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微积分代写Calculus代考2023

Posted on 2023年9月28日2023年10月6日 by statistics-lab

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statistics-lab™ 为您的留学生涯保驾护航在代写代考微积分Calculus方面已经树立了自己的口碑, 保证靠谱, 高质量且原创的统计Statistics和数学Math代写服务。我们的专家在代考微积分Calculus代写方面经验极为丰富，各种代写微积分Calculus相关的作业也就用不着说。

微积分代写Calculus代考

微积分或微积分学是数学领域之一，是分析研究的基础部分。微积分由两个支柱组成：微分和积分，前者捕捉局部变化，后者处理局部量的全局聚合，虽然很难确定该领域的范围，但一般包括与多元实值函数的微分和积分有关的事项（包括反函数定理和向量分析）。

微分是考虑函数在某一点的切线或切面的操作。用其他数学术语来说，它基本上是通过线性近似来捕捉复变函数的思想。因此，导数是一种线性映射。然而，将多元函数的导数视为线性映射的想法直到 20 世纪才出现。微分方程是这一思想的自然延伸。

相比之下，积分在几何上等同于求曲线或曲面与坐标轴之间区域的面积（体积）。黎曼（Bernhard Riemann）将定积分（单变量）的值直接定义为矩形近似的极限，并证明了连续函数有积分。根据他的定义，积分被称为黎曼积分。

导数和积分是完全不同的概念，但又密切相关，在单变量的情况下，它们与另一变量的逆运算具有相同的含义（微积分基本定理）。导数是斜率，积分是面积。

微积分包含几个不同的主题，列举如下：

多元微积分Multivariable calculus代写代考

多变量微积分（也称为多元微积分）是一变量微积分到多变量函数微积分的扩展：涉及多个变量（多变量）的函数的微分和积分，而不仅仅是一个变量。

多元微积分可以被认为是高级微积分的基本部分。对于高级微积分，请参阅欧几里得空间上的微积分。三维空间中微积分的特殊情况通常称为向量微积分。

偏微分方程Partial Differential Equations代写代考

微分方程通常有很多解，常常添加边界条件来限制解集。在常微分方程的情况下，每个解都有一系列由某些参数的值表征的解，但在偏微分方程的情况下，将参数视为取函数值更有用。除非方程组是超定的，否则这通常是正确的。

偏微分方程作为描述与流体、引力场和电磁场等场相关的自然现象的模型出现在自然科学领域。这些领域是在飞行模拟、计算机图形学或天气预报等处理中发挥重要作用的工具。广义相对论和量子力学的基本方程也是偏微分方程。它也是经济学尤其是金融工程中的一个重要概念。

其他相关科目课程代写：

常微分方程Ordinary Differential Equations
微分几何学Differential Geometry

微积分Calculus近代史

在欧洲，博纳文图拉-卡瓦列里在他的论文中讨论了将面积和体积确定为极精细区域的面积和体积之和的方法，从而奠定了微分学和积分学的基础。

他在微积分表述方面的工作促使卡瓦列里的微积分与大约同时在欧洲出现的有限差分法相结合。约翰-沃利斯、艾萨克-巴罗和詹姆斯-格里高利进行了这一整合，巴罗和格里高利在 1675 年左右证明了微积分基本定理的第二定理。

艾萨克-牛顿以独特的符号引入了乘积微分定律、链式法则、高阶微分符号、泰勒级数和解析函数等概念，并用它们解决了数学物理中的问题。在出版时，牛顿用等效的几何科目取代了微分，以适应当时的数学术语并避免受到指责。牛顿在《自然哲学的数学原理》中用微分和积分的方法讨论了各种问题，包括天体的轨道、旋转流体表面的形状、地球的偏心率和重物在摆线上滑动的运动。除此之外，牛顿还发展了函数的级数展开，显然他了解泰勒级数的原理。

戈特弗里德-莱布尼兹最初被怀疑剽窃牛顿未发表的论文，但现在被公认为是微积分发展的原始贡献者之一。
正是戈特弗里德-莱布尼茨将这些思想系统化，并将微积分确立为一门严谨的学科。当时，他被指责剽窃牛顿，但今天，他已被公认为建立和发展微分学和积分学的最初贡献者之一。莱布尼茨明确定义了微量的操作规则，使二阶和高阶导数的计算成为可能，并定义了莱布尼茨法则和链式法则。与牛顿不同，莱布尼茨非常注重形式主义，他花了很多天来苦苦思索用什么符号来表示每个概念。

In Europe, Bonaventure Cavalieri laid the foundations of differential and integral calculus by discussing in his dissertation the method of determining the area and volume as the sum of the areas and volumes of very fine regions.

His work on the formulation of the calculus led to the combination of Cavalieri’s calculus with the finite difference method, which appeared in Europe at about the same time. This integration was carried out by John Wallis, Isaac Barrow, and James Gregory, with Barrow and Gregory proving the Second Theorem of the Fundamental Theorem of Calculus around 1675.

Isaac Newton introduced the concepts of the law of product differentiation, the chain rule, higher-order differential notation, Taylor series, and analytic functions in a unique notation, and used them to solve problems in mathematical physics. At the time of publication, Newton replaced differentiation with equivalent geometric subjects to accommodate the mathematical terminology of the time and to avoid censure. In Mathematical Principles of Natural Philosophy, Newton used differential and integral methods to discuss a variety of problems, including the orbits of celestial bodies, the shapes of the surfaces of rotating fluids, the eccentricity of the earth, and the motion of a heavy object sliding on a pendulum. In addition to this, Newton developed the series expansion of functions, and it is clear that he understood the principles of Taylor’s series.

Gottfried Leibniz was initially suspected of plagiarizing Newton’s unpublished papers, but is now recognized as one of the original contributors to the development of calculus.
It was Gottfried Leibniz who systematized these ideas and established calculus as a rigorous discipline. At the time, he was accused of plagiarizing Newton, but today he is recognized as one of the original contributors to the establishment and development of differential and integral calculus. Leibniz explicitly defined the rules for the operation of differentials, made possible the computation of second- and higher-order derivatives, and defined Leibniz’s law and the chain rule. Unlike Newton, Leibniz was very much a formalist and spent many days agonizing over what symbols to use for each concept.

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微积分Calculus的重难点

什么是极限和无穷小Limits and infinitesimals？

微积分通常是通过处理非常小的量而发展起来的。从历史上看，第一种方法就是无穷小。这些对象可以像实数一样处理，但在某种意义上是 “无限小 “的。例如，一个无穷小数可能大于 0，但小于序列 1,1 / 2,1 / 3, \ldots$中的任何数，因此小于任何正实数。从这个角度看，微积分就是处理无穷小数的一系列技术。符号 $d x$ 和 $d y$ 被认为是无穷小数，导数 $d y / d x$ 是它们的比值。${ }^{[37]}$

什么是差分微积分Differential calculus？

数学中的微积分（微分；微积分）是微分和积分微积分的一个分支，其研究重点是量的变化。微积分与积分微积分是一个历史领域，将微积分分为两个分支。

微积分的主要研究方向是函数微分（微分商、微分系数）和相关概念，如无穷小及其应用。函数在所选输入值中的微分商描述了函数在输入值附近的变化率。求微分商的过程也称为微分。从几何学角度看，图形上某一点的微分系数是函数图形切线在该点的斜率（如果存在并定义在该点上）。对于单变量实值函数，函数在某一点的导数通常定义了函数在该点的最佳线性近似值。

什么是莱布尼兹符号Leibniz’s notation？

在微积分中，莱布尼兹符号是为了纪念17世纪德国哲学家和数学家戈特弗里德-威廉-莱布尼兹（Gottfried Wilhelm Leibniz）而命名的，它使用符号$d x$和$d y$分别表示$x$和$y_1$的无限小（或无穷小）增量，正如$\Delta x$和$\Delta y$分别表示$x$和$y$的有限增量一样。

将 $y$ 视为变量 $x$ 的函数，即 $y=f(x)$。如果是这种情况，那么 $y$ 相对于 $x$ 的导数，也就是后来的极限
$$
\lim {Delta x \rightarrow 0} \frac{Delta y}{Delta x}=\lim {Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{Delta x}、
$$
根据莱布尼茨的说法，是$y$的无穷小增量与$x$的无穷小增量之商，或
$$
\frac{d y}{d x}=f^{prime}(x)、
$$
其中右边是约瑟夫-路易-拉格朗日关于 $f$ 在 $x$ 处导数的符号。无穷小的增量称为微分。与此相关的是将无穷小增量相加的积分（例如，将长度、面积和体积作为微小部分的总和来计算），莱布尼茨也为此提供了一个涉及相同微分的密切相关的符号，这种符号的效率在欧洲大陆数学的发展中被证明是决定性的。

微积分Calculus的相关课后作业范例

这是一篇关于微积分Calculus的作业

问题 1.

