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| # Huajie's Blog | ||
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| 如果你发现你想访问[我的博客](https://liubinfighter.github.io/)却看到了这个页面，也许博客正在维护中，使用了默认的README文件。 | ||
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| 同时欢迎来我的[Github](https://github.com/LIUBINfighter)主页看看。 | ||
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| 本博客参考了前辈的博客[Ben's Blog (benx.dev)](https://blog.benx.dev/)。 |
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| --- | ||
| date: 2024-08-16 | ||
| title: "How can freshmen carve out their own path at SUSTech? - Preface " | ||
| categories: | ||
| - articles | ||
| tags: | ||
| - SUSTecher | ||
| summary: 敢闯敢试，敢为人先。 | ||
| draft: false | ||
| --- | ||
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| >He who dares, wins. | ||
|  | ||
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| --- | ||
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| ### Preface | ||
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| I believe that every new SUSTecher has experienced some level of confusion about how to navigate college life. Even during the summer vacation, many still worry about missing any lecture or opportunity provided by SUSTech. | ||
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| Unlike that in our past 12 years only for education, we now need to shift our mindset from doing what others deem right to pursuing what we believe is right, while courageously investing our energy and taking on responsibility. Mastering the art of directing your attention, and using that to gain insights into your surroundings and yourself, will be among the most crucial tasks you face. | ||
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| This is a condensed and revised version of my speech from the summer gathering in Sichuan and the Yueqi live broadcast, intended to offer some help to everyone. There will be a series of articles offering guidance from various perspectives. | ||
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| Obviously, no single article can solve everyone's lifelong challenges once for good. However, as we each strive for our own goals, we also open up new possibilities for others to consider. | ||
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| --- | ||
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| ### 前言 | ||
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| 我想新南科人都或多或少地迷惘过自己应当如何面对大学生活，即使在暑假也担惊受怕，放不下错过的某一场讲座。 | ||
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| 和前12年的教育经历不同的是，我们的心态要从**做大家认为正确的事**，转变为**做自己认为正确的事**，并且勇敢地投入精力和承担责任。正确控制自己注意力的投射，并以此为基础，理解环境和自己，将会是最重要的课题之一。 | ||
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| 这是我在暑假川聚线下活动和南科一家亲-悦启直播中演讲内容的精简再创作，希望能帮助到大家。将会有一系列文章陆续更新，从不同侧面为大家提供参考。 | ||
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| 虽然没有哪篇文章能一劳永逸地解决大家的人生问题，不过每个人在为自己拼搏的同时，也提供给了他人不同可能性的思考。 | ||
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| --- | ||
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| ### English Corner (ChatGPT) | ||
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| **Phrase:** Carve out | ||
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| **Definition:** To create or establish something through effort or determination. | ||
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| **Usage:** Often used metaphorically to describe building a career, niche, or path. | ||
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| **Synonyms:** Establish, create, forge. | ||
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| **Key Idea:** Implies effort and resourcefulness in making one’s own way. | ||
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| --- | ||
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| 顺带一提，网页上的阅读顺序和写作时间顺序是反着的，你懂吧？下一篇文章请按左键Prev. |
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| --- | ||
| date: 2024-08-17 | ||
| title: "How can freshmen carve out their own path at SUSTech? - 1 Your Map " | ||
| categories: | ||
| - articles | ||
| tags: | ||
| - SUSTecher | ||
| summary: 迷失在未知深山的一片大雾中————想去山顶观光？想去山底取水？ | ||
| draft: true | ||
| --- | ||
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| >Rome wasn't built in a day. |
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| --- | ||
| date: "2024-05-24" | ||
| title: "Golden Pheasant Award" | ||
| categories: | ||
| - articles | ||
| tags: | ||
| - SUSTecher | ||
| - Life | ||
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| --- | ||
| title: "Row Lost in Vandermonde" | ||
| date: "2023-11-27" | ||
| summary: "一道线性代数习题，Vandermonde Determinant" | ||
| math: true | ||
| draft: false | ||
| categories: | ||
| - math | ||
| tags: | ||
| - notes | ||
| - linear_algebra | ||
| --- | ||
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| 2023.11.27 week-11 TA class | ||
