# 微积分与解析几何代写 Calculus With Analytic Geometry I|MATH 124 University of Washington Assignment

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Let $W$ be a function which assigns to every function $F$ and any interval $[a, b]$ over which $F$ is integrable a real number $W(F, a, b)$ such that (5.1), (5.2), and (5.3) hold. Then
$$W(F, a, b)=\int_a^b F(x) d x$$

Proof. Let $F$ be a function, and $[a, b]$ an interval over which $F$ is integrable. We shall first show that, for every partition $\sigma$ of $[a, b]$, the upper and lower sums, $U_\sigma$ and $L_\sigma$, satisfy the inequalities
$$L_\sigma \leq W(F, a, b) \leq U_\sigma$$
To do this, we let $\sigma=\left{x_0, \ldots, x_n\right}$ and assume the usual ordering:
$$a=x_0 \leq x_1 \leq \cdots \leq x_n=b .$$
As we have done in the past, we denote by $M_i$ the least upper bound of the values of $F$ on the ith subinterval $\left[x_{i-1}, x_i\right]$, and by $m_i$ the greatest lower bound. Then
$$m_i \leq F(x) \leq M_i, \quad \text { whenever } x_{i-1} \leq x \leq x_i$$
The two constant functions with values Mi and mi, respectively, are certainly integrable over the subinterval $\left[x_{i-1}, x_i\right]$. Following the common practice of denoting a constant function by its value, we know, as a result of (5.2), that
$$W\left(m_i, x_{i-1}, x_i\right) \leq W\left(F, x_{i-1}, x_i\right) \leq W\left(M_i, x_{i-1}, x_i\right)$$
Using (5.3), we obtain
\begin{aligned} & W\left(m_i, x_{i-1}, x_i\right)=m_i\left(x_i-x_{i-1}\right) \ & W\left(M_i, x_{i-1}, x_i\right)=M_i\left(x_i-x_{i-1}\right) \end{aligned}
Hence
$$m_i\left(x_i-x_{i-1}\right) \leq W\left(F, x_{i-1}, x_i\right) \leq M_i\left(x_i-x_{i-1}\right) .$$
Adding these inequalities for $i=1, \ldots, n$, we get
$$\sum_{i=1}^n m_i\left(x_i-x_{i-1}\right) \leq \sum_{i=1}^n W\left(F, x_{i-1}, x_i\right) \leq \sum_{i=1}^n M_i\left(x_i-x_{i-1}\right) .$$
The left and right sides of the inequalities in the preceding equation are precisely $L_\sigma$ and $U_\sigma$, respectively. It follows from repeated use that
$$\sum_{i=1}^n W\left(F, x_{i-1}, x_i\right)=W(F, a, b)$$
and we have therefore proved that the inequalities do hold.
now essentially complete. Let $\sigma$ and $\tau$ be two arbitrary partitions of $[a, b]$. The union $U \cup T$ is the partition which is the common refinement of both. It is shown in the last line of the proof of Proposition (1.1), page 168, that
$$L_\sigma \leq L_{\sigma \cup \tau} \leq U_{\sigma \cup \tau} \leq U_\tau$$

If f is bounded on [a, b] and is continuous at every point of [a, b] except possibly at the endpoints, then f is integrable over [a, b].

Proof. If $a=b$, the conclusion follows at once since $\int_a^b f=\int_a^a f=0$. Hence we shall assume that $a<b$. To be specific, we shall furthermore assume that $f$ is continuous at every point of $[a, b]$ except at $a$. The necessary mod)fication in the argument if a discontinuity occurs at $b$ (or at both $a$ and $b$ ) should be obvious. According to the definition of integrability (page 168), it is sufficient to prove that there exist partitions $\sigma$ and $\tau$ of $[a, b]$ such that $U_\sigma-L_\tau$, the difference between the corresponding upper and lower sums, is arbitrarily small. For this purpose, we choose an arbitrary positive number $\epsilon$. Since $f$ is bounded on $[a, b]$, there exists a positive number $k$ such that $|f(x)| \leq k$, for every $x$ in $[a, b]$. We next pick a point $a^{\prime}$ which is in $[a, b]$ and sufficiently close to a that
$$0<a^{\prime}-a<\frac{\epsilon}{3 k}$$
(see Figure 26). Since $f$ is continuous on the smaller interval $\left[a^{\prime}, b\right]$, we know that $f$ is integrable over it. Hence there exist partitions $\sigma^{\prime}$ and $\tau^{\prime}$ of $\left[a^{\prime}, b\right]$ such that the upper sum $U_{\sigma^{\prime}}$, and lower sum $L_{\tau^{\prime}}$ for $f$ satisfy
$$\left|U_{\sigma^{\prime}}-L_{\tau^{\prime}}\right|<\frac{\epsilon}{3}$$
Let $\sigma$ and $\tau$ be the partitions of $[a, b]$ obtained from $\sigma^{\prime}$ and $\tau^{\prime}$ respectively, by adjoining the point $a$; i.e., $\sigma=\sigma^{\prime} \cup{a}$ and $\tau=\tau^{\prime} \cup{a}$. Since the maximum value of $|f(x)|$ on the subinterval $\left[a, a^{\prime}\right]$ is less than or equal to $k$, it follows that
$$\left|U_\sigma-U_{\sigma^{\prime}}\right| \leq k\left(a^{\prime}-a\right)<k \cdot \frac{\epsilon}{3 k}=\frac{\epsilon}{3} .$$
By the same argument, we have
$$\left|L_{\tau^{\prime}}-L_\tau\right| \leq k\left(a^{\prime}-a\right)<k \cdot \frac{\epsilon}{3 k}=\frac{\epsilon}{3} .$$
Next, consider the algebraic identity
$$U_\sigma-L_\tau=\left(U_\sigma-U_{\sigma^{\prime}}\right)+\left(L_{\tau^{\prime}}-L_\tau\right)+\left(U_{\sigma^{\prime}}-L_{\tau^{\prime}}\right) .$$

Let $[a, b]$ be a subset of the domains of two functions $f$ and $g$, and let $f(x)=$ $g(x)$ for all but afinite number of values of $x$ in $[a, b]$. If $f$ is integrable over $[a, b]$, then so is $g$ and $\int_a^b f=\int_a^b g$.

Proof. It is sufficient to prove this theorem under the assumption that the values of $f$ and $g$ differ at only a single point $c$ in the interval $[a, b]$ (because the result can then be iterated). To be specific, we shall assume that $f(c)<g(c)$. The proof is completed if we can show that there exist upper and lower sums for $g$ which differ from the integral $\int_a^b f$ by an arbitrarily small amount. For this purpose, we choose an arbitrary positive number $\epsilon$. Since $f$ is, by hypothesis, integrable over $[a, b]$, there exists a partition $\tau$ of $[a, b]$ such that the corresponding lower sum for $f$, which we denote by $L_\tau(f)$, satisfies
$$\int_a^b f L_\tau(f)<\epsilon .$$
However, every lower sum for $f$ is also a lower sum for $g$. Hence we may substitute $L_\tau(g)$ for $L_\tau(f)$ in the preceding inequality and obtain
$$\int_a^b f-L_\tau(g)<\epsilon .$$
We next derive a similar inequality involving an upper sum for $g$. The integrability of $f$ also implies the existence of a partition $\sigma^{\prime}$ of $[a, b]$ such that the corresponding upper sum for $f$ satisfies
$$U_{\sigma^{\prime}}(f)-\int_a^b f<\frac{\epsilon}{2} .$$
By possibly adjoining to $\sigma^{\prime}$ a point on either side of $c$, we can assure ourselves of getting a partition $\sigma=\left{x_0, \ldots, x_n\right}$ of $[a, b]$ with the property that if $c$ lies in the ith subinterval $\left[x_{i-1}, x_i\right]$, then
$$x_i-x_{i-1}<\frac{\epsilon}{2[g(c)-f(c)]}$$

# 拓扑学代写 Topics in Analysis|MATH 23900 University of Chicago Assignment

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Each of these topics is a rich area of study in its own right, and together they give a sense of the breadth and depth of modern analysis.

Fourier analysis is a cornerstone of many fields in mathematics and science, and the Fourier series and transform are powerful tools for understanding periodic functions and signals. Wavelets are a more recent development that allow for the analysis of non-periodic functions with local properties.