Let $P=(0,1,0), Q=(2,1,3), R=(1,-1,2)$. Compute $\overrightarrow{P Q} \times \overrightarrow{P R}$ and find the equation of the plane through $P, Q$, and $R$, in the form $a x+b y+c z=d$.

$\overrightarrow{P Q}=\langle 2,0,3\rangle ; \overrightarrow{P R}=\langle 1,-2,2\rangle ; \overrightarrow{P Q} \times \overrightarrow{P R}=\left|\begin{array}{ccc}\hat{\imath} & \hat{\jmath} & k \ 2 & 0 & 3 \ 1 & -2 & 2\end{array}\right|=6 \hat{\imath}-\hat{\jmath}-4 \hat{\boldsymbol{k}}$
Equation of the plane: $6 x-y-4 z=d$. Plane passing through $P: 6 \cdot 0-1-4 \cdot 0=d$.
Equation of the plane: $6 x-y-4 z=-1$.

最后的总结：

通过对微积分Calculus各方面的介绍，想必您对这门课有了初步的认识。如果你仍然不确定或对这方面感到困难，你仍然可以依靠我们的代写和辅导服务。我们拥有各个领域、具有丰富经验的专家。他们将保证你的 essay、assignment或者作业都完全符合要求、100%原创、无抄袭、并一定能获得高分。需要如何学术帮助的话，随时联系我们的客服。

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

时间序列分析代写Time Series Analysis代考2023

Posted on 2023年9月28日2023年10月6日 by statistics-lab

如果你也在时间序列分析Time Series Analysis这个学科遇到相关的难题，请随时添加vx号联系我们的代写客服。我们会为你提供专业的服务。

statistics-lab™ 为您的留学生涯保驾护航在代写代考时间序列分析Time Series Analysis方面已经树立了自己的口碑, 保证靠谱, 高质量且原创的统计Statistics和数学Math代写服务。我们的专家在代考时间序列分析Time Series Analysis代写方面经验极为丰富，各种代写时间序列分析Time Series Analysis相关的作业也就用不着说。

时间序列分析代写Time Series Analysis代考

时间序列是通过连续（或以固定间隔不连续）观察某一现象随时间的变化而获得的一系列数值。例如，在统计或信号处理中，它是按照时间测量的数据序列，以一定的时间间隔（通常是恒定的）测量。如果时间间隔不均匀，则称为点过程。

时间序列分析或时间序列分析是一种解释此类时间序列的方法，目的是找出数据序列背后的理论（为什么时间序列会出现这样的结果？） )或进行预测。时间序列预测包括根据已知的过去事件建立一个未来模型，并在测量之前预测未来可能出现的数据点。例如，根据某只股票过去的价格趋势预测其未来的价格。

时间序列包含几个不同的主题，列举如下：

线性模型general linear model代写代考

时间序列数据有多种形式的模型。经典的著名线性模型是自回归移动平均模型（ARMA），它将自回归（autoregressive；AR）模型与移动平均（moving average；MA）模型相结合。此外，还有自回归求和移动平均模型（ARIMA），它结合了求和模型（综合；I）。这些模型都线性依赖于过去的数据序列和噪声。对过去数据的非线性依赖很有意思，因为它可能会产生混乱的时间序列。

状态空间（控制论）State Space代写代考

状态空间模型是按以下方式表示时间序列 y_t$ 的模型：x_t$ 为状态（不可观测），y_t$ 为观测值，v_t$ 为系统噪声（状态转换噪声），w_t$ 为观测噪声。

$$
\begin{aligned}
x_t & =f_t\left(x_{t-1}, v_t\right) \
y_t & =h_t\left(x_t, w_t\right)
\end{aligned}
$$

此模型可与粒子过滤器（蒙特卡罗方法）一起使用，以找到状态 $x_t$ 的概率分布。对 $f_t$ 和 $h_t$ 函数没有限制，但 $h_t$ 必须能够根据观测值反向计算可能性（概率密度或概率质量）。x_t$ 和 y_t$ 不必是实向量，可以是任何数据结构。
如果状态和值都是实数列向量，函数 $f_t$ 和 $h_t$ 是线性的（矩阵相乘），系统噪声 $v_t$ 和观测噪声 $w_t$ 遵循多元正态分布，则可得到以下结果。

$$
\begin{aligned}
& x_t=F_t x_{t-1}+G_t v_t \
& y_t=H_t x_t+w_t
\end{aligned}
$$

状态 $x_t$ 的概率分布（多元正态分布）可以通过卡尔曼滤波得到精确解；ARMA 和 ARIMA 也可以用这个线性模型来处理。

其他相关科目课程代写：

Exploratory data analysis探索性数据分析
Curve fitting曲线拟合

时间序列分析Time Series Analysis定义

一般来说，数列指的是将对某一现象的若干观察结果按质量特征进行分类。如果该特征是时间，则该数列称为历史数列或时间数列。
观察到的现象称为变量，可以在给定的时间瞬间观察到（状态变量：公司员工人数、证券交易所股票收盘价、利率水平等），也可以在规定长度的周期结束时观察到（流量变量：公司年销售额、季度国内生产总值等）。
我们用 $Y$ 表示现象，用 $Y_t$ 表示时间 $t$ 的观测值，用 $t$ 表示从 1 到 $T$ 的整数，其中 $T$ 是所考虑的时间间隔或时间段的总数。因此，时间序列表示为 $Y_t=left{Y_1，Y_2，Y_3，\ldots，Y_T\right}$，在这种情况下，长度为 $T$。

例如，如果要调查 1981 年第一季度至 2008 年第二季度以百万欧元为单位的环比季度国内生产总值（参考年份：2000 年；原始数据），则有 $T=110$ 的观测值，包括： ${ }^{[1]}$

$Y_1$：1981 年第一季度末的国内生产总值（193 505）；
$Y_{12}$：1983 年第 4 季度末的本地生產總值 (215 584)；
$Y_{55}$：1994 年第三季度末的本地生產總值（263 660）。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

时间序列分析Time Series Analysis的重难点

什么是线性近似linear approximation？

数学中的线性逼近是指用线性函数（或更准确地说，用仿射映射）逼近一般函数。
例如，根据泰勒定理，两次微分的单项式函数 $\mathrm{f}$ 在 $n=1$ 时的值为
$$
f(x)=f(a)+f^{\prime}(a)(x-a)+R_2
$$
可表示为 $R_2$ 是均值项。线性近似丢掉了均值项。
$$
f(x) \approx f(a)+f^{\prime}(a)(x-a)
$$
即 $$f(x)+f^{\prime}(a)(x-a) $$。如果 $x$ 与 $a$ 足够接近，则该近似值成立。这个等式的右边只是 $f$ 图形在 $(a, f(a))$ 处的切线的表格表达式，因此也称为切线近似。
$$
f(x) \approx f(a)
$$
称为 f 在 a 处的标准线性近似值，$\mathrm{x}=\mathrm{a}$$ 称为中心。

