Just for fun

Some one ask me to count the number of element in the set $  \{\sigma\in S_5:|\sigma|=2\}$, after doing that I follow the same idea to count $  |\{\sigma\in S_n:|\sigma|=2\}|$, and a step further I count $  |\{\sigma\in S_n:|\sigma|=p\}|$, $  p\leq n$, and I get \[\sum_{k=1}^{[\frac{n}{p}]}\frac{\big((p-1)!\big)^{k}}{(k)!}\prod_{r=0}^{k-1}\binom{n-pr}{p}.\]

The result is summarized in the following document (I hope the result is true):
http://ihome.ust.hk/~cclee/document/sthgen.pdf

To ``verify" the result, let's count $  L(n,3)$ in another way, we count this by considering $  a_n$ defined by
\[a_n=\frac{1}{3}\times \underbrace{\binom{n}{2}}_{\text{choose 2 elements}\atop \text{to form permutation}}\underbrace{\binom{2}{1}}_{\text{choose 1 of the first two}\atop\text{ chosen numbers}}\underbrace{\binom{n-2}{1}}_{\text{choose another 1 to form permutation}\atop\text{with the number in 2C1 }}.\]

The factor $  \frac{1}{3}$ is left there becasue we observe that every length 3 permutation can be written as product of 2 transpositions in exactly 3 ways.

Then clearly the number of ways to form order $  3$ permutation by multiplying $  k$ disjoint length 3 cycles is given by \[ \frac{\prod_{r=0}^{k-1}a_{n-3r}}{(k)!}=\frac{2^{k}}{(k)!}\prod_{r=0}^{k-1}\binom{n-3r}{3},
\] exactly the same summand appear.

305 愈讀愈頭痛……

冇得返轉頭……，我應該要開始將成本嚴教授嘅 topology 啃左佢。(好彩佢個網有齊 homework 答案) 摺拉時數將創我自己新高，得閒記住探我。(睇黎又會好似 year 1 咁淨係顧 204 唔理其他科)

Some counting

Math 110 students are learning counting, now here is a suitable problem from HKgolden.

Problem. There are 2 guys, A and B, and there is a box containing 9000 balls numbered 1000, 1001, ..., 9999 respectively. Suppose A has chosen a ball with replacement, what is the probability that B chooses a ball with no digit in common with that of A? (Say if A has chosen 1234, then B will choose a number that does not contain any digit in {1, 2, 3, 4})

開學

期待嘅 fall sem 又開始喇，呢幾日叫做教得幾快 (以第一堂黎計) 嘅 course 可以叫 MATH321 及 MATH305。

(除非特別標明，以下所有 course code 均指 MATH)

星期三. 因為冇野做去 sit 下 217。全埸爆滿...，好多人 sit，原來 A 教授有一種規則，就係你喺佢個 first quiz 到 ``do well" 就可以去 reg 佢個 course (不需 al pure A 或 024 (或其他) 最少 A-)，難怪咁多人 sit 喇。不過聽講 203 一樣好多人 sit，啲學生未免太過份喇，搞到原本班學生冇位……。

今年 203 textbook 係用 rudin，呢屆真係好幸福，第一年就由 metric space 開始講起。有好有壞喇，以嚴教授果份 notes 黎教嘅話我最後學得好少 point-set topology on metric space (大部分相關知識都係李教授嘅 370 到學返黎的)，但相對地我開始接觸 305 嘅野 (204 的 multivariable differentiation 是煉獄的開始，linear algebra 不斷湧現...)。其實我懷疑 204 本身係 introduction to 305 (類似 202 係 introduction to 301)。

呢一日去到中午上 PHYS121，重點都係要知道交功課果個流程，其他都係一堆唔駛理嘅野...。

星期四. 如上一篇 entry 所見...，比較多堂嘅一日。321、305 李教授 (2號) 做左少少比較 deep 嘅 introduction。321 人真係好多...，305 反而好少人 sit (奇怪)。

305 第一堂李教授 (2號) 就恐嚇我地 ``Unlike MATH321, this is not a course for everyone, this course is for advanced student or student with strong background", ``This course is for student who wants to be a professional mathematician in the future" 云云。我諗...，呢啲說話只會令到學生鬥志更加激昂！