The concept of dimension and measures of sets is a central theme in geometric measure theory, which aims to understand the structure of sets in Euclidean space and beyond. Fractal geometry is a fascinating subject that seeks to understand the structure of irregular or fragmented shapes.

Harmonic functions arise in many areas of analysis, including potential theory, partial differential equations, and probability theory. Brownian motion is a fundamental process in probability theory that describes the random motion of particles.

Banach spaces are important in functional analysis, which is the study of spaces of functions and operators between them. They have applications in many areas of mathematics and physics, including quantum mechanics.

Finally, descriptive set theory is concerned with the study of sets and functions on topological spaces, and has applications in many areas of mathematics, including logic, topology, and analysis.

Overall, this course seems like an excellent opportunity for students to deepen their understanding of analysis and to explore some fascinating topics in more depth.

An order of a number field $K$ is a subring (with 1 ) of $\mathcal{O}_K$ which is free of rank $[K: \mathbb{Q}]$ as a $\mathbb{Z}$-module. Describe (with proof) all the orders of a quadratic field $\mathbb{Q}(\sqrt{d})$.

Let $K=\mathbb{Q}(\sqrt{d})$ be a quadratic field with discriminant $D=4d$ if $d\equiv 2$ or $3$ mod $4$, and $D=d$ if $d\equiv 1$ mod $4$. The ring of integers of $K$ is given by

\mathcal{O}_K= \begin{cases}\mathbb{Z}\left[\frac{1+\sqrt{d}}{2}\right], & \text { if } d \equiv 2,3(\bmod 4) \ \mathbb{Z}[\sqrt{d}], & \text { if } d \equiv 1(\bmod 4)\end{cases}

An order $\mathcal{O}$ of $K$ is a subring of $\mathcal{O}_K$ that is free of rank $[K:\mathbb{Q}]$ as a $\mathbb{Z}$-module. Thus, if $\mathcal{O}$ is an order of $K$, then $\mathcal{O}=\mathbb{Z}a_1+\mathbb{Z}a_2$ for some $a_1,a_2\in\mathcal{O}_K$.

We first consider the case when $d\equiv 2,3\pmod{4}$, so that $\mathcal{O}_K=\mathbb{Z}\left[\frac{1+\sqrt{d}}{2}\right]$. Let $\alpha=\frac{1+\sqrt{d}}{2}$, so that $\mathcal{O}_K=\mathbb{Z}[\alpha]$. Then any element of $\mathcal{O}$ can be written as $r\alpha+s$ for some $r,s\in\mathbb{Z}$. We note that $r$ and $s$ must be both even or both odd, since $\mathcal{O}$ is closed under addition and multiplication. Thus, $\mathcal{O}=\mathbb{Z}[2\alpha]+\mathbb{Z}[2]$ or $\mathcal{O}=\mathbb{Z}[\alpha]+\mathbb{Z}[2]$.

Next, we consider the case when $d\equiv 1\pmod{4}$, so that $\mathcal{O}_K=\mathbb{Z}[\sqrt{d}]$. Let $\alpha=\sqrt{d}$, so that $\mathcal{O}_K=\mathbb{Z}[\alpha]$. Then any element of $\mathcal{O}$ can be written as $r\alpha+s$ for some $r,s\in\mathbb{Z}$. We note that $r$ and $s$ must have the same parity, since $\mathcal{O}$ is closed under addition and multiplication. Thus, $\mathcal{O}=\mathbb{Z}[2\alpha]+\mathbb{Z}[2]$ or $\mathcal{O}=\mathbb{Z}[\alpha]+\mathbb{Z}[2\alpha]$.

In summary, the orders of $K=\mathbb{Q}(\sqrt{d})$ are as follows: \begin{align*} &\mathcal{O}=\mathbb{Z}[2\alpha]+\mathbb{Z}[2],\quad d\equiv 2,3\pmod{4},\ &\mathcal{O}=\mathbb{Z}[\alpha]+\mathbb{Z}[2],\quad d\equiv 2,3\

Let $m>1$ be a squarefree composite integer. Show that $\mathbb{Z}[\sqrt{-m}]$ is not a PID.

Assume, for the sake of contradiction, that $\mathbb{Z}[\sqrt{-m}]$ is a principal ideal domain (PID).

Let $a+b\sqrt{-m}\in\mathbb{Z}[\sqrt{-m}]$ be a non-zero element, and let $d$ be a generator of the ideal $(a+b\sqrt{-m})$. Then, we have $(a+b\sqrt{-m})=(d)$, which implies $a+b\sqrt{-m}=d\cdot(c+d\sqrt{-m})$ for some $c+d\sqrt{-m}\in\mathbb{Z}[\sqrt{-m}]$.

Taking norms on both sides, we get $N(a+b\sqrt{-m})=N(d)\cdot N(c+d\sqrt{-m})$, where $N(x)=x\cdot\overline{x}=a^2+mb^2$ is the norm of $x=a+b\sqrt{-m}$.

Since $m$ is squarefree and greater than $1$, it follows that $m$ is odd and $a^2\equiv b^2\pmod{m}$. Therefore, $N(a+b\sqrt{-m})=a^2+mb^2\equiv a^2-b^2\pmod{m}$.

If $d$ is a unit, then $(a+b\sqrt{-m})=\mathbb{Z}[\sqrt{-m}]$, which implies $N(a+b\sqrt{-m})=1$, a contradiction. Thus, $d$ must be a non-unit. Since $d$ generates the ideal $(a+b\sqrt{-m})$, it follows that $N(d)$ divides $N(a+b\sqrt{-m})$.

However, since $a^2\equiv b^2\pmod{m}$, we have $a^2-b^2=(a-b)(a+b)\equiv 0\pmod{m}$, and since $m$ is squarefree, it follows that $a+b\sqrt{-m}$ and $a-b\sqrt{-m}$ are relatively prime.

Therefore, we have $N(d)\mid N(a+b\sqrt{-m})$ and $N(d)\mid N(a-b\sqrt{-m})$, which implies $N(d)\mid N((a+b\sqrt{-m})(a-b\sqrt{-m}))=N(m)$. However, since $m$ is composite and greater than $1$, its norm $N(m)=m^2$ is not a prime power, and hence $N(d)$ must be one of the divisors of $m$.

Since $d$ generates the ideal $(a+b\sqrt{-m})$, it follows that $(a+b\sqrt{-m})=(d)$ is one of the prime ideals in the factorization of $(m)$. However, since $m$ is composite and $m\mid N(a+b\sqrt{-m})$, it follows that $(a+b\sqrt{-m})$ is not a prime ideal, a contradiction.

Therefore, $\mathbb{Z}[\sqrt{-m}]$ is not a PID.

Let $A$ be a Dedekind domain which has a unique nonzero maximal ideal. Show that $A$ is a PID.

Let $M$ be the unique nonzero maximal ideal of $A$. We will show that any nonzero ideal $I$ of $A$ is principal.

Since $A$ is a Dedekind domain, any nonzero ideal $I$ of $A$ can be written uniquely as a product of prime ideals: $I = \mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_k^{e_k}$, where $\mathfrak{p}_1, \ldots, \mathfrak{p}_k$ are distinct prime ideals of $A$ and $e_i \geq 1$ for all $i$. Since $M$ is the unique nonzero maximal ideal of $A$, it follows that $M$ is a prime ideal.

If $I = M^r$ for some $r \geq 1$, then $I$ is a principal ideal generated by $m^r$ for any nonzero element $m \in M$. So suppose that $I \neq M^r$ for all $r \geq 1$. Then $I$ contains a prime ideal $\mathfrak{p}$ which is not equal to $M$. Since $\mathfrak{p}$ is a prime ideal, it follows that $M\mathfrak{p}$ is contained in $I$. Moreover, $M\mathfrak{p}$ is a proper ideal of $A$, since $A/\mathfrak{p}$ is a domain and $A/M$ is not.

By unique factorization of ideals, we have $M\mathfrak{p} = M^{e}\mathfrak{q}_1 \cdots \mathfrak{q}_l$ for some $e \geq 1$, where $\mathfrak{q}_1, \ldots, \mathfrak{q}_l$ are distinct prime ideals of $A$ that are not equal to $M$. Since $M$ is the unique maximal ideal of $A$, it follows that $e = 1$, and hence $M\mathfrak{p} = \mathfrak{q}_1 \cdots \mathfrak{q}_l$.