什么是信号估算Signal processing？

信号处理（英语：signal processing）是使用数学方法处理（分析和处理）信号（光学、音频、图像信号等）的研究和技术的总称。

它可分为模拟信号处理和数字信号处理。信号处理的基础领域也称为信号理论。

从根本上说，它是从信号到信号的转换，不包括以信号以外的其他形式产生信息的任何工作（例如，产生推理信息的分类和联想、识别和理解）。压缩通常也不包括在内。然而，作为识别、理解和压缩的第一步，信号的转换被称为信号处理。因此，信号处理对这些技术非常重要，也与这些技术息息相关。如果输入和输出是同一类型（物理量）的信号（例如输入和输出具有相同的声压），也称为滤波。

什么是相关系数（Correlation）？

相关系数是两个数据或随机变量之间线性关系强度的度量。相关系数是一个无量纲量，取值介于 -1 和 1 之间的实数。当相关系数为正值时，随机变量具有正相关性；当相关系数为负值时，随机变量具有负相关性。当相关系数为零时，随机变量被认为是不相关的。

例如，发达国家的失业率和实际经济增长率呈强负相关，相关系数接近-1。

只有当两个数据（随机变量）呈线性关系时，相关系数的值才为±1。此外，如果两个随机变量相互独立，则相关系数为 0，反之则不成立。

时间序列分析Time Series Analysis的相关课后作业范例

这是一篇关于时间序列分析Time Series Analysis的作业

问题 1.

For an $A R(1)$ process the optimal forecast $k$ periods in the future is:
$$
E_t\left[Y_{t+k}\right]=\phi^k Y_t
$$

Proof. For an $\mathrm{AR}(1)$ process we have shifting (2.6) $k$ periods in the future that:
$$
Y_{t+k}=\phi Y_{t+k-1}+a_{t+k} .
$$
Applying $E_t$ to both sides we obtain:
$$
E_t\left[Y_{t+k}\right]=\phi E_t\left[Y_{t+k-1}\right]+\underbrace{E_t\left[a_{t+k}\right]}{=0} $$ where: $E_t\left[a{t+k}\right]=0$ since $a_{t+k}$ is $i . i . d$. and hence independent of the information at time $t$. Hence:
$$
E_t\left[Y_{t+k}\right]=\phi \underbrace{E_t\left[Y_{t+k-1}\right]}{=\phi E_t\left[Y{t+k-2}\right]} .
$$
Continuing this process of substitution we have:
$$
E_t\left[Y_{t+k}\right]=\phi^k E_t\left[Y_t\right] .
$$
Since $Y_t$ is observed at time $t$ and is thus in the information set, it follows that:
$$
E_t\left[Y_t\right]=Y_t
$$

最后的总结：

通过对时间序列分析Time Series Analysis各方面的介绍，想必您对这门课有了初步的认识。如果你仍然不确定或对这方面感到困难，你仍然可以依靠我们的代写和辅导服务。我们拥有各个领域、具有丰富经验的专家。他们将保证你的 essay、assignment或者作业都完全符合要求、100%原创、无抄袭、并一定能获得高分。需要如何学术帮助的话，随时联系我们的客服。

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

MATH2099｜Mathematics 2B数学2B 新南威尔士大学

Posted on 2023年9月28日2023年10月6日 by statistics-lab

statistics-lab^TM为您提供新南威尔士大学（The University of New South Wales）Mathematics 2B数学2B澳洲代写代考和辅导服务！

课程介绍：

Statistics: The primary objective of the statistics stream is to enable students to apply and interpret statistical methods in an Engineering context, and to build foundations for future courses in their UG degree programs.

Linear Algebra: Linear algebra is a key tool in all of mathematics and its applications. For example, the output of many electrical circuits depends linearly on the input (over moderate ranges of input), and successfully correcting the trajectory of a space probe involves repeatedly solving systems of linear equations in hundreds of variables. Linear methods are vital in ecological population models, and in mathematics itself. You have begun to understand systems of linear equations and matrices, vector spaces and linear transformations in rst year mathematics courses. In MATH2099, you will learn about geometric transformations: projections (which can also be viewed as least squares approximations), rotations and re ections. You will see how to view many linear transformations as being made up of \stretches” in various directions, (the diagonalisation process), and the more general Jordan form. This will allow you to calculate functions of matrices (such as the exponential of a matrix) and hence to solve systems of linear differential equations.

Mathematics 2B数学2B案例

问题 1.

Let $F$ be bounded increasing with $F(-\infty)=0$, and let $F^{\prime}$ denote its derivative wherever existing. Then the following assertions are true.

(a) If $S$ denotes the set of all $x$ for which $F^{\prime}(x)$ exists with $0 \leq F^{\prime}(x)<$ $\infty$, then $m\left(S^c\right)=0$.
(b) This $F^{\prime}$ belongs to $L^1$, and we have for every $x<x^{\prime}$ :
$$
\int_x^{x^{\prime}} F^{\prime}(t) d t \leq F\left(x^{\prime}\right)-F(x) .
$$
(c) If we put
$$
\forall x: F_{a c}(x)=\int_{-\infty}^x F^{\prime}(t) d t, \quad F_s(x)=F(x)-F_{a c}(x),
$$
then $F_{a c}^{\prime}=F^{\prime}$ a.c. so that $F_s^{\prime}=F^{\prime}-F_{a c}^{\prime}=0$ a.e. and consequently $F_s$ is singular if it is not identically zero.

DEFINTIION. Any positive function $f$ that is equal to $F^{\prime}$ a.e. is called a density of $F . F_{a c}$ is called the absolutely continuous part, $F_s$ the singular part of $F$. Note that the previous $F_d$ is part of $F_s$ as defined here.

It is clear that $F_{a c}$ is increasing and $F_{a c} \leq F$. From (4) it follows that if $x<x^{\prime}$
$$
F_s\left(x^{\prime}\right)-F_s(x)=F\left(x^{\prime}\right)-F(x)-\int_x^{x^{\prime}} f(t) d t \geq 0 .
$$
Hence $F_s$ is also increasing and $F_s \leq F$. We are now in a position to announce the following result

问题 2.

The axioms of finite additivity and of continuity together are equivalent to the axiom of countable additivity.

PROOF. Let $E_n \downarrow$. We have the obvious identity:
$$
E_n=\bigcup_{k=n}^{\infty}\left(E_k \backslash E_{k+1}\right) \cup \bigcap_{k=1}^{\infty} E_k .
$$
If $E_n \downarrow \varnothing$, the last term is the empty set. Hence if (ii) is assumed, we have
$$
\forall n \geq 1: \mathscr{P}\left(E_n\right)=\sum_{k=n}^{\infty} \mathscr{P}\left(E_k \backslash E_{k+1}\right) ;
$$
the series being convergent, we have $\lim {n \rightarrow \infty} \mathscr{P}\left(E_n\right)=0$. Hence (1) is true. Conversely, let $\left{E_k, k \geq 1\right}$ be pairwise disjoint, then $$ \bigcup{k=n+1}^{\infty} E_k \downarrow \varnothing
$$
(why?) and consequently, if (1) is true, then
$$
\lim {n \rightarrow \infty} \mathscr{P}\left(\bigcup{k=n+1}^{\infty} E_k\right)=0 .
$$
Now if finite additivity is assumed, we have
$$
\begin{aligned}
\mathscr{P}\left(\bigcup_{k=1}^{\infty} E_k\right) & =\mathscr{P}\left(\bigcup_{k=1}^n E_k\right)+\mathscr{P}\left(\bigcup_{k=n+1}^{\infty} E_k\right) \
& =\sum_{k=1}^n \mathscr{P}\left(E_k\right)+\mathscr{P}\left(\bigcup_{k=n+1}^{\infty} E_k\right) .
\end{aligned}
$$
This shows that the infinite series $\sum_{k=1}^{\infty} \mathscr{P}\left(E_k\right)$ converges as it is bounded by the first member above. Letting $n \rightarrow \infty$, we obtain

$$
\begin{aligned}
\mathscr{P}\left(\bigcup_{k=1}^{\infty} E_k\right) & =\lim {n \rightarrow \infty} \sum{k=1}^n \mathbb{T}\left(E_k\right)+\lim {n \rightarrow \infty}\left(\bigcup{k=n+1}^{\infty} E_k\right) \
& =\sum_{k=1}^{\infty} \mathscr{P}\left(E_k\right) .
\end{aligned}
$$
Hence (ii) is true.