星期五. 冇乜特別，number theory...，有見識過出面嘅數學比賽都知道初等數論係非常困難...，論證雖然優美、簡單，但愈簡單嘅野對智力需求愈高 (追求初等證明是好習慣)，希望應付得黎。

Confirmed enrollment

	Mon	Tue	Wed	Thu	Fri
09:00 - 09:20		MATH321 L1 (1504)		MATH321 L1 (1504)	MATH311 T1C (2463)
09:30 - 09:50
10:00 - 10:20
10:30 - 10:50	PHYS121 LA2 (6137)	MATH305 L1 (1504)		MATH305 L1 (1504)
11:00 - 11:20
11:30 - 11:50
12:00 - 12:20					MATH315 L1 (4504)
12:30 - 12:50
13:00 - 13:20		LANG208 T12 (5561)		LANG208 T12 (5561)
13:30 - 13:50			PHYS121 L1 (4504)		PHYS121 L1 (4504)
14:00 - 14:20		PHYS121 T1 (2306)
14:30 - 14:50
15:00 - 15:20	MATH315 T1B (3584)	MATH311 L1 (2465)		MATH311 L1 (2465)
15:30 - 15:50
16:00 - 16:20
16:30 - 16:50	MATH315 L1 (4504)
17:00 - 17:20
17:30 - 17:50
18:00 - 18:20	MATH321 T1A 4620			MATH305 T1A (4503)
18:30 - 18:50

Record of some solved inequalities

可嘗試以下問題 (1, 2, 4, 6, 7, 8 都是 AL 知識範圍內)，請不要使用暴力的方法 (暴力通分) 解決問題。

Problem 1. Let $  a,b,c>0$ and $  abc=1$. Prove that  \[\frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}\geq 1.\]
Problem 2. Let $ x,y,z>0$; $  x+y+z=1$ prove that \[\sqrt{\frac{x}{yz}}+\sqrt{\frac{y}{zx}}+\sqrt{\frac{z}{xy}}\ge 2\left(\sqrt{\frac{x}{(x+y)(x+z)}}+\sqrt{\frac{y}{(y+z)(y+x)}}+\sqrt{\frac{z}{(z+x)(z+y)}}\right).\]
以下雖然不難，卻十分漂亮，值得牢記！經驗告訊我 $  ab+bc+ca$ 和 $  a+b+c$ 也是十分常見的因子。

Problem 3. Prove that for any $a,b,c\ge 0$, we always have \[
9(a+b)(b+c)(c+a)\ge 8(a+b+c)(ab+bc+ca)
\] and \[ (a+b+c)(a^{2}+b^{2}+c^{2})+9abc\ge 2(a+b+c)(ab+bc+ca).
\]
Problem 4. When $  a+b+c=3$, $  a,b,c\ge 0$, prove that \[\frac{a+3}{3a+bc}+\frac{b+3}{3b+ca}+\frac{c+3}{3c+ab}\ge 3.\]
Problem 5. Let $ a,b,c>0$, show that \[a^{2}+b^{2}+c^{2}+2abc+1\ge 2(ab+bc+ac).\]
Problem 6. Let $  x,y,z>0$, prove that \[\frac{xy}{x^{2}+y^{2}+2z^{2}}+\frac{yz}{y^{2}+z^{2}+2x^{2}}+\frac{zx}{z^{2}+x^{2}+2y^{2}}\leq\frac{3}{4}.\]
Problem 7. Let $  a,b,c$ be positive real numbers such that $  abc=1$. Prove that \[\frac{1}{a+b^{2}+c^{3}}+\frac{1}{b+c^{2}+a^{3}}+\frac{1}{c+a^{2}+b^{3}}\leq 1.\]
Problem 8. $  a,b,c$ are real positive numbers, prove that \[\frac{ab}{c(c+a)}+\frac{bc}{a(a+b)}+\frac{ca}{b(b+c)} \geq \frac{a}{c+a}+\frac{b}{a+b}+\frac{c}{b+c}.\]

有關 topology 及 D.G. 的 workshop ...