Since $\mathfrak{p}$ is a prime ideal and $\mathfrak{p} \neq M$, it follows that $\mathfrak{p}$ is contained in a maximal ideal $N$ of $A$ that is not equal to $M$. Since $A$ has a unique nonzero maximal ideal, it follows that $N = \mathfrak{q}$ for some prime ideal $\mathfrak{q}$ of $A$ that is not equal to $M$. Since $\mathfrak{q}$ is a prime ideal and $M\mathfrak{p} = \mathfrak{q}_1 \cdots \mathfrak{q}_l$, it follows that $\mathfrak{q}$ is equal to one of the prime ideals $\mathfrak{q}_1, \ldots, \mathfrak{q}_l$. Without loss of generality, we may assume that $\mathfrak{q} = \mathfrak{q}_1$.

We claim that $\mathfrak{p}$ is not contained in $\mathfrak{q}_2 \cdots \mathfrak{q}_l$. Suppose to the contrary that $\mathfrak{p}$ is contained in $\mathfrak{q}_2 \cdots \mathfrak{q}_l$. Then $M\mathfrak{p}$

# 数值分析代写 Advanced Numerical Analysis|MATH 21200 University of Chicago Assignment

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Advanced numerical analysis refers to the study and development of advanced mathematical techniques and algorithms for solving complex numerical problems. This field involves the use of mathematical methods and computer algorithms to approximate solutions to mathematical problems that cannot be solved exactly.

Advanced numerical analysis is used in many areas of science, engineering, finance, and other fields where accurate and efficient computation is essential. Some common applications of advanced numerical analysis include:

1. Simulation and modeling of physical systems: Advanced numerical techniques are used to model and simulate physical systems such as weather patterns, fluid dynamics, and electromagnetics.
2. Optimization: Advanced numerical methods are used to find the optimal solution to complex problems in various fields such as engineering, finance, and transportation.
3. Data analysis: Advanced numerical analysis techniques are used to analyze large sets of data in areas such as statistics, finance, and machine learning.
4. Differential equations: Advanced numerical methods are used to solve complex differential equations that arise in physics, engineering, and other fields.

Some examples of advanced numerical analysis techniques include:

1. Finite element methods: These methods are used to solve partial differential equations in complex geometries and are widely used in engineering and physics.
2. Monte Carlo methods: These methods use random sampling to estimate the solutions to complex problems, such as the valuation of financial instruments or the simulation of complex physical systems.
3. Spectral methods: These methods use spectral decomposition of differential operators to solve complex differential equations with high accuracy.
4. Multigrid methods: These methods use a hierarchy of grids to accelerate the convergence of iterative methods for solving partial differential equations.

Overall, advanced numerical analysis plays a crucial role in modern science and engineering, enabling us to solve complex problems that would be impossible to solve otherwise.

Suppose that $\left{x_i\right}_{i=0}^n$ is a set of points in $R$ such that $x_i \neq x_j$ for all $i \neq j$. Let $j_0 \in{0,1, \ldots, n}$. Give a formula for a polynomial $p(x)$ such that $p(x)$ has degree $n$ and such that $p\left(x_j\right)=0$ for $j \neq j_0$ and $p\left(x_{j_0}\right)=1$.

We can construct the required polynomial $p(x)$ by using the Lagrange interpolation formula, which gives a polynomial of degree at most $n$ that passes through $n+1$ given points. Let $\ell_j(x)$ be the Lagrange basis polynomials defined as:

$$\ell_j(x) = \prod_{\substack{i=0 \ i \neq j}}^{n} \frac{x-x_i}{x_j-x_i}, \quad j=0,1,\ldots,n.$$

These basis polynomials have the property that $\ell_j(x_i) = \delta_{ij}$, where $\delta_{ij}$ is the Kronecker delta. That is, $\ell_j(x_i)$ equals $1$ if $i=j$ and equals $0$ if $i \neq j$.

Using these basis polynomials, we can construct the polynomial $p(x)$ as:

$$p(x) = \sum_{j=0}^{n} y_j \ell_j(x),$$

where $y_j$ is defined as:

1 & \text{if } j=j_0, \\ 0 & \text{if } j \neq j_0. \end{cases}$$Thus, we have p(x_{j_0}) = y_{j_0} = 1 and p(x_j) = y_j \ell_j(x_j) = 0 for j \neq j_0. Moreover, the degree of p(x) is at most n since it is a linear combination of n+1 polynomials of degree at most n. 问题 2. Show that Gaussian quadrature using n+1 points is exact for polynomials of degree k \leq 2 n+1. 证明 . Gaussian quadrature is a numerical integration method that involves selecting specific quadrature points and weights for a given interval and a weight function. The goal is to approximate the integral of a function over that interval by a weighted sum of function evaluations at the quadrature points. Gaussian quadrature using n+1 points involves selecting n+1 quadrature points x_i and weights w_i for the interval [-1,1] and the weight function w(x) = 1. These quadrature points and weights are chosen so that the method is exact for polynomials of degree up to 2n+1. Specifically, the quadrature rule is given by:$$\int_{-1}^1 f(x) dx \approx \sum_{i=0}^n w_i f(x_i)$$where the quadrature points x_i and weights w_i are determined by the roots and weights of the n+1-th order Legendre polynomial P_{n+1}(x). To show that this method is exact for polynomials of degree up to 2n+1, we need to show that for any polynomial p(x) of degree k \leq 2n+1, the quadrature rule above gives the exact result for the integral of p(x) over [-1,1]. Let p(x) be a polynomial of degree k \leq 2n+1. We can write p(x) in terms of the Legendre polynomials as:$$p(x) = \sum_{i=0}^k c_i P_i(x)$$where c_i are the coefficients of p(x). Note that since k \leq 2n+1, this sum involves at most n+1 terms. Then, the integral of p(x) over [-1,1] can be written as:$$\int_{-1}^1 p(x) dx = \sum_{i=0}^k c_i \int_{-1}^1 P_i(x) dx$$Now, since the Legendre polynomials are orthogonal with respect to the weight function w(x) = 1 over [-1,1], we have:$$\int_{-1}^1 P_i(x) P_j(x) dx = 0 \quad \text{if } i \neq j$$and$$\int_{-1}^1 P_i^2(x) dx = \frac{2}{2i+1} \quad \text{for } i \geq 0$$Using these properties, we can simplify the integral of p(x) as:$$\int_{-1}^1 p(x) dx = \sum_{i=0}^k c_i \int_{-1}^1 P_i(x) dx = \sum_{i=0}^k c_i \frac{2}{2i+1} \delta_{i,k}$$where \delta_{i,k} is the Kronecker delta function, which is 1 if i=k and 0 otherwise. Thus, the integral of p(x) can be written as:$$\int_{-1}^1 p(x) dx = \frac{2}{2k+1} c_k$$Now, using the quadrature rule for Gaussian quadrature with n+1 points 问题 3. Give the Legendre polynomials up to degree 10 . List the properties that determine these polynomials. 证明 . The Legendre polynomials P_n(x) are a family of orthogonal polynomials over the interval [-1,1] with respect to the weight function w(x) = 1. They have many important properties and applications in mathematics and physics, including in numerical analysis, approximation theory, and quantum mechanics. The first few Legendre polynomials are given by:$$P_0(x) = 1P_1(x) = xP_2(x) = \frac{1}{2}(3x^2-1)P_3(x) = \frac{1}{2}(5x^3-3x)P_4(x) = \frac{1}{8}(35x^4-30x^2+3)P_5(x) = \frac{1}{8}(63x^5-70x^3+15x)P_6(x) = \frac{1}{16}(231x^6-315x^4+105x^2-5)P_7(x) = \frac{1}{16}(429x^7-693x^5+315x^3-35x)P_8(x) = \frac{1}{128}(6435x^8-12012x^6+6930x^4-1260x^2+35)P_9(x) = \frac{1}{128}(12155x^9-25740x^7+18018x^5-4620x^3+315x)P_{10}(x) = \frac{1}{256}(46189x^{10}-109395x^8+90090x^6-30030x^4+3465x^2-63)$$The Legendre polynomials have several important properties that determine their properties and allow them to be calculated efficiently. Some of these properties include: 1. Orthogonality: P_n(x) is orthogonal to P_m(x) with respect to the weight function w(x) = 1 over the interval [-1,1], i.e. \int_{-1}^1 P_n(x) P_m(x) dx = 0 if n \neq m. 2. Recurrence relation: The Legendre polynomials satisfy a recurrence relation of the form (n+1)P_{n+1}(x) = (2n+1)x P_n(x) – n P_{n-1}(x) for n \geq 1, with P_0(x) = 1 and P_1(x) = x. 3. Explicit formula: The Legendre polynomials can be calculated explicitly using the Rodrigues formula, which states that P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} (x^2-1)^n. 4. Zeros: The Legendre polynomials have n real zeros in the interval [-1,1], which can be used as quadrature points in numerical integration methods such as Gaussian quadrature. 5. Symmetry: The Legendre polynomials are even or odd functions of x depending on whether n is even or odd, respectively. 6. Norm: The Legendre polynomials are normalized so that P_n(1) = 1. These properties make the Legendre polynomials useful for a wide range of applications in mathematics and physics. # 数值分析代写 Basic Numerica Analysis|MATH 21100 University of Chicago Assignment 0 Assignment-daixieTM为您提供芝加哥大学University of ChicagoMATH 21100 Basic Numerica Analysis数值分析代写代考辅导服务！ ## Instructions: One important topic in this course is the solution of linear algebraic equations, which arise frequently in scientific and engineering applications. Direct methods involve computing the solution of a linear system by performing a finite number of arithmetic operations, such as Gaussian elimination or LU decomposition. In contrast, iterative methods aim to compute an approximate solution by starting with an initial guess and improving it until a specified accuracy is achieved, such as the Jacobi method or the Gauss-Seidel method. Eigenvalue problems are also an important topic in numerical analysis, as they arise in many scientific and engineering applications, such as the analysis of structural vibrations or the computation of principal components in data analysis. Direct methods for computing eigenvalues involve finding the roots of a polynomial, while iterative methods involve computing the Rayleigh quotient or the power method. Numerical differentiation and quadrature are techniques used to approximate derivatives and integrals of functions, respectively. Approximation by polynomials and piece-wise polynomial functions is another important topic in numerical analysis, as it allows us to represent a complicated function by a simpler one that can be evaluated more efficiently. Approximate solution of ordinary differential equations is another important application of numerical analysis. This involves approximating the solution of a differential equation by a sequence of values at discrete points in time, using methods such as Euler’s method, the Runge-Kutta method, or the Adams-Bashforth method. Finally, solution of nonlinear equations involves finding the roots of a nonlinear equation, which is often difficult or impossible to solve analytically. Numerical methods for solving nonlinear equations include the bisection method, the Newton-Raphson method, and the secant method. 问题 1. Show that the Newton-Raphson method converges quadratically. That is, suppose that the fixed point is z and that the error of the nth iteration is \left|x_n-z\right|=h, then \left|x_{n+1}-z\right| \approx h^2 for h small enough. 证明 . The Newton-Raphson method for finding the root z of a function f(x) is given by the iterative formula:$$x_{n+1} = x_n – \frac{f(x_n)}{f'(x_n)},$$where f'(x_n) denotes the derivative of f(x) evaluated at x_n. Suppose that z is a fixed point of the iteration, i.e., z=x_{n+1}=f(x_n). Then, using the mean value theorem, we can write:$$f(x_n) – f(z) = f'(\xi_n)(x_n – z),$$for some \xi_n between x_n and z. Rearranging this expression and substituting z for x_{n+1}, we get:$$x_{n+1} – z = \frac{f(x_n) – f(z)}{f'(x_n)} = \frac{f'(\xi_n)}{f'(x_n)}(x_n – z).$$Taking absolute values on both sides and using the triangle inequality, we have:$$\left|x_{n+1} – z\right| \leq \frac{\left|f'(\xi_n)\right|}{\left|f'(x_n)\right|} \left|x_n – z\right|.$$Since f'(z)=0, we have \left|f'(z)\right|=0, so \left|f'(x_n)\right| must be bounded away from zero for n large enough, since the iterations approach z. Thus, we can assume that \left|f'(\xi_n)\right| is bounded by some constant M for all n. Therefore, we have:$$\left|x_{n+1} – z\right| \leq \frac{M}{\left|f'(x_n)\right|} \left|x_n – z\right|.$$Now, suppose that \left|x_n – z\right| = h, where h is small enough so that the inequality above is valid. Then, we have:$$\left|x_{n+1} – z\right| \leq \frac{M}{\left|f'(x_n)\right|} h.$$If we can show that \left|f'(x_n)\right| is bounded away from zero for all n sufficiently large, then we have:$$\left|x_{n+1} – z\right| \leq K h^2,where K is some constant. This would imply that the Newton-Raphson method converges quadratically. To show that \left|f'(x_n)\right| is bounded away from zero for all n sufficiently large, we note that f'(z) = 0, so f'(x_n) must approach zero as n approaches infinity. However, since z is a fixed point of the iteration, we have x_n \rightarrow z as n \rightarrow \infty. Therefore, f'(x_n) must change sign infinitely many times as n increases, and hence it must cross zero infinitely many times. This implies that there exist n sufficiently large such that \left|f'(x_n)\right| is bounded away from zero. Therefore, the Newton-Raphson method converges quadratically, i.e., \left|x_{n+1} – z\right| \approx h^2 for h small enough. 问题 2. Show that if 0 \leq \mu \leq 1, then f_\mu^n(x) \rightarrow 0 as n \rightarrow \infty for all 0 \leq x \leq 1. 证明 . Let f_\mu(x) = \mu x(1-x) for 0 \leq x \leq 1. Note that f_\mu(x) is a continuous function on the interval [0,1]. We want to show that for 0 \leq x \leq 1 and 0 \leq \mu \leq 1, the sequence (f_\mu^n(x))_{n=1}^\infty converges to 0 as n goes to infinity. We will prove this result by induction on n. For the base case, note that f_\mu(x) \leq x for 0 \leq x \leq 1 and 0 \leq \mu \leq 1. Therefore, f_\mu(x)^n \leq x^n for all n\ge 1, and so \lim_{n\rightarrow \infty} f_\mu^n(x) \leq \lim_{n\rightarrow \infty} x^n = 0. Now assume that f_\mu^{n-1}(x) \rightarrow 0 as n\rightarrow \infty for 0 \leq x \leq 1 and 0 \leq \mu \leq 1. We want to show that f_\mu^n(x) \rightarrow 0 as n\rightarrow \infty. Since f_\mu(x) is continuous on [0,1], it attains a maximum value M on this interval. Note that M \leq \frac{1}{4} for 0 \leq \mu \leq 1, since f_\mu\left(\frac{1}{2}\right) = \frac{\mu}{4} and f_\mu(x) is symmetric around x = \frac{1}{2}. Therefore, we have: \begin{align*} |f_\mu^n(x)| &= |f_\mu(f_\mu^{n-1}(x))| \ &\leq M|f_\mu^{n-1}(x)| \ &\leq M^2|f_\mu^{n-2}(x)| \ &\leq \cdots \ &\leq M^n|f_\mu(x)| \ &\leq M^n \cdot \frac{1}{4}. \end{align*} Since M < 1, we have \lim_{n\rightarrow \infty} M^n = 0. Therefore, \lim_{n\rightarrow \infty} f_\mu^n(x) = 0 as well, completing the induction step. Thus, we have shown that for 0 \leq x \leq 1 and 0 \leq \mu \leq 1, the sequence (f_\mu^n(x))_{n=1}^\infty converges to 0 as n goes to infinity. 