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

金融工程是使用数学技术来解决金融问题。金融工程使用计算机科学、统计学、经济学和应用数学领域的工具和知识来解决当前的金融问题，以及设计新的和创新的金融产品。

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

术语广义线性模型（GLM）通常是指给定连续和/或分类预测因素的连续响应变量的常规线性回归模型。它包括多元线性回归，以及方差分析和方差分析（仅含固定效应）。

有限元方法代写

有限元方法（FEM）是一种流行的方法，用于数值解决工程和数学建模中出现的微分方程。典型的问题领域包括结构分析、传热、流体流动、质量运输和电磁势等传统领域。

有限元是一种通用的数值方法，用于解决两个或三个空间变量的偏微分方程（即一些边界值问题）。为了解决一个问题，有限元将一个大系统细分为更小、更简单的部分，称为有限元。这是通过在空间维度上的特定空间离散化来实现的，它是通过构建对象的网格来实现的：用于求解的数值域，它有有限数量的点。边界值问题的有限元方法表述最终导致一个代数方程组。该方法在域上对未知函数进行逼近。[1] 然后将模拟这些有限元的简单方程组合成一个更大的方程系统，以模拟整个问题。然后，有限元通过变化微积分使相关的误差函数最小化来逼近一个解决方案。

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

随机分析代写

随机微积分是数学的一个分支，对随机过程进行操作。它允许为随机过程的积分定义一个关于随机过程的一致的积分理论。这个领域是由日本数学家伊藤清在第二次世界大战期间创建并开始的。

时间序列分析代写

随机过程，是依赖于参数的一组随机变量的全体，参数通常是时间。随机变量是随机现象的数量表现，其时间序列是一组按照时间发生先后顺序进行排列的数据点序列。通常一组时间序列的时间间隔为一恒定值（如1秒，5分钟，12小时，7天，1年），因此时间序列可以作为离散时间数据进行分析处理。研究时间序列数据的意义在于现实中，往往需要研究某个事物其随时间发展变化的规律。这就需要通过研究该事物过去发展的历史记录，以得到其自身发展的规律。

回归分析代写

多元回归分析渐进（Multiple Regression Analysis Asymptotics）属于计量经济学领域，主要是一种数学上的统计分析方法，可以分析复杂情况下各影响因素的数学关系，在自然科学、社会和经济学等多个领域内应用广泛。

MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习和应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

MTH3011｜Partial differential equations偏微分方程莫纳什大学

Posted on 2023年9月28日2023年10月6日 by statistics-lab

statistics-lab^TM为您提供蒙纳士大学（Monash University）Partial differential equations偏微分方程澳洲代写代考和辅导服务！

课程介绍：

Introduction to PDEs; first-order PDEs and characteristics, the advection equation, nonlinear equations. Classes of second-order PDEs; boundary and/or initial conditions for well-posed problems. The wave equation: exact solutions on infinite and finite spatial domains, other hyperbolic PDEs, reflection of waves. The heat equation: exact solutions on infinite domain, separation of variables for fixed and/or insulating boundary conditions. Finite-difference methods for ODEs, truncation error. Forward, backward and Crank-Nicolson numerical methods for the heat equation, truncation errors and stability analysis. Numerical methods for the advection equation; upwind differencing. Exact solutions of Laplace’s equation in various domains. Numerical methods for Laplace’s and Poisson’s equation.

MTH3011｜Partial differential equations偏微分方程莫纳什大学

Partial differential equations偏微分方程案例

问题 1.

II. Let $u \in C^{\infty}\left(\mathbb{R}^3\right)$ be a harmonic function on $\mathbb{R}^3$ :
$$
\Delta u(x)=0, \quad x \in \mathbb{R}^3 .
$$
Assume that $|u(x)| \leq \sqrt{|x|}$ holds for all $x$, where $x \stackrel{\text { def }}{=}\left(x^1, x^2, x^3\right)$, and $|x| \stackrel{\text { def }}{=} \sqrt{\left(x^1\right)^2+\left(x^2\right)^2+\left(x^3\right)^2}$. Show that $u(x)=0$ for all $x \in \mathbb{R}^3$.

Define $v(x)=u(x)+\sqrt{R}$. Then $\Delta v=0$, and $v(x) \geq 0$ for $x \in B_R(0)$. Thus by Harnack’s inequality
$$
\frac{R(R-|x|)}{(R+|x|)^2} v(0) \leq v(x) \leq \frac{R(R+|x|)}{(R-|x|)^2} v(0)
$$
holds for $x \in B_R(0)$. Thus, for $x \in B_R(0)$, we have
$$
\left{\frac{R(R-|x|)}{(R+|x|)^2}-1\right} \sqrt{R} \leq u(x) \leq\left{\frac{R(R+|x|)}{(R-|x|)^2}-1\right} \sqrt{R} .
$$
For a fixed $x$, we let $R \rightarrow \infty$ and apply L’Hôpital’s rule to conclude that
$$
0 \leq u(x) \leq 0
$$
Thus,
$$
u(x)=0
$$

问题 2.

Let $u(t, x) \in C^{1,2}([0,2] \times[0,1])$ be a solution to the following initial/boundary-value problem:
$$
\begin{aligned}
\partial_t u-\partial_x^2 u & =-u, \quad(t, x) \in[0,2] \times[0,1], \
u(0, x) & =f(x), \quad x \in[0,1], \
u(t, 0) & =g(t), \quad u(t, 1)=h(t), \quad t \in[0,2] .
\end{aligned}
$$
Assume that $f(x) \leq 0$ for $x \in[0,1]$ and that $g(t) \leq 0, h(t) \leq 0$ for $t \in[0,2]$. Prove that $u(t, x) \leq 0$ holds for all $(t, x) \in[0,2] \times[0,1]$.

Let $\left(t_0, x_0\right)$ be the point in $[0,2] \times[0,1]$ at which $u$ achieves its max. We will show $t$ $u\left(t_0, x_0\right) \leq 0$, which implies the desired conclusion.

If $t_0=0, x_0=0$, or $x_0=1$, then the conditions on $f, g, h$ immediately imply $\mathrm{t}$ $u\left(t_0, x_0\right) \leq 0$, and we are done. So let us assume that none of these cases occur.

If $t_0=2$, then by the above remarks we can assume that $x_0 \in(0,1)$. Then $\partial_t u\left(2, x_0\right)$ since otherwise we could slightly decrease $t_0$ and cause the value of $u$ to increase, which tradicts the definition of a max. Also, by standard calculus, we must have that $\partial_x u\left(2, x_0\right.$ 0 , and by Taylor expanding, we can conclude that $\partial_x^2 u\left(2, x_0\right) \leq 0$ at the $\max$. $\mathrm{Tl}$ $\partial_t u\left(2, x_0\right)-\partial_x^2 u\left(2, x_0\right) \geq 0$. Using the PDE, we thus conclude that $-u\left(2, x_0\right) \geq 0$, we are done.