原來這是一個我不應該去的 workshop =.=...。預備知識多，可見到講者很用心澄清每一個數學用語，但這樣便花了大部分時間 ...。到最後我幾乎還未認識甚麼是 manifold :o。

Record of solved math202/3-level problems

首兩條的解題方法在 202 我們多次看到，作為回憶從某論壇 copy 下來。

Problem 1. Let $  f:[a,+\infty)\to \mathbb{R}$ a twice differentiable function on the interval $  (a,+\infty)$  with $  \lim_{n\to\infty}f(x)=l\in\mathbb{R}$  and $  |f''(x)|\leq M,\forall x \in [a,+\infty)$. Prove that $  \displaystyle \lim_{x\to\infty}f'(x)=0$.

Problem 2. Let $  f$ be differentiable on $  (a,+\infty)$. If $  \lim_{x\to\infty}\big(f(x)+xf'(x)\ln x\big)=l$, then show that $  \lim_{x\to\infty}f(x)=l$.

以下這一條取自 2010 Department of Mathematics, Graduate School of Science, Kyoto University subjects of tests : Mathematics I

Problem 3*. Let $  f$ be continuous real valued function on $  [0,1]$ and $  f(0)=0, f(1)=1$. Find \[\lim_{n\to\infty}n\int_0^1f(x)x^{2n}\,dx.\]
基本上 $  f(0)=0, f(1)=1$ 是一多餘的 condition。把它們保留下來作為 unknown 可得到一般結果。我想多餘的資料旨在 confuse 解題的人。把 2 換成 $  k$，結果應該差不多！

以下在 analysis section 裏看到這條，可能與 Lagrange multiplier 有關吧。其實只需中學知識。

Problem 4. Let $  x_1,x_2,\dots,x_n$ be positive real number satisfying $  \displaystyle \sum_{k=1}^n\frac{1}{x_k}=n$. Find the minimum of \[
\sum_{k=1}^n\frac{(x_k)^k}{k}.\]
Problem 5. Find $  \displaystyle\lim_{n\to\infty}\left\{n^{2}\left(\int_{0}^{1}\sqrt[n]{1+x^{n}}\,dx-1\right)\right\} $.

Problem 6. Suppose that $  f:\mathbb{R}-\{0,1\}\to \mathbb{R}$ satisfies the equation
$  \displaystyle f(x)+f\left(\frac{x-1}{x}\right)=1+x$, find $  f(x)$.

Problem 7. Let $  f:[0,1]\to\mathbb{R}$ be differentiable on $  [0,1]$ with $  f(1)=0$, show that
$  \displaystyle \lim_{n\to+\infty}n^{2}\int_{0}^{1}f(x)x^{n}\, dx =-f'(1)$.


都是些 202 基本概念便能夠解決的問題。

!!!

Workshop on Differential Topology & Differential Geometry

The workshop would be based on the lecture notes of MATH323 by Prof. Yan Min, MATH305 by Prof. Li Wei-Ping and MATH321 by Prof. Li Wei-Ping & Prof. Hu Jishan. Some review on Linear Algebra will be covered at the beginning.

The workshop would contain some exercises and discussions, the details are as follows:

• Date: Aug 5, 2010 (Thu) & Aug 6, 2010 (Fri)


• Time: 1:00pm-4:00pm


• Venue: Room 3209D (near Lifts 19 & 21, via Room 3209B)

Please contact Hoi (hoien14@gmail.com) for the soft copy of notes above.

今天的星期日生活，有關動漫。

我愛日本漫畫﹕全球化浪潮下 再閱讀日本動漫

【明報專訊】過去幾十年，日本動漫對香港讀者有什麼影響？這個問題，很「通識」，難答，但要急切處理。

通識，是因為多年來，日本動漫在香港累積過百萬讀者，經典動漫如《小甜甜》、《機動戰士》、《美少女戰士》及《鋼之鍊金術師》等亦製造了幾代人的集體回憶；換句話說，日本動漫已紮根港人生活，問動漫對香港讀者的影響，其實自然不過，很「通識」。難答——主流論述動輒把動漫的色情及暴力成分，看成年輕人越軌行為的主因（有媒體就把近日的家庭慘劇和《多啦A夢》掛鹇），而近年亦常把動漫迷等同電車男。這些答案，稍有批判力的人都知道扭曲事實；然而沒有更適切的角度，要另外提供有說服力的答案又很困難。急切——如此「通識」的問題還未有答案，其實要急切處理。