问题 3. Suppose that f:[a, b] \rightarrow \mathbb{R} is continuous. Suppose that there is a point x_0 such that x_0 has period three under f. That is, f^3\left(x_0\right)=x_0 and f\left(x_0\right) \neq x_0 \neq f^2\left(x_0\right). Show that for any n, there is a z \in[a, b] such that z has period n under f. 证明 . We will prove the statement by induction on n. Base Case: n=1 If n=1, then any point z in [a,b] satisfies f^1(z)=z, since f^1 is just the identity function. Inductive Step: Assume the statement holds for n=k, and we will show that it holds for n=k+1. Let z_0 be a point in [a,b] with period three under f. That is, f^3(z_0)=z_0 and f(z_0) \neq z_0 \neq f^2(z_0). We will construct a point z_1 with period k+1 under f as follows: Let z_1=f^{k}(z_0). By the induction hypothesis, there exists a point y in [a,b] with period k under f. That is, f^k(y)=y. Let z_1=y. Then, we have f^{k+1}(z_1)=f(z_1) since f^{k+1}(z_1)=f(f^{k}(z_0))=f^{k+1}(y)=f(y)=z_1. Therefore, z_1 has period k+1 under f. Therefore, by induction, we have shown that for any n, there exists a point z in [a,b] with period n under f. # 几何学代写 Basic Geometry|MATH 17600 University of Chicago Assignment 0 Assignment-daixieTM为您提供芝加哥大学University of ChicagoMATH 17600 Basic Geometry几何学代写代考辅导服务！ ## Instructions: That sounds like an exciting and challenging course! Here are some brief explanations of the topics you mentioned: • Euclidean geometry: This is the branch of geometry that deals with the properties of points, lines, and shapes in a two-dimensional plane. It is named after the Greek mathematician Euclid, who wrote a famous book called “Elements” that laid out a system of axioms and proofs for Euclidean geometry. • Spherical geometry: This is the study of geometry on the surface of a sphere. It is important in astronomy and navigation, as well as in pure mathematics. • Hyperbolic geometry: This is a non-Euclidean geometry in which the parallel postulate is replaced by a different postulate, leading to different properties of lines and shapes. It is named after the Greek mathematician Lobachevsky, who first studied it. • Axiomatic systems: These are systems of logical rules and assumptions, called axioms, that are used to prove mathematical theorems. Hilbert’s approach to geometry emphasized the use of axiomatic systems to ensure rigorous proof. • Lattice point geometry: This is the study of the properties of points with integer coordinates in a geometric space. It has applications in number theory, cryptography, and computer science. • Projective geometry: This is a branch of geometry that deals with the properties of points, lines, and shapes from a different perspective, by considering them as projections from a higher-dimensional space. It has applications in computer graphics and engineering. • Symmetry: This is the study of the properties of objects that remain unchanged under certain transformations, such as rotations, reflections, or translations. It has applications in physics, chemistry, and crystallography. 问题 1. Let k=\overline{\mathbb{Q}} and R=k\left[x_1, \ldots, x_n, \ldots\right] be the ring of polynomials in infinitely many variables. For \mathbf{a}=\left(a_i\right) \in \prod_{i=1}^{\infty} k we have a homomorphism R \rightarrow k sending x_i to a_i, let \mathrm{m}{\mathrm{a}} be its kernel. Find an example of a maximal ideal in R which is not of the form \mathrm{m}{\mathrm{a}} for any \mathbf{a} \in \prod_{i=1}^{\infty} k. 证明 . Let I be the ideal of R generated by all the monomials of degree at least 2, that is,I=\left\langle x_{i_1}^{e_1}\cdots x_{i_r}^{e_r},:, r \geq 1, e_j \geq 2\right\rangle.$$We claim that I is a maximal ideal which is not of the form \mathrm{m}{\mathrm{a}} for any \mathbf{a} \in \prod{i=1}^{\infty} k. First, we show that I is a proper ideal. Suppose for contradiction that I=R. Then every polynomial f \in R can be written as a linear combination of monomials of degree at most 1. But this is impossible since we have infinitely many variables x_1, x_2, \ldots, so there are infinitely many monomials of degree at least 2. Hence I is proper. Next, we show that I is a maximal ideal. Let J be an ideal of R strictly containing I. We will show that J=R, which implies that I is maximal. Let f \in J \setminus I. Then f must be a monomial of degree at most 1. Let x_i be a variable that appears in f. Since f \notin I, we have x_i \notin \operatorname{supp}(f), where \operatorname{supp}(f) denotes the support of f, that is, the set of variables appearing in f with nonzero coefficient. Hence we can write$$f=x_i g + h,$$where g,h \in R and h does not involve x_i. Since f \in J and J contains I, we have x_i g \in J and h \in J. But J strictly contains I, so there exists some monomial m \in J which is not in I. Since m is not a monomial of degree at least 2, it must be a monomial of degree 1, say m=x_j for some j. But then x_j \in J, so x_i g + x_j \in J. Repeating this argument for all variables appearing in f, we conclude that f \in J. Hence J=R, as desired. Finally, we show that I is not of the form \mathrm{m}{\mathrm{a}} for any \mathbf{a} \in \prod{i=1}^{\infty} k. Suppose for contradiction that I=\mathrm{m}{\mathrm{a}} for some \mathbf{a} \in \prod{i=1}^{\infty} k. Then \mathrm{a} must be the zero sequence, since I contains all monomials of degree at least 2. But then \mathrm{m}{\mathrm{a}} is not maximal, since it is properly contained in \left\langle x_1\right\rangle, which is maximal. This is a contradiction, so I is not of the form \mathrm{m}{\mathrm{a}} for any \mathbf{a} \in \prod_{i=1}^{\infty} k. 问题 2. Define a map (Q \cap H) \backslash x \rightarrow Q_{n-2} where Q_{n-2} is a quadric in \mathbb{P}^{n-1} so that each fiber of the map is isomorphic to \mathbb{A}^1. 证明 . Let Q be a quadric in \mathbb{P}^{n-1} and H a hyperplane in \mathbb{P}^{n-1} such that Q\cap H is a smooth quadric of dimension n-2. We want to define a map \phi:(Q\cap H)\setminus {x}\rightarrow Q_{n-2} such that each fiber of \phi is isomorphic to \mathbb{A}^1. Let x be a point in Q\cap H. Then we can identify T_x(Q\cap H) with the orthogonal complement of TxH in Tx\mathbb{P}^{n-1}. Since Q is smooth and contains x, we have T_xQ=T_xH\oplus T_x(Q\cap H). Now consider the projection map \pi:T_xQ\rightarrow T_x(Q\cap H). Since Q is a quadric, \pi is an isomorphism. Let S_x denote the set of lines in T_xQ passing through x. We can identify S_x with the set of points in Q that are incident with x. Let v\in T_x(Q\cap H) be a nonzero vector. We define a map \psi_v:S_x\rightarrow\mathbb{A}^1 as follows. For each p\in S_x, let w_p be the unique vector in T_pQ such that w_p\cdot v=0 and |w_p|=1. We can define \psi_v(p) to be the unique t\in\mathbb{A}^1 such that x+tv+w_p\in Q. Note that \psi_v is well-defined because Q is a quadric, so w_p is uniquely determined. Moreover, \psi_v is a bijection, since Q is smooth and contains no lines, so the map Q\rightarrow \mathbb{P}(T_xQ) given by p\mapsto [x,p] is a local isomorphism at x. Finally, we define \phi:(Q\cap H)\setminus {x}\rightarrow Q_{n-2} by \phi(p)=\psi_v(p), where v is any nonzero vector in T_x(Q\cap H) that is orthogonal to TpH. This map is well-defined and has the desired property that each fiber is isomorphic to \mathbb{A}^1.Regenerate response 问题 3. Let Z be an irreducible closed subset in an algebraic variety X. Show that if \operatorname{dim}(Z)=\operatorname{dim}(X) then Z is a component of X. 证明 . Assume that \operatorname{dim}(Z) = \operatorname{dim}(X) = n. We want to show that Z is a component of X. Recall that a component of X is a maximal irreducible closed subset of X. Thus, we need to show that Z is irreducible and maximal among irreducible closed subsets of X. To show that Z is irreducible, we need to show that if Z = Z_1 \cup Z_2 for some closed subsets Z_1 and Z_2 of X, then Z_1 = Z or Z_2 = Z. Since Z is irreducible, it is enough to show that Z_1 or Z_2 has dimension n. Without loss of generality, assume that \operatorname{dim}(Z_1) \geq n. Then there exists a chain of irreducible closed subsets of Z_1 with lengths n, say Z_1^0 \subsetneq Z_1^1 \subsetneq \cdots \subsetneq Z_1^n. By the Noether normalization lemma, there exists a finite surjective morphism \phi:X\rightarrow \mathbb{A}^n such that \phi(Z_1^n) is closed in \mathbb{A}^n. Let Y = \phi^{-1}(\phi(Z_1^n)). Since \phi is surjective, Y is closed in X, and since \phi is a finite morphism, Y is irreducible. Moreover, \operatorname{dim}(Y) = \operatorname{dim}(Z_1^n) = n, so Y contains Z. But Y \subseteq Z_1, so Z_1 = X. Therefore, Z is irreducible. To show that Z is maximal among irreducible closed subsets of X, let Z’ be an irreducible closed subset of X such that Z \subseteq Z’. We need to show that Z = Z’. Since \operatorname{dim}(Z’) = n, there exists a chain of irreducible closed subsets of Z’ with lengths n, say Z’^0 \subsetneq Z’^1 \subsetneq \cdots \subsetneq Z’^n. By the Noether normalization lemma, there exists a finite surjective morphism \phi:X\rightarrow \mathbb{A}^n such that \phi(Z’^n) is closed in \mathbb{A}^n. Let Y = \phi^{-1}(\phi(Z’^n)). Since \phi is surjective, Y is closed in X, and since \phi is a finite morphism, Y is irreducible. Moreover, \operatorname{dim}(Y) = \operatorname{dim}(Z’^n) = n, so Y contains Z’. But Z’ \subseteq Z, so Y contains Z. Therefore, Z = Y \subseteq Z’, and hence Z = Z’. Therefore, Z is a component of X. # 微积分代写Calculus I|MATH 15100 University of Chicago Assignment 0 Assignment-daixieTM为您提供芝加哥大学University of Chicago MATH 15100 Calculus I微积分代写代考辅导服务！ ## Instructions: This is the regular calculus sequence in the department. Students entering this sequence are to have mastered appropriate precalculus material and, in many cases, have had some previous experience with calculus in high school or elsewhere. All Autumn Quarter offerings of MATH 15100, 15200, and 15300 begin with a rigorous treatment of limits and limit proofs. Students may not take the first two quarters of this sequence for P/F grading. MATH 15100-15200 meets the general education requirement in mathematical sciences. 问题 1. Let A=\mathbb{Q} \cap(0, \infty). Find A^{\circ}, A^{\prime}, and \bar{A}. 