For the final case, we assume that $0<t_0<2$ and $x_0 \in(0,1)$. Then by standard calcu $\partial_t u\left(t_0, x_0\right)=0$ at the max. Also, as above, by standard calculus $\partial_x^2 u\left(t_0, x_0\right) \leq 0$ at the $\mathrm{m}$ Thus, $\partial_t u\left(t_0, x_0\right)-\partial_x^2 u\left(t_0, x_0\right) \geq 0$. Using the PDE, we thus conclude that $-u\left(t_0, x_0\right) \geq$ and we are again done.

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

MTH3051｜Introduction to computational mathematics计算数学入门莫纳什大学

Posted on 2023年9月28日2023年10月6日 by statistics-lab

statistics-lab^TM为您提供蒙纳士大学（Monash University）Introduction to computational mathematics计算数学入门澳洲代写代考和辅导服务！

课程介绍：

When mathematics is used in real-world applications, it almost always involves the use of computers. This unit provides an introduction to numerical methods for solving maths-related problems on computers. Topics covered include an introduction to Matlab programming; error analysis; methods for solving linear systems, methods for finding roots of nonlinear equations; polynomial interpolation; numerical differentiation and integration; and numerical methods for ordinary differential equations. You will receive a solid introduction to the theory of the numerical methods (with derivations of the methods and some proofs) and will learn to implement the computational methods efficiently in Matlab. The methods and techniques learned have broad applicability in areas that include the natural sciences, engineering, the biomedical sciences, finance, business, machine learning, and data science.

MTH3051｜Introduction to computational mathematics计算数学入门莫纳什大学

Introduction to computational mathematics计算数学入门案例

问题 1.

Let $V$ be the vector space of row vectors on which $\mathrm{GL}(d, q)$ acts, let $G$ be a subgroup of $\mathrm{GL}(d, q)$, and let $Z$ be the subgroup of scalar matrices of $G$. Then one of the following is true:

Gacts reducibly.
G acts imprimitively: $G$ preserves a decomposition of $V$ as a direct sum $V_1 \oplus V_2 \oplus \cdots \oplus V_r$ of $r>1$ subspaces of dimension $s$, which are permuted transitively by $G$, and so $G \leq \mathrm{GL}(s, q)$ ? $\operatorname{Sym}(r)$.

Gacts on Vas a group of semilinear automorphisms of a space of dimensional d/e over the extension field $\mathbb{F}{q e}$ for some e $>1$, and so $G$ embeds in $\Gamma \mathrm{L}\left(d / e, q^{\mathrm{e}}\right)$. (This covers the class of ‘absolutely reducible’ matrix groups, where G embeds in $\mathrm{GL}\left(d / e, q^e\right)$.) $G$ preserves a decomposition of $V$ as a tensor product $U \otimes W$ of spaces of dimensions $d_1, d_2>1$ over $F$. Then $G$ is a subgroup of $a$ central product of $\mathrm{GL}\left(d_1, q\right)$ and $\mathrm{GL}\left(d_2, q\right)$. $G$ is definable modulo scalars over a subfield: for some proper subfield $\mathbb{F}{q^{\prime}}$ of $\mathbb{F}_{\varphi} G^g \leq \mathrm{GL}\left(d, q^{\prime}\right) . Z$ for some $g \in \mathrm{GL}(d, q)$.

For some prime $r, d=r^m$ and $G / Z$ is contained in the normalizer of an extraspecial group of order $r^{2 \mathrm{~m}+1}$, or of a group of order $2^{2 \mathrm{~m}+2}$ and symplectic type.
$G$ is tensor-induced: it preserves a decomposition of $V$ as $V_1 \otimes V_2 \otimes \cdots \otimes V_m$, where each $V_i$ has dimension $r>1$ and the set of $V_i$ is permuted by $G$, and so $G / Z \leq \operatorname{PGL}(r, q)$ ? $\operatorname{Sym}(m)$.
G normalizes a classical group in its natural representation.
$G$ is almost simple modulo scalars.

问题 2.

Let $G:=\langle(1,2,3,4),(1,3)\rangle \cong D_8$ the dihedral group of order 8 . We shall introduce a variety of polycyclic sequences for this group $G$. Using this example, we note that the length of a polycyclic sequence for $G$ is not uniquely defined by $G$. Also, we observe that different polycyclic sequences of $G$ may define the same polycyclic series.

a) Let $G_2:=\langle(1,2,3,4)\rangle \cong C_4$. Then the chain $G=G_1 \geq G_2 \geq G_3=1$ is a polycyclic series for $G$. This series allows several different polycyclic sequences. For example, the sequences $X:=[(1,3),(1,2,3,4)]$ and $Y:=[(2,4),(1,4,3,2)]$ are polycyclic sequences defining this series. They have the relative orders $R(X)=R(Y)=(2,4)$ and the finite orders set $I(X)=I(Y)={1,2}$.
b) Let $G_2:=\langle(1,2)(3,4),(1,3)(2,4)\rangle \cong V_4$ and $G_3:=\langle(1,3)(2,4)\rangle \cong C_2$. Then the chain $G=G_1 \geq G_2 \geq G_3 \geq G_4=1$ is a polycyclic series for $G$. This series also allows several different polycyclic sequences. For example, the sequences $X:=[(2,4),(1,2)(3,4),(1,3)(2,4)]$ and $Y:=[(1,2,3,4)$, $(1,2)(3,4),(1,3)(2,4)]$ are polycyclic sequences defining this series. They have the relative orders $R(X)=R(Y)=(2,2,2)$ and the finite orders $I(X)=I(Y)={1,2,3}$.

We note that $8=|G|=2 \cdot 2 \cdot 2=2 \cdot 4$ is the product of the relative orders in all

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

MATH2831｜Linear models线性模型新南威尔士大学

Posted on 2023年9月28日2023年10月6日 by statistics-lab

statistics-lab^TM为您提供新南威尔士大学（英语：The University of New South Wales，简称UNSW）Linear Models线性模型澳洲代写代考和辅导服务！

课程介绍：

This course introduces students to statistical model building using the important class of linear models. Topics covered in the course include how to estimate parameters in linear models, how to compare models using hypothesis testing, how to select a good model or models when prediction of the
response is the goal, and how to detect violations of model assumptions and observations which have undue influence on decisions of interest. Concepts are illustrated with applications from finance, economics, medicine, environmental science and engineering. Linear models are a fundamental component of statistical practice and the course is a solid background for more advanced statistical courses.

Linear models线性模型案例

问题 1.

(a) $A$ and $B$ are any matrices with the same number of rows. What can you say (and explain why it is true) about the comparison of
$$
\text { rank of } A \quad \text { rank of the block matrix }\left[\begin{array}{ll}
A & B
\end{array}\right]
$$
(b) Suppose $B=A^2$. How do those ranks compare? Explain your reasoning.
(c) If $A$ is $m$ by $n$ of rank $r$, what are the dimensions of these nullspaces?
$$
\text { Nullspace of } A \quad \text { Nullspace of }\left[\begin{array}{ll}
A & A
\end{array}\right]
$$

(a) All you can say is that $\operatorname{rank} A \leq \operatorname{rank}[A B]$. ( $A$ can have any number $r$ of pivot columns, and these will all be pivot columns for $[A B]$; but there could be more pivot columns among the columns of $B$.)
(b) Now $\operatorname{rank} A=\operatorname{rank}\left[A A^2\right]$. (Every column of $A^2$ is a linear combination of columns of $A$. For instance, if we call $A$ ‘s first column $a_1$, then $A a_1$ is the first column of $A^2$. So there are no new pivot columns in the $A^2$ part of $\left[A A^2\right]$.)
(c) The nullspace $N(A)$ has dimension $n-r$, as always. Since $[A A]$ only has $r$ pivot columns – the $n$ columns we added are all duplicates – $[A A]$ is an $m$-by- $2 n$ matrix of rank $r$, and its nullspace $N([A A])$ has dimension $2 n-r$.

问题 2.

