Statistics-lab™可以为您提供harvard.edu AM221 Convex optimization凸优化课程的代写代考和辅导服务！

AM221 Convex optimization课程简介

This is a graduate-level course on optimization. The course covers mathematical programming and combinatorial optimization from the perspective of convex optimization, which is a central tool for solving large-scale problems. In recent years convex optimization has had a profound impact on statistical machine learning, data analysis, mathematical finance, signal processing, control, theoretical computer science, and many other areas. The first part will be dedicated to the theory of convex optimization and its direct applications. The second part will focus on advanced techniques in combinatorial optimization using machinery developed in the first part.

PREREQUISITES 

Instructor: Yaron Singer (OH: Wednesdays 4-5pm, MD239)
Teaching fellows:

• Thibaut Horel (OH: Tuesdays 4:30-5:30pm, MD’s 2nd floor lounge)
• Rajko Radovanovic (OH: Mondays 5:30-6:30pm, MD’s 2nd floor lounge)
  Time: Monday \& Wednesday, 2:30pm-4:00pm
  Room: MD119
  Sections: Wednesdays 5:30-7:00pm in MD221

AM221 Convex optimization HELP（EXAM HELP， ONLINE TUTOR）

To find the distance between two parallel hyperplanes, we first note that any vector in one hyperplane is orthogonal to the normal vector $a$ of the other hyperplane. Let $x_1$ and $x_2$ be points in the two hyperplanes respectively, such that $x_1 – x_2$ is orthogonal to $a$. Then, we have:

$$a^T (x_1 – x_2) = 0$$

Solving for $x_2$, we get:

$$x_2 = x_1 – \frac{a^T x_1 – b_2}{|a|_2^2}a$$

where $b_2$ is a constant such that $a^Tx = b_2$ defines the second hyperplane. Similarly, we can find a point $x_1$ in the first hyperplane such that $x_1 – x_2$ is orthogonal to $a$. Then, we have:

$$x_1 = x_2 + \frac{a^T x_2 – b_1}{|a|_2^2}a$$

where $b_1$ is a constant such that $a^Tx = b_1$ defines the first hyperplane.

Now, the distance between the two hyperplanes is the length of the vector $x_1 – x_2$, which is given by:

\begin{align*} \left|x_1 – x_2\right|_2 &= \left|\frac{a^T x_2 – b_1}{|a|_2^2}a – \frac{a^T x_1 – b_2}{|a|_2^2}a\right|_2 \ &= \frac{\left|b_1 – b_2\right|}{|a|_2} \end{align*}

Therefore, the distance between the two parallel hyperplanes $\left{x \in \mathbf{R}^n \mid a^T x=b_1\right}$ and ${x \in$ $\left.\mathbf{R}^n \mid a^T x=b_2\right}$ is $\left|b_1-b_2\right| /|a|_2$.

Solution. Let $\mathcal{H}=\left{x \mid a^T x \leq b\right}$ and $\tilde{\mathcal{H}}=\left{x \mid \tilde{a}^T x \leq \tilde{b}\right}$. The conditions are:

• $\mathcal{H} \subseteq \tilde{\mathcal{H}}$ if and only if there exists a $\lambda>0$ such that $\tilde{a}=\lambda a$ and $\tilde{b} \geq \lambda b$.
• $\mathcal{H}=\tilde{\mathcal{H}}$ if and only if there exists a $\lambda>0$ such that $\tilde{a}=\lambda a$ and $\tilde{b}=\lambda b$.
  Let us prove the first condition. The condition is clearly sufficient: if $\tilde{a}=\lambda a$ and $\tilde{b} \geq \lambda b$ for some $\lambda>0$, then
  $$
  a^T x \leq b \Longrightarrow \lambda a^T x \leq \lambda b \Longrightarrow \tilde{a}^T x \leq \tilde{b}
  $$
  i.e., $\mathcal{H} \subseteq \tilde{\mathcal{H}}$.
  To prove necessity, we distinguish three cases. First suppose $a$ and $\tilde{a}$ are not parallel. This means we can find a $v$ with $\tilde{a}^T v=0$ and $a^T v \neq 0$. Let $\hat{x}$ be any point in the intersection of $\mathcal{H}$ and $\tilde{\mathcal{H}}$, i.e., $a^T \hat{x} \leq b$ and $\tilde{a}^T x \leq \tilde{b}$. We have $a^T(\hat{x}+t v)=a^T \hat{x} \leq b$ for all $t \in \mathbf{R}$. However $\tilde{a}^T(\hat{x}+t v)=\tilde{a}^T \hat{x}+t \tilde{a}^T v$, and since $\tilde{a}^T v \neq 0$, we will have $\tilde{a}^T(\hat{x}+t v)>\tilde{b}$ for sufficiently large $t>0$ or sufficiently small $t<0$. In other words, if $a$ and $\tilde{a}$ are not parallel, we can find a point $\dot{x}+t v \in \mathcal{H}$ that is not in $\tilde{\mathcal{H}}$, i.e., $\mathcal{H} \nsubseteq \tilde{\mathcal{H}}$. Next suppose $a$ and $\bar{a}$ are parallel, but point in opposite directions, i.e., $\bar{a}=\lambda a$ for some $\lambda<0$. Let $\hat{x}$ be any point in $\mathcal{H}$. Then $\hat{x}-t a \in \mathcal{H}$ for all $t \geq 0$. However for $t$ large enough we will have $\tilde{a}^T(\hat{x}-t a)=\tilde{a}^T \hat{x}+t \lambda|a|_2^2>\tilde{b}$, so $\hat{x}-t a \notin \tilde{\mathcal{H}}$. Again, this shows $\mathcal{H} \notin \tilde{\mathcal{H}}$.
• Finally, we assume $\tilde{a}=\lambda a$ for some $\lambda>0$ but $\tilde{b}<\lambda b$. Consider any point $\hat{x}$ that satisfies $a^T \hat{x}=b$. Then $\tilde{a}^T \hat{x}=\lambda a^T \hat{x}=\lambda b>\tilde{b}$, so $\hat{x} \notin \tilde{\mathcal{H}}$.
• The proof for the second part of the problem is similar.

Textbooks

• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.

Statistics-lab™可以为您提供harvard.edu AM221 Convex optimization现代代数课程的代写代考和辅导服务！ 请认准Statistics-lab™. Statistics-lab™为您的留学生涯保驾护航。