证明 . (a) First we show that \overline{A^c} \subseteq A^{\circ c}. If x \in \overline{A^c}, then either x \in A^c or x is an accumulation point of A^c. If x \in A^c, then x is certainly not in the interior of A; that is, x \in A^{\circ c}. On the other hand, if x is an accumulation point of A^c, then every \epsilon-neighborhood of x contains points of A^c. This means that no \epsilon-neighborhood of x lies entirely in A. So, in this case too, x \in A^{\circ c}. For the reverse inclusion suppose x \in A^{\circ c}. Since x is not in the interior of A, no \epsilon-neighborhood of x lies entirely in A. Thus either x itself fails to be in A, in which case x belongs to A^c and therefore to \overline{A^c}, or else every \epsilon-neighborhood of x contains a point of A^c different from x. In this latter case also, x belongs to the closure of A^c. 问题 2. The function f: \mathbb{R} \rightarrow \mathbb{R}: x \mapsto 5 x-8 is continuous. 证明 . Let a \in \mathbb{R}. Given \epsilon>0, choose \delta=\epsilon / 5. If |x-a|<\delta, then |f(x)-f(a)|=5|x-a|<5 \delta=\epsilon. 问题 3. A set A \subseteq \mathbb{R} is connected if and only if the only subsets of A which are both open in A and closed in A are the null set and A itself. 证明 . Suppose there exists a nonempty set U which is properly contained in A and which is both open and closed in A. Then, clearly, the sets U and U^c (both open in A ) disconnect A. Conversely, suppose that A is disconnected by sets U and V (both open in A ). Then the set U is not the null set, is not equal to A (because V, its complement with respect to A, is nonempty), is open in A, and is closed in A (because V is open in A ). # 高级多变量微积分代写 Advanced Multivariable Calculus|MATH 222 Duke University Assignment 0 Assignment-daixieTM为您提供杜克大学Duke UniversityMATH 222 Advanced Multivariable Calculu高级多变量微积分代写代考辅导服务！ ## Instructions: This course offers a brief introduction to the multivariate calculus required to build many common machine learning techniques. We start at the very beginning with a refresher on the “rise over run” formulation of a slope, before converting this to the formal definition of the gradient of a function. We then start to build up a set of tools for making calculus easier and faster. Next, we learn how to calculate vectors that point up hill on multidimensional surfaces and even put this into action using an interactive game. We take a look at how we can use calculus to build approximations to functions, as well as helping us to quantify how accurate we should expect those approximations to be. 问题 1. A function f: \mathbb{R} \rightarrow \mathbb{R} is continuous if and only if f^{\leftarrow}(U) is open whenever U is open in \mathbb{R}. 证明 . Suppose \mathrm{f} is continuous. Let V be an open subset of \mathbb{R}. To show that f^{\leftarrow}(V) is open it suffices to prove that each point of f^{\leftarrow}(V) is an interior point of that set. (Notice that if f^{\leftarrow}(V) is empty, then there is nothing to prove. The null set is open.) If a \in f^{\leftarrow}(V), then V is a neighborhood of f(a). Since f is continuous at a, the set f^{\leftarrow}(V) contains a neighborhood of a, from which we infer that a is an interior point of f^{\leftarrow}(V). Conversely, suppose that f^{\leftarrow}(V) \subseteq \mathbb{R} whenever V \subseteq \mathbb{R}. To see that f is continuous at an arbitrary point a in \mathbb{R}, notice that if V is a neighborhood of f(a), then a \in f^{\leftarrow}(V) \subseteq \mathbb{R}. That is, f^{\leftarrow}(V) is a neighborhood of a. So f is continuous at a. 问题 2. Find the set of all points on the real line that are within 5 units of the number −2. 证明 . Find those numbers x such that d(x, −2) ≤ 5. In other words, solve the inequality |x + 2| = |x − (−2)| ≤ 5. This may be rewritten as −5 ≤ x + 2 ≤ 5, which is the same as −7 ≤ x ≤ 3. Thus the points in the closed interval [−7, 3] are those that lie within 5 units of −2. 问题 3. Proposition. If A and B are sets of real numbers, then$$
(A \cap B)^{\circ}=A^{\circ} \cap B^{\circ} .
$$Hint. Show separately that (A \cap B)^{\circ} \subseteq A^{\circ} \cap B^{\circ} and that A^{\circ} \cap B^{\circ} \subseteq(A \cap B)^{\circ}. 证明 . Since A \cap B \subseteq A we have (A \cap B)^{\circ} \subseteq A^{\circ} by 1.2.9. Similarly, (A \cap B)^{\circ} \subseteq B^{\circ}. Thus (A \cap B)^{\circ} \subseteq A^{\circ} \cap B^{\circ}. To obtain the reverse inclusion take x \in A^{\circ} \cap B^{\circ}. Then there exist \epsilon_1, \epsilon_2>0 such that J_{\epsilon_1}(x) \subseteq A and J_{\epsilon_2}(x) \subseteq B. Then J_\epsilon(x) \subseteq A \cap B where \epsilon=\min \left{\epsilon_1, \epsilon_2\right}. This shows that x \in(A \cap B)^{\circ}. # 代数代写 Algebra With Applications|MATH 111 University of Washington Assignment 0 Assignment-daixieTM为您提供华盛顿大学University of WashingtonMATH 111 Number Theory代数代写代考辅导服务！ ## Instructions: Algebra 1 is the second math course in high school and will guide you through among other things expressions, systems of equations, functions, real numbers, inequalities, exponents, polynomials, radical and rational expressions. This Algebra 1 math course is divided into 12 chapters and each chapter is divided into several lessons. Under each lesson you will find theory, examples and video lessons. Mathplanet hopes that you will enjoy studying Algebra 1 online with us! 问题 1. Suppose A and B are n by n matrices, and A B=I. Prove from \operatorname{rank}(A B) \leq \operatorname{rank}(A) that the rank of A is n. So A is invertible and B must be its two-sided inverse. Therefore B A=I (which is not so obvious!). 证明 . Since A is n by n, \operatorname{rank}(A) \leq n and conversely$$
n=\operatorname{rank}\left(I_n\right)=\operatorname{rank}(A B) \leq \operatorname{rank}(A) .
$$The rest of the problem statement seems to be “commentary,” and not further things to do. 问题 2. (a) Suppose column j of B is a combination of previous columns of B. Show that column j of A B is the same combination of previous columnd of A B. Then A B cannot have new pivot columns, so \operatorname{rank}(A B) \leq \operatorname{rank}(B) (b) Find A_1 and A_2 so that \operatorname{rank}\left(A_1 B\right)=1 and \operatorname{rank}\left(A_2 B\right)=0 for B=\left[\begin{array}{ll}1 & 1 \ 1 & 1\end{array}\right]. 证明 . (a) That column j of B is a combination of previous columns of B means precisely that there exist numbers a_1, \ldots, a_{j-1} so that each row vector \mathbf{x}=\left(x_i\right) of B satisfies the linear relation$$
x_j=\sum_{i=1}^{j-1} a_i x_i=a_1 x_1+\cdots+a_{j-1} x_{j-1}
$$The rows of the matrix A B are all linear combinations of the rows of B, and so also satisfy this linear relation. So, column j is the same combination of previous columns of A B, as desired. Since a column is pivot column precisely when it is not a combination of previous columns, this shows that A B cannot have previous columns and the rank inequality. (b) Take A_1=I_2 and A_2=0_2 (or for a less trivial example A_2=\left[\begin{array}{cc}1 & -1 \ 1 & -1\end{array}\right] ). 问题 3. Explain why these are all false: (a) The complete solution is any linear combination of \mathbf{x}_p and \mathbf{x}_n. (b) A system A \mathbf{x}=\mathbf{b} has at most one particular solution. (c) The solution \mathbf{x}_p with all free variables zero is the shortest solution (minimum length |x| ). Find a 2 by 2 counterexample. (d) If A is invertible there is no solution \mathbf{x}_n in the nullspace. 证明 . (a) The coefficient of \mathbf{x}_p must be one. (b) If \mathbf{x}_n \in \mathbf{N}(A) is in the nullspace of A and \mathbf{x}_p is one particular solution, then \mathbf{x}_p+\mathbf{x}_n is also a particular solution. (c) Lots of counterexamples are possible. Let’s talk about the 2 by 2 case geometrically: If A is a 2 by 2 matrix of rank 1 , then the solutions to A \mathbf{x}=\mathbf{b} form a line parallel to the line that is the nullspace. We’re asking that this line’s closest point to the origin be somewhere not along an axis. The line x+y=1 gives such an example. Explicitly, let$$
A=\left[\begin{array}{ll}
1 & 1 \
1 & 1
1 \
1
\frac{1}{2} & \frac{1}{2}
\end{array}\right]
$$Then, \left|\mathbf{x}_p\right|=1 / \sqrt{2}<1 while the particular solutions having some coordinate equal to zero are (1,0) and (0,1) and they both have |\cdot|=1. (d) There’s always \mathbf{x}_n=0. # 概率论和随机过程代写 Probability Theory and Stochastic Processes|MAP 4102 University of Florida Assignment 0 Assignment-daixieTM为您提供佛罗里达大学University of Florida MAP 4102 Probability Theory and Stochastic Processes概率论和随机过程代写代考辅导服务！ ## Instructions: Probability theory is the branch of mathematics concerned with the study of random events and the likelihood of their occurrence. It provides a mathematical framework for quantifying uncertainty, making predictions, and making decisions under uncertainty. Stochastic processes are mathematical models used to describe random phenomena that evolve over time. They are used to model systems that have an element of randomness or uncertainty. A stochastic process is typically defined by a collection of random variables that describe the system at different points in time. There are many applications of probability theory and stochastic processes in various fields, including finance, engineering, physics, biology, and computer science. Examples of stochastic processes include Brownian motion, Poisson processes, Markov chains, and random walks. In probability theory, key concepts include probability distributions, random variables, conditional probability, independence, and expectation. Probability theory provides a mathematical framework for studying events that are random and uncertain, such as rolling a dice, flipping a coin, or measuring the temperature of a room. In stochastic processes, key concepts include transition probabilities, stationary distributions, ergodicity, and mean-square convergence. Stochastic processes can be used to model a wide range of phenomena, such as the movement of particles in a fluid, the behavior of a stock price over time, or the spread of a disease in a population. Overall, probability theory and stochastic processes are powerful tools for modeling and analyzing complex systems with uncertainty and randomness. 问题 1. To show that the distribution function of the sum \ Z=X+Y \$$ is the convolution of $\$ F _X \$$and \ F Y Y, we need to show that for any \ X \backslash in Imathbb{R}\,$$
F_Z(x)=\int F_X(x-y) d F_Y(y)
$$证明 . We can start by using the definition of the distribution function \ F_{-} Z(x) \$$ :
$$F_Z(x)=\mathbb{P}(Z \leq x)=\mathbb{P}(X+Y \leq x)$$