Suppose $A$ is a 5 by 3 matrix and $A x$ is never zero (except when $x$ is the zero vector).
(a) What can you say about the columns of $A$ ?
(b) Show that $A^{\mathrm{T}} A x$ is also never zero (except when $x=0$ ) by explaining this key step:

If $A^{\mathrm{T}} A x=0$ then obviously $x^{\mathrm{T}} A^{\mathrm{T}} A x=0$ and then (WHY ?) $A x=0$.
(c) We now know that $A^{\mathrm{T}} A$ is invertible. Explain why $B=\left(A^{\mathrm{T}} A\right)^{-1} A^{\mathrm{T}}$ is a one-sided inverse of $A$ (which side of $A$ ?). $B$ is NOT a 2 -sided inverse of $A$ (explain why not).

(a) $N(A)=0$ so $A$ has full column rank $r=n=3$ : the columns are linearly independent.
(b) $x^{\mathrm{T}} A^{\mathrm{T}} A x=(A x)^{\mathrm{T}} A x$ is the squared length of $A x$. The only way it can be zero is if $A x$ has zero length (meaning $A x=0$ ).
(c) $B$ is a left inverse of $A$, since $B A=\left(A^{\mathrm{T}} A\right)^{-1} A^{\mathrm{T}} A=I$ is the (3-by-3) identity matrix. $B$ is not a right inverse of $A$, because $A B$ is a 5 -by-5 matrix but can only have rank 3 . (In fact, $B A=A\left(A^{\mathrm{T}} A\right)^{-1} A^{\mathrm{T}}$ is the projection onto the (3-dimensional) column space of $A$.)

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

MATH2521｜Higher Linear Models高等线性模型新南威尔士大学

Posted on 2023年9月28日2023年10月6日 by statistics-lab

statistics-lab^TM为您提供新南威尔士大学（英语：The University of New South Wales，简称UNSW）Higher Linear Models高等线性模型澳洲代写代考和辅导服务！

课程介绍：

This course aims to extend our understanding of differential and integral calculus from functions of a single real variable to functions of a complex variable. The differences between the two are often unexpected and very surprising. The theory of complex valued functions will give us many new insights into the real variable theory.

MATH2521｜Higher Linear Models高等线性模型新南威尔士大学

Higher Linear Models高等线性模型案例

问题 1.

$1(\mathbf{4}+\mathbf{7}=\mathbf{1 1}$ pts. $) \quad$ Suppose $A$ is 3 by 4 , and $A x=0$ has exactly 2 special solutions:
$$
x_1=\left[\begin{array}{l}
1 \
1 \
1 \
0
\end{array}\right] \quad \text { and } \quad x_2=\left[\begin{array}{r}
-2 \
-1 \
0 \
1
\end{array}\right]
$$
(a) Remembering that $A$ is 3 by 4 , find its row reduced echelon form $R$.
(b) Find the dimensions of all four fundamental subspaces $C(A), N(A)$, $C\left(A^{\mathrm{T}}\right), N\left(A^{\mathrm{T}}\right)$.

You have enough information to find bases for one or more of these subspaces – find those bases.

(a) Each special solution tells us the solution to $R x=0$ when we set one free variable $=1$ and the others $=0$. Here, the third and fourth variables must be the two free variables, and the other two are the pivots: $R=\left[\begin{array}{cccc}1 & 0 & * & * \ 0 & 1 & * & * \ 0 & 0 & 0 & 0\end{array}\right]$ Now multiply out $R x_1=0$ and $R x_2=0$ to find the *’s: $R=\left[\begin{array}{rrrr}1 & 0 & -1 & 2 \ 0 & 1 & -1 & 1 \ 0 & 0 & 0 & 0\end{array}\right]$ (The *’s are just the negatives of the special solutions’ pivot entries.)
(b) We know the nullspace $N(A)$ has $n-r=4-2=2$ dimensions: the special solutions $x_1, x_2$ form a basis.

The row space $C\left(A^{\mathrm{T}}\right)$ has $r=2$ dimensions. It’s orthogonal to $N(A)$, so just pick two linearly-independent vectors orthogonal to $x_1$ and $x_2$ to form a basis: for example, $x_3=\left[\begin{array}{r}1 \ 0 \ -1 \ 2\end{array}\right], x_4=\left[\begin{array}{r}0 \ 1 \ -1 \ 1\end{array}\right]$.
(Or: $C\left(A^{\mathrm{T}}\right)=C\left(R^{\mathrm{T}}\right)$ is just the row space of $R$, so the first two rows are a basis. Same thing!)

The column space $C(A)$ has $r=2$ dimensions (same as $C\left(A^{\mathrm{T}}\right)$ ). We can’t write down a basis because we don’t know what $A$ is, but we can say that the first two columns of $A$ are a basis.

The left nullspace $N\left(A^{\mathrm{T}}\right)$ has $m-r=1$ dimension; it’s orthogonal to $C(A)$, so any vector orthogonal to the first two columns of $A$ (whatever they are) will be a basis.

问题 2.

(a) Find the inverse of a 3 by 3 upper triangular matrix $U$, with nonzero entries $a, b, c, d, e, f$. You could use cofactors and the formula for the inverse. Or possibly Gauss-Jordan elimination.
Find the inverse of $U=\left[\begin{array}{ccc}a & b & c \ 0 & d & e \ 0 & 0 & f\end{array}\right]$.
(b) Suppose the columns of $U$ are eigenvectors of a matrix $A$. Show that $A$ is also upper triangular.
(c) Explain why this $U$ cannot be the same matrix as the first factor in the Singular Value Decomposition $A=U \Sigma V^{\mathrm{T}}$.

(a) By elimination: (We keep track of the elimination matrix $E$ on one side, and the product $E U$ on the other. When $E U=I$, then $E=U^{-1}$.)
$$
\begin{aligned}
& {\left[\begin{array}{rrrrrr}
a & b & c & 1 & 0 & 0 \
0 & d & e & 0 & 1 & 0 \
0 & 0 & f & 0 & 0 & 1
\end{array}\right] } \rightsquigarrow\left[\begin{array}{rrrrrr}
1 & b / a & c / a & 1 / a & 0 & 0 \
0 & 1 & e / d & 0 & 1 / d & 0 \
0 & 0 & 1 & 0 & 0 & 1 / f
\end{array}\right] \
& \rightsquigarrow\left[\begin{array}{rrrrrr}
1 & 0 & 0 & 1 / a & -b / a d & (b e-c d) / a d f \
0 & 1 & 0 & 0 & 1 / d & -e / d f \
0 & 0 & 1 & 0 & 0 & 1 / f
\end{array}\right]=\left[\begin{array}{ll}
I & U^{-1}
\end{array}\right]
\end{aligned}
$$
By cofactors: (Take the minor, then “checkerboard” the signs to get the cofactor matrix, then transpose and $\operatorname{divide}$ by $\operatorname{det}(U)=a d f$.)
$$
\begin{aligned}
& {\left[\begin{array}{lll}
a & b & c \
0 & d & e \
0 & 0 & f
\end{array}\right] \rightsquigarrow\left[\begin{array}{rrr}
d f & 0 & 0 \
b f & a f & 0 \
b e-c d & a e & a d
\end{array}\right] \rightsquigarrow\left[\begin{array}{rrr}
d f & 0 & 0 \
-b f & a f & 0 \
b e-c d & -a e & a d
\end{array}\right] \rightsquigarrow\left[\begin{array}{rrr}
d f & -b f & b e-c d \
0 & a f & -a e \
0 & 0 & a d
\end{array}\right] \rightsquigarrow} \
& {\left[\begin{array}{rrr}
1 / a & -b / a d & (b e-c d) / a d f \
0 & 1 / d & -e / d f \
0 & 0 & 1 / f
\end{array}\right]=U^{-1}}
\end{aligned}
$$
(b) We have a complete set of eigenvectors for $A$, so we can diagonalize: $A=U \Lambda U^{-1}$. We know $U$ is upper-triangular, and so is the diagonal matrix $\Lambda$, and we’ve just shown that $U^{-1}$ is upper-triangular too. So their product $A$ is also upper-triangular.
(c) The columns aren’t orthogonal! (For example, the product $u_1^{\mathrm{T}} u_2$ of the first two columns is $a b+0 d+0 \cdot 0=a b$, which is nonzero because we’re assuming all the entries are nonzero.)