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| Supple10 Q3 | ||
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| ### Problem | ||
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| 求行列式 $D_{n}$的值（代数表达式）。 | ||
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| $$D_{n} = | ||
| \left| | ||
| \begin{array}{cccc} | ||
| 1 & 1 & \dots & 1 \\\\ | ||
| x_{1} & x_{2} & \dots & x_{n} \\\\ | ||
| x_{1}^{2} & x_{2}^{2} & \dots & x_{n}^{2} \\\\ | ||
| \vdots & \vdots & & \vdots \\\\ | ||
| x_{1}^{n-2} & x_{2}^{n-2} & \dots & x_{n}^{n-2} \\\\ | ||
| x_{1}^n & x_{2}^n & \dots & x_{n}^n | ||
| \end{array} | ||
| \right| | ||
| $$ | ||
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| 思想：补全缺失的行和列 | ||
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| ### A simplified example | ||
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| $$ \begin{vmatrix} | ||
| 1& 1 & 1 & 1 \\\\ | ||
| a& b & c & d \\\\ | ||
| a^{2}& b^{2} &c^{2} &d^{2} \\\\ | ||
| a^{4}& b^{4} &c^{4} &d^{4} | ||
| \end{vmatrix}$$ | ||
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| Solution:Construct complete *Vandermonde* and compare coefficient. | ||
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| 构造完整范德姆行列式并比较（二项式）系数 | ||
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| $$ \det A= | ||
| \begin{vmatrix} | ||
| 1& 1 & 1 & 1 & 1 \\\\ | ||
| a& b & c & d &x \\\\ | ||
| a^{2}& b^{2} &c^{2} &d^{2} & x^2 \\\\ | ||
| a^{3}& b^{3} &c^{3} &d^{3} & x^3 \\\\ | ||
| a^{4}& b^{4} &c^{4} &d^{4} & x^4 | ||
| \end{vmatrix}$$ | ||
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| Now we want to find the minor $M_{25}$ of Matrix A. | ||
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| Define constant $S = (d-c)(b-d)(d-a)(c-b)(c-a)(b-a)$. | ||
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| Calculate the Vandermonde determinant and co-factor expansion by column n; | ||
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| 计算范德姆行列式和列 n 的代数余子式 | ||
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| $$\begin{align} | ||
| |A| & =S(x-d)(x-c)(x-b)(x-a) \\\\ | ||
| & =C_{15}+C_{25}x+C_{35}x^{2}+C_{45}x^{3}+C_{55}x^{4} \\\\ | ||
| \end{align}$$ | ||
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| Compare the coefficient of x: | ||
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| 比较x的系数 | ||
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| $$C_{25}=(-abc-abd-acd-bcd)S$$ | ||
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| So, the original determinant is | ||
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| $$M_{25}=-C_{25}=(+abc+abd+acd+bcd)S$$ | ||
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| ### Solution | ||
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| $$D_{expansion}= | ||
| \begin{vmatrix} 1 & 1 & \dots& 1 &1\\\\ | ||
| x_{1} &x_{2} &\dots &x_{n} &y \\\\ | ||
| x_{1}^{2} &x_{2}^{2} &\dots &x_{n}^{2} &y^2 \\\\ | ||
| \dots&\dots& &\dots &\dots \\\\ | ||
| x_{1}^{n-2} &x_{2}^{n-2} &\dots &x_{n}^{n-2} &y^{n-2} \\\\ | ||
| x_{1}^{n-1} &x_{2}^{n-1} &\dots &x_{n}^{n-1} &y^{n-1} \\\\ | ||
| x_{1}^n &x_{2}^n &\dots &x_{n}^n &y^{n} | ||
| \end{vmatrix} | ||
| $$ | ||
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| 然后按最后一列展开，我们所需要求的$D_{n}$应该会和 n 行 n+1 列（即补充的行和列，对应 $y^{n-1}$ ）的代数余子式有关，因此只看这一项。 | ||
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| 将 $D_{e}$ 算两次即可比较得出 $D_{n}$ | ||
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| 第一步，按最后一列展开,只看 $y^{n-1}$ | ||
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| $$ | ||
| (-1)^{n+(n+1)}D_{n}y^{n-1}=-D_{n}y^{n-1} | ||
| $$ | ||
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| 第二步，根据 **Vandermonde Determinant** 的性质: | ||
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| （为了方便书写，我们记 $y=x_{n+1}$ ） | ||
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| $$ | ||
| D_{e}=\prod_{1\le j < i \le n+1} (x_{i}-x_{j}) | ||
| $$ | ||
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| 我们想要$y^{n-1}$的系数，于是改写为 | ||
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| $$ | ||
| D_{e}=\left[ \prod_{1\le j < i \le n} (x_{i}-x_{j}) \right] \left[ \prod_{1\le i \le n} (y-x_{i}) \right] | ||
| $$ | ||
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| 左边方括号为$D_{n}$ ，右边运用二项式定理，得$y^{n-1}$项为 | ||
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| $$ | ||
| \sum_{i=1}^{n}(-x_i)y^{n-1} | ||
| $$ | ||
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| 比较即得 | ||
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| $$ | ||
| -D_{n}=\left[ \prod_{1\le j < i \le n} (x_{i}-x_{j}) \right] \left[ \sum_{i=1}^{n}(-x_i) \right] | ||
| $$ | ||
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| $$ | ||
| D_{n}=\left[ \prod_{1\le j < i \le n} (x_{i}-x_{j}) \right] \left[ \sum_{i=1}^{n}x_i \right] | ||
| $$ |
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