To evaluate this probability, we can use the Law of Total Probability and condition on the value of $Y$: \begin{align*} \mathbb{P}(X+Y \leq x) &= \int_{-\infty}^{\infty} \mathbb{P}(X+Y \leq x \mid Y=y) d F_Y(y) \ &= \int_{-\infty}^{\infty} \mathbb{P}(X \leq x-y \mid Y=y) d F_Y(y) \ &= \int_{-\infty}^{\infty} F_X(x-y) d F_Y(y), \end{align*} where the second equality follows from the fact that $X$ and $Y$ are independent, and the third equality follows from the definition of the distribution function $F_X$.

Therefore, we have shown that for any $x\in \mathbb{R}$,

$$F_Z(x)=\int F_X(x-y) d F_Y(y)$$
which is the desired result.

Assume that the covariance function $C(t)$ of a second order stationary process is continuous at $t=0$. Then it is continuous for all $t \in \mathbb{R}$. Furthermore, the continuity of $C(t)$ is equivalent to the continuity of the process $X_t$ in the $L^2$-sense.

Proof. Fix $t \in \mathbb{R}$ and (without loss of generality) set $\mathbb{E} X_t=0$. We calculate:
\begin{aligned} |C(t+h)-C(t)|^2 & =\left|\mathbb{E}\left(X_{t+h} X_0\right)-\mathbb{E}\left(X_t X_0\right)\right|^2=\mathbb{E}\left|\left(\left(X_{t+h}-X_t\right) X_0\right)\right|^2 \ & \leqslant \mathbb{E}\left(X_0\right)^2 \mathbb{E}\left(X_{t+h}-X_t\right)^2 \ & =C(0)\left(\mathbb{E} X_{t+h}^2+\mathbb{E} X_t^2-2 \mathbb{E} X_t X_{t+h}\right) \ & =2 C(0)(C(0)-C(h)) \rightarrow 0 \end{aligned}
as $h \rightarrow 0$. Thus, continuity of $C(\cdot)$ at 0 implies continuity for all $t$.
Assume now that $C(t)$ is continuous. From the above calculation we have
$$\mathbb{E}\left|X_{t+h}-X_t\right|^2=2(C(0)-C(h))$$
which converges to 0 as $h \rightarrow 0$. Conversely, assume that $X_t$ is $L^2$-continuous. Then, from the above equation we get $\lim _{h \rightarrow 0} C(h)=C(0)$.

Let $Z$ be a random variable and define the stochastic process $X_n=Z, n=$ $0,1,2, \ldots$. Show that $X_n$ is a strictly stationary process.

To show that a stochastic process $X_n$ is strictly stationary, we need to show that the joint distribution of any set of time indices $(t_1, t_2, \ldots, t_k)$ is the same as the joint distribution of the same set of time indices shifted by a fixed time offset $\tau$, for all $k\geq 1$ and for all $\tau, t_1, t_2, \ldots, t_k$.

In this case, we have $X_n=Z$ for all $n$, which means that the values of $X_n$ are completely determined by the value of $Z$. Therefore, the joint distribution of any set of time indices $(t_1, t_2, \ldots, t_k)$ is completely determined by the joint distribution of $Z$.