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

MATH2521｜Complex analysis复杂分析新南威尔士大学

Posted on 2023年9月28日2023年10月6日 by statistics-lab

statistics-lab^TM为您提供新南威尔士大学（英语：The University of New South Wales，简称UNSW）Complex analysis复杂分析澳洲代写代考和辅导服务！

课程介绍：

Complex analysis复杂分析案例

问题 1.

Evaluate the following integrals, justifying your procedures. For c) and d) you should also state why the integral is well defined (i.e., independent of the path taken).
(a) $\int_C \frac{2 d z}{z^2-1}$, where $C$ is the circle with radius $1 / 2$, centre 1 , positively oriented;
(b) $\int_C\left(e^z+\frac{1}{z}\right) d z$, where $\mathrm{C}$ is the lower half of the circle with radius 1 , centre 0 , negatively oriented;
(c) $\int_C z e^{z^2} d z$
(d) $\int_C^C \cosh z d z$.

Solution. (a) Notice that
$$
\frac{2}{z^2-1}=\frac{1}{z-1}-\frac{1}{z+1}
$$
Thus
$$
\int_C \frac{2}{z^2-1} d z=\int_C \frac{1}{z-1} d z-\int_C \frac{1}{z+1} d z
$$
On the one hand,
$$
\int_C \frac{1}{z-1} d z=2 \pi i
$$
by Cauchy integral formula, $f\left(z_0\right)=\frac{1}{2 \pi i} \int_C \frac{f(z)}{z-z_0} d z$.
On the other hand, since $1 /(z+1)$ is analytic on and inside $C$, then
$$
\int_C \frac{1}{z+1} d z=0
$$
by Cauchy’s Theorem. Therefore,
$$
\int_C \frac{2}{z^2-1} d z=2 \pi i
$$
(b) Notice that the integrand $f(z)=e^z-1 / z$ is analytic on $C$. The function
$$
F(z)=e^z-\log z
$$
serves as an antiderivative of $f(z)$. Here $\log z$ is a branch of the logarithm chosen with the branch cut on the positive imaginary axis. That is,
$$
\log z=\ln r+i \theta, \quad\left(r>0, \frac{\pi}{2}<\theta<\frac{5 \pi}{2}\right) .
$$
Thus
$$
\int_C\left(e^z-\frac{1}{z}\right) d z=\left.\left(e^z-\log z\right)\right|_1 ^{-1}=\frac{1}{e}-\pi i-e+2 \pi i=\frac{1}{e}-e+\pi i .
$$

(c) Since the integrand $f(z)=z e^{z^2}$ is analytic, the integral is path independent. An antiderivative of $f(z)$ is
$$
F(z)=\frac{e^{z^2}}{2} .
$$
Thus
$$
\int_C z e^{z^2} d z=\left.\frac{e^{z^2}}{2}\right|0 ^i=\frac{1}{2 e}-\frac{1}{2} . $$ (d) Since the integrand $f(z)=\cosh z$ is analytic, the integral is path independent. An antiderivative of $f(z)$ is $$ F(z)=\sinh z $$ Thus $$ \int_C \cosh z d z=\left.\sinh z\right|{\pi i} ^{2 \pi i}=\sinh (2 \pi i)-\sinh (\pi i)=0 .
$$

问题 2.

Let $C_R$ be the circle with radius $R$, centre 0 , positively oriented. Show that
$$
\lim {R \rightarrow \infty} \int{C_R} \frac{z^2+4 z+7}{\left(z^2+4\right)\left(z^2+2 z+2\right)} d z=0 .
$$
Use this fact to prove that
$$
\int_C \frac{z^2+4 z+7}{\left(z^2+4\right)\left(z^2+2 z+2\right)} d z=0 .
$$
where $C$ is the circle with radius 5 , centre 2 , positively oriented.

Solution. Recall that for a contour $C$ of length $L$ and a piecewise continuous $f(z)$ on $C$, if $M$ is a nonnegative constant such that $|f(z)|<M$ for all points $z$ on $C$ at which $f(z)$ is defined, then
$$
\left|\int_C f(z) d z\right|<M L .
$$
Now, consider the function
$$
f(z)=\frac{z^2+4 z+7}{\left(z^2+4\right)\left(z^2+2 z+2\right)} .
$$
For $|z|$ large, we have that
$$
|f(z)|=\frac{\left|1+\frac{4}{z}+\frac{7}{z^2}\right|}{\left|1+\frac{4}{z^2}\right|\left|z^2+2 z+2\right|} \approx \frac{1}{|z|^2}
$$

On the circle $C_R$ we have that $|z|=R$. Thus, for $R$ large we have that
$$
|f(z)|<\frac{2}{R^2} $$ Since the length of $C_R$ is $2 \pi R$, then $$ \left|\int_{C_R} f(z) d z\right|7$, then
$$
\int_C f(z) d z=\int_{C_R} f(z) d z .
$$
Thus
$$
\lim {R \rightarrow \infty} \int_C f(z) d z=\lim {R \rightarrow \infty} \int_{C_R} f(z) d z=0 .
$$
Hence
$$
\int_C f(z) d z=0 .
$$

问题 3.

Let $C_1$ denote the positively oriented boundary of the curve given by $|x|+|y|=2$ and $C_2$ be the positively oriented circle $|z|=4$. Apply Cauchy Integral Theorem to show that
$$
\int_{C_1} f(z) d z=\int_{C_2} f(z) d z
$$
when
(a) $f(z)=\frac{z+1}{z^2+1}$;
(b) $f(z)=\frac{z+2}{\sin (z / 2)}$;
(c) $f(z)=\frac{\sin (z)}{z^2+6 z+5}$.

Solution. By Cauchy Integral Theorem, $\int_{C_1} f(z) d z=\int_{C_2} f(z) d z$ if $f(z)$ is analytic on and between $C_1$ and $C_2$. Hence it is enough to show that $f(z)$ is analytic in ${|x|+|y| \geq 2,|z| \leq 4}$.
(a) $f(z)$ is analytic in ${z \neq \pm i}$. Since $\pm i \in{|x|+|y|<2}, f(z)$ is analytic in ${|x|+|y| \geq 2,|z| \leq 4}$. (b) $f(z)$ is analytic in ${z: \sin (z / 2) \neq 0}={z \neq 2 n \pi: n \in \mathbb{Z}}$. Since $2 n \pi \in{|x|+$ $|y|<2}$ for $n=0$ and $|2 n \pi|>4$ for $n \neq 0$ and $n \in \mathbb{Z}, f(z)$ is analytic in ${|x|+|y| \geq 2,|z| \leq 4}$.
(c) $f(z)$ is analytic in ${z \neq-1,-5}$. Since $-1 \in{|x|+|y|<2}$ for $n=0$ and $|-5|>4, f(z)$ is analytic in ${|x|+|y| \geq 2,|z| \leq 4}$.

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

MAST20022｜Group Theory and Linear Algebra群论与线性代数墨尔本大学

Posted on 2023年9月28日2023年10月6日 by statistics-lab

statistics-lab^TM为您墨尔本大学（The University of Melbourne，简称UniMelb，中文简称“墨大”）Group Theory and Linear Algebra群论与线性代数澳洲代写代考和辅导服务！

课程介绍：

This subject introduces the theory of groups, which is at the core of modern algebra, and which has applications in many parts of mathematics, chemistry, computer science and theoretical physics. It also develops the theory of linear algebra, building on material in earlier subjects and providing both a basis for later mathematics studies and an introduction to topics that have important applications in science and technology.

