# 线性代数网课代修|交换代数代写Commutative Algebra代考|MATH483

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• 数值分析
• 高等线性代数
• 矩阵论
• 优化理论
• 线性规划
• 逼近论

## 线性代数作业代写linear algebra代考|Implicitization

In Chapter 1 , we saw that a variety $V$ can sometimes be described using parametric equations. The basic idea of the implicitization problem is to convert the parametrization into defining equations for $V$. The name “implicitization” comes from Chapter 1 , where the equations defining $V$ were called an “implicit representation” of $V$. However, some care is required in giving a precise formulation of implicitization. The problem is that the parametrization need not fill up all of the variety $V$-an example is given by equation (4) from Chapter 1 , $\S$ 3. So the implicitization problem really asks for the equations defining the smallest variety $V$ containing the parametrization. In this section, we will use the elimination theory developed in $\S \S 1$ and 2 to give a complete solution of the implicitization problem.

Furthermore, once the smallest variety $V$ has been found, two other interesting questions arise. First, does the parametrization fill up all of $V$ ? Second, if there are missing points, how do we find them? As we will see, Groebner bases and the Extension Theorem give us powerful tools for studying this situation.

To illustrate these issues in a specific case, let us look at the tangent surface to the twisted cubic in $\mathbb{R}^{3}$, first studied in Chapter $1, \S 3$. Recall that this surface is given parametrically by
(1)
\begin{aligned} &x=t+u, \ &y=t^{2}+2 t u, \ &z=t^{3}+3 t^{2} u . \end{aligned}
In $\S 8$ of Chapter 2 , we used these equations to show that the tangent surface lies on the variety $V$ in $\mathbb{R}^{3}$ defined by
$$x^{3} z-(3 / 4) x^{2} y^{2}-(3 / 2) x y z+y^{3}+(1 / 4) z^{2}=0 .$$
However, we do not know if $V$ is the smallest variety containing the tangent surface and, thus, we cannot claim to have solved the implicitization problem. Furthermore, even if $V$ is the smallest variety, we still do not know if the tangent surface fills it up completely. So there is a lot of work to do.

## 线性代数作业代写linear algebra代考|Singular Points and Envelopes

In this section, we will discuss two topics from geometry:

• the singular points on a curve,
• the envelope of a family of curves.
Our goal is to show how geometry provides interesting equations that can be solved by the techniques studied in $\S \S 1$ and 2 .

We will introduce some of the basic ideas concerning singular points and envelopes, but our treatment will be far from complete. One could write an entire book on these topics [see, for example, BRUCE and GIBLIN (1992)]. Also, our discussion of envelopes will not be fully rigorous. We will rely on some ideas from calculus to justify what is going on.

Suppose that we have a curve in the plane $k^{2}$ defined by $f(x, y)=0$, where $f \in$ $k[x, y]$. We expect that $\mathbf{V}(f)$ will have a well-defined tangent line at most points, although this may fail where the curve crosses itself or has a kink. Here are two examples:

If we demand that a tangent line be unique and follow the curve on both sides of the point, then each of these curves has a point where there is no tangent. Intuitively, a singular point of $\mathbf{V}(f)$ is a point such as above where the tangent line fails to exist.

To make this notion more precise, we first must give an algebraic definition of tangent line. We will use the following approach. Given a point $(a, b) \in \mathbf{V}(f)$, a line $L$ through $(a, b)$ is given parametrically by
\begin{aligned} &x=a+c t, \ &y=b+d t . \end{aligned}
This line goes through $(a, b)$ when $t=0$. Notice also that $(c, d) \neq(0,0)$ is a vector parallel to the line. Thus, by varying $(c, d)$, we get all lines through $(a, b)$. But how do we find the one that is tangent to $\mathbf{V}(f)$ ? Can we find it without using calculus?
Let us look at an example. Consider the line $L$
( 2$)$
\begin{aligned} &x=1+c t, \ &y=1+d t, \end{aligned}through the point $(1,1)$ on the parabola $y=x^{2}$ .

## 线性代数作业代写linear algebra代考|Implicitization

(1)参数化给出的
$$x=t+u, \quad y=t^{2}+2 t u, z=t^{3}+3 t^{2} u .$$

$$x^{3} z-(3 / 4) x^{2} y^{2}-(3 / 2) x y z+y^{3}+(1 / 4) z^{2}=0 .$$

## 线性代数作业代写linear algebra代考|Singular Points and Envelopes

• 曲线上的奇异点，
• 曲线族的包络线。
我们的目标是展示几何如何提供有趣的方程，这些方程可以通过 $\S \S 1$ 和 2 。
我们将介绍一些关于奇异点和包络的基本概念，但我们的处理还远末完成。人们可以就这些 主题写一整本书 [例如，参见 BRUCE 和 GIBLIN (1992)]。此外，我们对信封的讨论不会完 全严格。我们将依靠微积分的一些想法来证明正在发生的事情。
假设我们在平面上有一条曲线 $k^{2}$ 被定义为 $f(x, y)=0$ ， 在哪里 $f \in k[x, y]$. 我们期望 $\mathbf{V}(f)$ 在大多数点处会有一条明确的切线，尽管这可能会在曲线与自身相交或有扭结的地方失败。 这里有两个例子:
如果我们要求一条切线是唯一的并且沿着该点两侧的曲线，那么这些曲线中的每一条都有一 个没有切线的点。直观地说，一个奇异点 $\mathbf{V}(f)$ 是一个点，例如上面切线不存在的点。
为了使这个概念更准确，我们首先必须给出切线的代数定义。我们将使用以下方法。给定一 个点 $(a, b) \in \mathbf{V}(f)$, 条线 $L$ 通过 $(a, b)$ 由参数给出
$$x=a+c t, \quad y=b+d t .$$
这条线经过 $(a, b)$ 什么时候 $t=0$. 另请注意 $(c, d) \neq(0,0)$ 是平行于直线的向量。因此，通过 改变 $(c, d)$ ，我们让所有的线路都通过 $(a, b)$. 但是我们如何找到相切的那个 $\mathbf{V}(f)$ ? 我们可以 不使用微积分找到它吗?
让我们看一个例子。考虑线 $L$
(2)
$$x=1+c t, \quad y=1+d t,$$
通过点 $(1,1)$ 在抛物线上 $y=x^{2}$.

# 计量经济学代写

## 在这种情况下，如何学好线性代数？如何保证线性代数能获得高分呢？

1.1 mark on book

【重点的误解】划重点不是书上粗体，更不是每个定义，线代概念这么多，很多朋友强迫症似的把每个定义整整齐齐用荧光笔标出来，然后整本书都是重点，那期末怎么复习呀。我认为需要标出的重点为

A. 不懂，或是生涩，或是不熟悉的部分。这点很重要，有的定义浅显，但证明方法很奇怪。我会将晦涩的定义，证明方法标出。在看书时，所有例题将答案遮住，自己做，卡住了就说明不熟悉这个例题的方法，也标出。

B. 老师课上总结或强调的部分。这个没啥好讲的，跟着老师走就对了

C. 你自己做题过程中，发现模糊的知识点

1.2 take note

1.3 understand the relation between definitions