# 线性代数网课代修|数值线性代数代写Numerical linear algebra代考|MATH3204

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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|The Buckling of a Beam

Consider a horizontal beam of length $L$ located between 0 and $L$ on the $x$-axis of the plane. We assume that the beam is fixed at $x=0$ and $x=L$ and that a force $F$ is applied at $(L, 0)$ in the direction towards the origin. This situation can be modeled by the boundary value problem
$$R y^{\prime \prime}(x)=-F y(x), \quad y(0)=y(L)=0,$$
where $y(x)$ is the vertical displacement of the beam at $x$, and $R$ is a constant defined by the rigidity of the beam. We can transform the problem to the unit interval $[0,1]$ by considering the function $u:[0,1] \rightarrow \mathbb{R}$ given by $u(t):=y(t L)$. Since $u^{\prime \prime}(t)=$ $L^{2} y^{\prime \prime}(t L)$, the problem $(2.24)$ then becomes
$$u^{\prime \prime}(t)=-K u(t), \quad u(0)=u(1)=0, \quad K:=\frac{F L^{2}}{R} .$$
Clearly $u=0$ is a solution, but we can have nonzero solutions corresponding to certain values of the $K$ known as eigenvalues. The corresponding function $u$ is called an eigenfunction. If $F=0$ then $K=0$ and $u=0$ is the only solution, but if the force is increased it will reach a critical value where the beam will buckle and maybe break. This critical value corresponds to the smallest eigenvalue of (2.25). With $u(t)=\sin (\pi t)$ we find $u^{\prime \prime}(t)=-\pi^{2} u(t)$ and this $u$ is a solution if $K=\pi^{2}$. It can be shown that this is the smallest eigenvalue of $(2.25)$ and solving for $F$ we find $F=\frac{\pi^{2} R}{L^{2}}$.

## 线性代数作业代写linear algebra代考|The Eigenpairs of the 1D Test Matrix

The second derivative matrix $T=\operatorname{tridiag}(-1,2,-1)$ is a special case of the tridiagonal matrix
$$\boldsymbol{T}{1}:=\operatorname{tridiag}(a, d, a)$$ where $a, d \in \mathbb{R}$. We call this the $1 \mathrm{D}$ test matrix. It is symmetric and strictly diagonally dominant if $|d|>2|a|$. We show that the eigenvectors are the columns of the sine matrix defined by $$\boldsymbol{S}=\left[\sin \frac{j k \pi}{m+1}\right]{j, k=1}^{m} \in \mathbb{R}^{m \times m}$$

For $m=3$,
$$\boldsymbol{S}=\left[s_{1}, s_{2}, s_{3}\right]=\left[\begin{array}{c} \sin \frac{\pi}{4} \sin \frac{2 \pi}{4} \sin \frac{3 \pi}{4} \ \sin \frac{2 \pi}{4} \sin \frac{4 \pi}{4} \sin \frac{6 \pi}{4} \ \sin \frac{3 \pi}{4} \sin \frac{6 \pi}{4} \sin \frac{9 \pi}{4} \end{array}\right]=\left[\begin{array}{ccc} t & 1 & t \ 1 & 0 & -1 \ t & -1 & t \end{array}\right], \quad t:=\frac{1}{\sqrt{2}} .$$

## 线性代数作业代写linear algebra代考|The Buckling of a Beam

$$R y^{\prime \prime}(x)=-F y(x), \quad y(0)=y(L)=0,$$

$$u^{\prime \prime}(t)=-K u(t), \quad u(0)=u(1)=0, \quad K:=\frac{F L^{2}}{R} .$$

## 线性代数作业代写linear algebra代考|The Eigenpairs of the 1D Test Matrix

$$\boldsymbol{T} 1:=\operatorname{tridiag}(a, d, a)$$

$$\boldsymbol{S}=\left[\sin \frac{j k \pi}{m+1}\right] j, k=1^{m} \in \mathbb{R}^{m \times m}$$

$$\boldsymbol{S}=\left[s_{1}, s_{2}, s_{3}\right]=\left[\sin \frac{\pi}{4} \sin \frac{2 \pi}{4} \sin \frac{3 \pi}{4} \sin \frac{2 \pi}{4} \sin \frac{4 \pi}{4} \sin \frac{6 \pi}{4} \sin \frac{3 \pi}{4} \sin \frac{6 \pi}{4} \sin \frac{9 \pi}{4}\right]=\left[\begin{array}{lllll} t & 1 & t & 0 & -1 t \end{array}\right.$$

# 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