Topics include: modular arithmetic and RSA cryptography; abstract groups, homomorphisms, normal subgroups, quotient groups, group actions, symmetry groups, permutation groups and matrix groups; theory of general vector spaces, inner products, linear transformations, spectral theorem for normal matrices, Jordan normal form.

MAST20022｜Group Theory and Linear Algebra群论与线性代数墨尔本大学

Group Theory and Linear Algebra群论与线性代数案例

问题 1.

Let $\mathbf{A}$ be a $n$ by $n$ matrix, then the following 4 statements are equivalent:
(a) The matrix $\mathbf{A}$ is invertible (non-singular).
(b) The linear system $\mathbf{A x}=\mathbf{O}$ only has the trivial solution $\mathbf{x}=\mathbf{O}$.

(c) The reduced row echelon form of the matrix $\mathbf{A}$ is the identity matrix $\mathbf{I}$.
(d) $\mathbf{A}$ is a product of elementary matrices.

We show (a) implies (b) and (b) implies (c) and (c) implies (d) and (d) implies (a). In symbolic notation, this is (a) $\Rightarrow(b) \Rightarrow(c) \Rightarrow(d) \Rightarrow(a)$. We first prove (a) implies (b).
For $(\mathbf{a}) \Rightarrow(\mathbf{b})$ :
We assume statement (a) to be true and deduce statement (b).
By assuming $\mathbf{A}$ is an invertible matrix we need to show that the linear system $\mathbf{A x}=\mathbf{O}$ only has the trivial solution $\mathbf{x}=\mathbf{O}$. Consider the linear system $\mathbf{A x}=\mathbf{O}$, because $\mathbf{A}$ is invertible, therefore there is a unique matrix $\mathbf{A}^{-1}$ such that $\mathbf{A}^{-1} \mathbf{A}=\mathbf{I}$. Multiplying both sides of $\mathbf{A x}=\mathbf{O}$ by $\mathbf{A}^{-1}$ yields
$$
\begin{aligned}
& \underbrace{\mathbf{A}^{-1} \mathbf{A} \mathbf{x}}_{=\mathbf{I}}=\mathbf{A}^{-1} \mathbf{O} \
& \qquad \mathbf{I x}=\mathbf{A}^{-1} \mathbf{O}=\mathbf{O} \
& \mathbf{I x}=\mathbf{O} \text { which gives } \mathbf{x}=\mathbf{O} \quad[\text { because } \mathbf{I x}=\mathbf{x}]
\end{aligned}
$$
The answer $\mathbf{x}=\mathbf{O}$ means that we only have the trivial solution $\mathbf{x}=\mathbf{O}$. Hence we have shown (a) $\Rightarrow$ (b). Next we prove (b) implies (c).
For $(b) \Rightarrow(c)$ :
The procedure to prove (b) $\Rightarrow$ (c) is to assume statement (b) and deduce (c). This time we assume that $\mathbf{A x}=\mathbf{O}$ only has the trivial solution $\mathbf{x}=\mathbf{O}$ and, by using this, prove that the reduced row echelon form of the matrix $\mathbf{A}$ is the identity matrix $\mathbf{I}$.

The reduced row echelon form of matrix A cannot have a row (equation) of zeros, otherwise we would have $n$ unknowns but less than $n$ non-zero rows (equations), which means that by Proposition (1.31): if $r<n$ then the linear system $\mathbf{A x}=\mathbf{O}$ has an infinite number of solutions.

We would have an infinite number of solutions. However, we only have a unique solution to $\mathbf{A x}=\mathbf{O}$, therefore there are no zero rows.

Question (7) of Exercises 1.7 claims: the reduced row echelon form of a matrix is either the identity $\mathbf{I}$ or it contains a row of zeros.
Hence the reduced row echelon form of $\mathbf{A}$ is the identity matrix. We have (b) $\Rightarrow$ (c). For $(c) \Rightarrow$ (d):
In this case, we assume part (c), that is ‘the reduced row echelon form of the matrix $\mathbf{A}$ is the identity matrix I’, which means that the matrix $\mathbf{A}$ is row equivalent to the identity matrix I. By definition (1.33):
$\mathbf{B}$ is row equivalent to a matrix $A$ if and only if $\mathbf{B}=\mathbf{E}n \mathbf{E}{n-1} \cdots \mathbf{E}2 \mathbf{E}_1 \mathbf{A}$. There are elementary matrices $\mathbf{E}_1, \mathbf{E}_2, \mathbf{E}_3, \ldots$ and $\mathbf{E}_k$ such that $$ \begin{aligned} \mathbf{A} & =\mathbf{E}_k \mathbf{E}{k-1} \cdots \mathbf{E}2 \mathbf{E}_1 \mathbf{I} \ & =\mathbf{E}_k \mathbf{E}{k-1} \cdots \mathbf{E}_2 \mathbf{E}_1
\end{aligned}
$$
[because $\mathbf{A}$ is row equivalent to $\mathbf{I}$ ]

This shows that matrix $\mathbf{A}$ is a product of elementary matrices. We have (c) $\Rightarrow$ (d). For $(\mathbf{d}) \Rightarrow$ (a):
In this last case, we assume that matrix $\mathbf{A}$ is a product of elementary matrices and deduce that matrix A is invertible. By Proposition (1.34): an elementary matrix is invertible.

We know that elementary matrices are invertible (have an inverse) and therefore the matrix multiplication $\mathbf{E}k \mathbf{E}{k-1} \cdots \mathbf{E}2 \mathbf{E}_1$ is invertible. In fact, we have $$ \begin{aligned} \mathbf{A}^{-1} & =\left(\mathbf{E}_k \mathbf{E}{k-1} \cdots \mathbf{E}2 \mathbf{E}_1\right)^{-1} \ & =\mathbf{E}_1^{-1} \mathbf{E}_2^{-1} \cdots \mathbf{E}{k-1}^{-1} \mathbf{E}_k^{-1} \quad\left[\text { because }(\mathbf{X Y Z})^{-1}=\mathbf{Z}^{-1} \mathbf{Y}^{-1} \mathbf{X}^{-1}\right]
\end{aligned}
$$
Hence the matrix $\mathbf{A}$ is invertible, which means that we have proven (d) $\Rightarrow$ (a). We have shown (a) $\Rightarrow$ (b) $\Rightarrow$ (c) $\Rightarrow$ (d) $\Rightarrow$ (a) which means that the four statements (a), (b), (c) and (d) are equivalent.

问题 2. Let $\mathbf{A}$ be a square matrix and $\mathbf{R}$ be the reduced row echelon form of $\mathbf{A}$. Then $\mathbf{R}$ has at least one row of zeros $\Leftrightarrow \mathbf{A}$ is non-invertible (singular).

The proof of this result can be made a lot easier if we understand some mathematical logic. Generally to prove a statement of the type $P \Leftrightarrow Q$ we assume $P$ to be true and then deduce $Q$. Then we assume $Q$ to be true and deduce $P$.
However, in mathematical logic this can also be proven by showing:
$$
(\operatorname{Not} P) \Leftrightarrow(\operatorname{Not} Q)
$$
Means that $(\operatorname{Not} P) \Rightarrow(\operatorname{Not} Q)$ and $(\operatorname{Not} Q) \Rightarrow(\operatorname{Not} P)$. This is because statements $P \Leftrightarrow Q$ and $(\operatorname{Not} P) \Leftrightarrow($ Not $Q)$ are equivalent. See website for more details.
The following demonstrates this proposition.

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COSC2002｜Computational Modelling计算建模悉尼大学

Posted on 2023年9月28日2024年1月7日 by statistics-lab

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