恭喜哂

Castar 同 yin 入左 ust math！

撞，撞，撞！

小喇叭 MATH306 撞哂 PHYS121, 111，無奈之下只好揀 MATH321，將會有兩日對住同一個 professor 連續 3 個鐘- -。

---

我 d linear algebra 真係好唔掂 (難怪我 B- = =)，證下面條 inequality 證左好耐 (仲要搵人幫拖)。($  \mathcal L(X,Y)$ denotes a collection of linear maps from $  X$ to $  Y$.)

Problem 1. Let $  U$ and $  V$ be finite dimensional vector spaces, $  S\in \mathcal L(V,W)$, $  T\in \mathcal L(U,V)$, prove that \[\dim \text{Nul}(ST)\leq \dim \text{Nul}(S)+\dim \text{Nul}(T).\]
遲啲自己啲 fact prove 得夠多嘅話試下順便整埋 collection，而家煩惱在 d 書冇答案。

一條 analysis problem，可視為 202 練習：

Problem 2. Let $  f$ be continuous on $  [0,\pi]$, $  n\in \mathbb{N}$. Prove that \[\lim_{n\to\infty}\int_{0}^{\pi}f(x)|\sin{nx}|\,dx=\frac{2}{\pi}\int_{0}^{\pi}f(x)\,dx.\]

冇記錯其實有 $  {\displaystyle \lim_{n\to\infty}} \int_0^\pi f(x)g(nx)\,dx$ 之類嘅一般結果，但詳細唔記得左。

New question(s)

I am waiting for the response to my proof to this problem. (actually waiting kunny...)

Problems 1. Let $  c$ be a real constant such that $\lim_{x\to 0} \frac{f(x)}{x}=c$. Find $ f(x)$ which satisfies $ f(x+y)\leq f(x)+f(y)$ for all real numbers $  x,y$.

I can't solve the following but already get a solution from Dr. Kin Li.

Problem 2. Let $  f$ be holomorphic on $  \{z:|z|<1\}$ such that $  |z|+|f(z)|\leq 1$, show that $  f\equiv 0$.

我永遠都只係識喺一點附近整個圓俾佢=.=，以後都係睇 integrand 做人。

原來有得睇 reg 得咩 course 喇

黎緊我會讀 MATH305、311、315、PHYS111(or 121), LANG2xx。可悲嘅係今個 sem 冇 combinatorial analysis，咁我仲有咩可以揀??

既然之前講過想讀返好 d PHYS，但無奈 UST 要 reg mechanics 同 EM 的話關卡重重，唯有揀返自己 MATH department 裏面疑似 PHYS 的物體。所以我決定行偽 Applied maths 路線，reg MATH306，再喺來年 take MATH308。咁岩又可以同自己識的 PHYS 巴打上堂了。

不過我 reg PHYS221、223、224 嘅夢想係唔會變的，
黎緊我會爭取 reg 入 PHYS121。

Minimum credit this sem: $4\times 5 + 1=21$，暫時 GEE 仲差一科 engine，一科 business 及一科 HUMA/SOSC。

**********

最近先知「嘅」嘅倉頡碼係 口竹心山，再睇返打出黎個字會併到出黎，但... 嘅呢個字唔係口字邊個「既」咩？點解會變到咁嘅樣=.=。

MI

最近有朋友幫佢朋友個朋友補習，學生係岩岩考完 ce，補習內容係預習 AL pure。(有冇搞錯 ...)

