principle, it has a smallest positive element, say, b+k a. But then b+(k-1) a is smaller since a>0, contradiction. 问题 2. Let a and b be positive integers whose ged is 1 . Find the largest positive integer n(a, b) which is not a non-negative integer linear combination of a and b. Prove your answer (i.e. show that n(a, b) cannot be represented as a x+b y with x, y \in \mathbb{N} \cup{0} and that every greater integer can be represented in such a way). 证明 . The largest such integer is a b-a-b. To see it’s not a nonnegative integer linear combination, suppose a b-a-b=a x+b y with x, y \in \mathbb{Z}_{\geq 0}. Then a(b-1-x)=b(y+1). And since (a, b)=1 we have a \mid y+1 (and b \mid b-1-x ). This forces y \geq a-1 because y+1 \geq 1. So$$
a x+b y \geq a \cdot 0+b(a-1)=a b-b>a b-a-b
$$contradicting a b-a-b=a x+b y. On the other hand, suppose n>a b-a-b. Since \operatorname{gcd}(a, b)=1 we can write n=a x+b y with x, y \in \mathbb{Z} (not necessarily nonnegative). Now note that n=a(x-b k)+b(y+a k) for any integer k. By the division algorithm, there exists an integer k such that 0 \leq x-b k<b. Let x^{\prime}=x-b k and y^{\prime}=y+a k. Then we have n=a x^{\prime}+b y^{\prime} with 0 \leq x^{\prime} \leq b-1, so$$
b y^{\prime}=n-a x^{\prime} \geq(a b-a-b+1)-a(b-1)=-(b-1) .
$$Therefore y^{\prime} \geq \frac{-(b-1)}{b}, and since y^{\prime} is an integer, we get y^{\prime} \geq 0. This shows that n=a x^{\prime}+b y^{\prime} is a nonnegative integer linear combination. 问题 3. Let a>1 be a positive integer, and m, n be natural numbers. Show that a^m-1 \mid a^n-1 if and only if m \mid n. Show that the ged of a^m-1 and a^n-1 is a^{(m, n)}-1. 证明 . One direction is clear: if m \mid n then n=m k for some positive integer k, and$$
a^n-1=a^{m k}-1=\left(a^m-1\right)\left(a^{m(k-1)}+a^{m(k-2)}+\cdots+a^m+1\right)
$$is divisible by a^m-1. Now if m \nmid n, we write n=m k+r with 0<r<m. Then$$
a^n-1=a^{m k+r}-1=a^{m k+r}-a^r+a^r-1=a^r\left(a^{m k}-1\right)+a^r-1

Now $a^m-1$ divides $a^{m k}-1$ but it doesn’t divide $a^r-1$, since $0<a^r-1<a^m-1$. So $a^m-1$ can’t divide $a^n-1$