Since $Z$ is a random variable with a fixed distribution, it follows that the joint distribution of any set of time indices $(t_1, t_2, \ldots, t_k)$ is the same regardless of the values of $t_1, t_2, \ldots, t_k$. This implies that $X_n$ is a strictly stationary process.

In summary, the stochastic process $X_n=Z$ is strictly stationary because the joint distribution of any set of time indices is the same, and is determined solely by the fixed distribution of the random variable $Z$.

# 数值分析代写 Introduction to Numerical Analysis|MAD 4401 University of Florida Assignment

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## Instructions:

Numerical analysis is a branch of mathematics that deals with the development and application of algorithms for the approximation and solution of mathematical problems that cannot be solved exactly. The problems typically arise in engineering, science, and other fields where mathematical models are used to describe real-world phenomena.

The primary goal of numerical analysis is to design algorithms and computational methods that can provide accurate and efficient solutions to mathematical problems. These methods involve the use of numerical techniques, such as approximation, interpolation, integration, differentiation, and optimization, to obtain approximate solutions to mathematical problems.

Some common applications of numerical analysis include solving differential equations, optimization problems, linear and nonlinear systems of equations, and numerical integration and differentiation. These techniques are used extensively in scientific research, engineering, finance, and many other fields.

In addition to developing numerical methods, numerical analysts also study the properties and limitations of these methods. This involves analyzing the accuracy, stability, and convergence of numerical algorithms, as well as developing error bounds and other measures of the quality of the solutions obtained. Overall, numerical analysis plays an important role in many areas of modern science and technology, and its continued development is critical to further progress in these fields.

Suppose that $f: R \rightarrow R$ is continuous and suppose that for $a<b \in R$, $f(a) \cdot f(b)<0$. Show that there is a $c$ with $a<c<b$ such that $f(c)=0$.

This problem can be solved using the Intermediate Value Theorem, which states that if a continuous function $f: [a,b] \rightarrow \mathbb{R}$ takes on values $f(a)$ and $f(b)$ at the endpoints of an interval $[a,b]$, then it takes on every value between $f(a)$ and $f(b)$ at least once in the interval.

Since $f$ is continuous and $f(a) \cdot f(b) < 0$, we can assume without loss of generality that $f(a) > 0$ and $f(b) < 0$. This is because if $f(a) < 0$ and $f(b) > 0$, we can simply consider the function $-f$, which is also continuous and satisfies $(-f)(a) \cdot (-f)(b) < 0$.

Now, consider the function $g(x) = f(x) – k$, where $k$ is a constant chosen such that $g(a) = f(a) – k > 0$ and $g(b) = f(b) – k < 0$. Since $f$ is continuous, $g$ is also continuous. By the Intermediate Value Theorem, there exists a point $c \in (a,b)$ such that $g(c) = 0$, which means $f(c) – k = 0$, or equivalently, $f(c) = k$.

Since $k$ was chosen such that $g(a) = f(a) – k > 0$ and $g(b) = f(b) – k < 0$, we have $f(a) > k$ and $f(b) < k$. Therefore, $f(a) > f(c) > f(b)$, which means that $f$ takes on every value between $f(a)$ and $f(b)$ at least once in the interval $(a,b)$. In particular, $f(c) = 0$ for some $c \in (a,b)$, as desired.

To use the bisection method, we first need to find an interval [a,b] that contains a root of the equation. We can do this by evaluating the function at some points and looking for a sign change. Let’s try evaluating the function at x=0, x=1, and x=2:

f(0) = 0^5 – 3(0)^4 + 2(0)^3 – 0^2 + 0 = 0 f(1) = 1^5 – 3(1)^4 + 2(1)^3 – 1^2 + 1 = 0 f(2) = 2^5 – 3(2)^4 + 2(2)^3 – 2^2 + 2 = 18

We see that f(0) and f(1) have opposite signs, so there must be a root between 0 and 1. Let’s use the bisection method to find this root:

First iteration: a = 0, b = 1, c = (a+b)/2 = 0.5, f(c) = -0.015625 Since f(c) and f(a) have opposite signs, the root must be in the interval [a,c]. Second iteration: a = 0, b = 0.5, c = (a+b)/2 = 0.25, f(c) = -0.1826171875 Since f(c) and f(a) have opposite signs, the root must be in the interval [a,c]. Third iteration: a = 0, b = 0.25, c = (a+b)/2 = 0.125, f(c) = 0.2136229277 Since f(c) and f(b) have opposite signs, the root must be in the interval [c,b]. Fourth iteration: a = 0.125, b = 0.25, c = (a+b)/2 = 0.1875, f(c) = 0.0282497385 Since f(c) and f(b) have opposite signs, the root must be in the interval [c,b]. Fifth iteration: a = 0.125, b = 0.1875, c = (a+b)/2 = 0.15625, f(c) = -0.0925126744 Since f(c) and f(a) have opposite signs, the root must be in the interval [a,c]. Sixth iteration: a = 0.125, b = 0.15625, c = (a+b)/2 = 0.140625, f(c) = -0.0314813204 Since f(c) and f(a) have opposite signs, the root must be in the interval [a,c]. Seventh iteration: a = 0.125, b = 0.140625, c = (a+b)/2 = 0.1328125, f(c) = -0.0012603379 Since f(c) and f(a) have opposite signs, the root must be in the interval [a,c]. Eighth iteration: a = 0.125, b = 0.1328125, c = (a+b)/2 = 0.12890625, f(c) = 0.0139563582 Since f(c) and f(b) have opposite signs, the root must be in the interval [c,b]. Ninth iteration: a = 0.12890625, b = 0.1328125, c = (a+b)/2

Solve the equation $x=\cos x$ by the Bisection method and by the NewtonRaphson method. How many solutions are there? Solve the equation $\sin (x)=\cos x$ by the Bisection method and by the Newton-Raphson method. How many solutions are there?

First equation: $x=\cos x$

Bisection Method: Let $f(x)=x-\cos x$. Then, $f$ is continuous and changes sign on the interval $[0,1]$ since $f(0)=1>0$ and $f(1)=\cos 1-1<0$. Starting with the interval $[a_0,b_0]=[0,1]$, we have:

• $c_0=\frac{a_0+b_0}{2}=\frac{1}{2}$, $f(c_0)=c_0-\cos c_0\approx 0.0707>0$,
• $a_1=a_0$, $b_1=c_0$ since $f(a_0)f(c_0)<0$,
• $c_1=\frac{a_1+b_1}{2}=\frac{1}{4}$, $f(c_1)\approx -0.3065<0$,
• $a_2=c_1$, $b_2=b_1$ since $f(c_1)f(b_1)<0$,
• $c_2=\frac{a_2+b_2}{2}=\frac{3}{8}$, $f(c_2)\approx -0.1207<0$,
• $a_3=c_2$, $b_3=b_2$ since $f(c_2)f(b_2)<0$,
• $c_3=\frac{a_3+b_3}{2}=\frac{7}{16}$, $f(c_3)\approx -0.0247<0$,
• $a_4=c_3$, $b_4=b_3$ since $f(c_3)f(b_3)<0$,
• $c_4=\frac{a_4+b_4}{2}=\frac{15}{32}$, $f(c_4)\approx 0.0230>0$,
• $a_5=a_4$, $b_5=c_4$ since $f(a_4)f(c_4)<0$,
• $c_5=\frac{a_5+b_5}{2}=\frac{23}{64}$, $f(c_5)\approx -0.0016<0$,
• $a_6=c_5$, $b_6=b_5$ since $f(c_5)f(b_5)<0$,
• $c_6=\frac{a_6+b_6}{2}=\frac{87}{256}$, $f(c_6)\approx 0.0105>0$,
• $a_7=a_6$, $b_7=c_6$ since $f(a_6)f(c_6)<0$,
• $c_7=\frac{a_7+b_7}{2}=\frac{169}{512}$, $f(c_7)\approx 0.0044>0$,
• $a_8=a_7$, $b_8=c_7$ since $f(a_7)f(c_7)<0$,
• $c_8=\frac{a_8+b_8}{2}=\frac{325}{1024}$, $f(c_8)\approx 0.0014>0$,
• $a_9=a_8$, $b_9=c_8$ since $f(a_8)f(c_8)<0$,
• \$c_9=\frac{a_9+b_9}{2}=\frac