一開頭係 breakthrough 第 1 課 (我估係)，教 ge 都係點用 summation 同 method of difference。提到呢到，即刻將 MATH190 week 1 tutorial notes send 過佢睇，我當然鼓勵佢將 d 問題交俾個學生做 (事不關己嘛)，而且知識上係足夠應付。

過左幾日，MI，之後 binomial theorem，我見識大左之後知道呢類型題目真係可以變態得好緊要 (唔係指 AL)。尤其是 MI，往往要自己「加強命題」，非常之考創意。

提到 MI，又可以俾中五、六、七學生做，又幾技巧性 ge...，我手頭上唔多，而其中一條係

Problem. Show that for all positive integer $  n$, $  a \neq 1$, \[\frac{1-a^n}{1-a}+\frac{(1-a^n)(1-a^{n-1})}{1-a^2}+\cdots + \frac{(1-a^n)(1-a^{n-1})\cdots (1-a)}{1-a^n}=n.\]
我自己個做法都幾轉折，題目來自小卒討論區，當時只有我回答，冇人應我。幫緊中學生補習 ge 同學可以試試

最後兩條

真係 orz ...， style 同前果幾年唔同左。因為技術問題喺 workpress 打冇咁方便，用 pdf 黎記返低 d 題目。

370 Final Examination Problems.

小結係俾 kin li 玩死，同大眾一樣，做到頭 2 條，但 2(b) 慘死。
係唔係死左呢？其實我真係唔知，我睇返自己證法同 kin 完全唔同，
不過我做錯機會真係大好多。

**********

今日上完 seminar，發現唔少師兄都係高登仔，
我唔係孤獨架！

*********

這個暑假真的很長.....

好攰...，但為左可恨嘅 SOSC，只有博。

UST 真好，明明冇興趣都要迫人讀 SOSC，制度真係英明。

久違的 math midterm

好耐冇考過 maths midterm 喇，唔知點解往往都係臨考試前知識增長得特別快。三點半左右喺拉巴開始行去 rm2465，沿路見到好多中學生喺學術大堂果到睇 report。我埋去睇睇原來係初中 ge 數學專題報告黎，研究內容有三四種，我只記得一種，就係 convex polygon 有幾多種 partition 方法。其實呢樣係組合數學裏面講用 recurrence relation 做 counting 時 ge 一個典型 example，不過中二中三就識呢 d 野，對我黎講真係幾好野= = (我果時都唔知做緊乜)。沿路撞到 macro，佢又撞到自己中學啊 sir，咁佢地就傾傾傾，我用佢啊 sir ge 權力笠左包菊花茶飲。然後上課室等 4 點正果堂 midterm。

原來佢啊 sir 十幾年前係 ust 人，佢 ge 年代就係 kin li 教 204 ge 時代。

三點九，去到 rm 2465 等運到；
四點正，奇怪，kin li 遲到？
四點半，kin li office 冇人；
四半九，終於見人，原來係去左向班中學生做演講，唔記得內容係咩了。

**********

Problem 1. (10 marks) Let $  h:\mathbb{R}^2\to \mathbb{R}$ satisfy for all $  x,y\in \mathbb{R}^2$, $  |h(x)-h(y)|\leq \frac{1}{3}d(x,y)$, where $  d$ is the usual metric on $  \mathbb{R}^2$. Prove that there exists a unique continuous function $  f:[0,1]\to \mathbb{R}$ such that for all $  x\in [0,1]$,
$  h\big(f(x),x\big)+h\big(x,f(x)\big)=f(x)$.

Problem 2. (a) (9 marks) For $  n=1,2,3,\dots$, let $  f_n:\mathbb{R}\to \mathbb{R}$ be differentiable such that $  f'_n(x)$ is continuous on $  I=[0,1]$. For every $  t\in I$, there exists at least one $  n$ such that $  f'_n(t)=0$. Prove that there exist a positive integer $  N$ and a nonempty subinterval $  J$ of $  I$ such that $  f_N$ is constant on $  J$.

(b) (1 mark) In part (a), if we replace $  I$ by the open interval $  (0,1)$, will the statement remain true? Please give a `yes' or `no' answer.

**********

其實兩條都係 standard 問題，我地成班人 ge 高低就係取決於成份卷裏面最神聖 ge 1 分題。yes？or no？我地大部分人都係做左九個字左右 (考試時間為兩個鐘)。每人大概都係用半個鐘就寫好頭兩條 ge 答案。(b) part 就算自己幾肯定都好，始終 metric spaces 呢個課題我仲係幾陌生，我唔敢擔保，自己做左少少證明，答左 yes。唯一令人猶疑 ge 地方就係 $  (0,1)$ 唔係 complete metric space，但佢只要 existence ...... 。

MC 永遠都係最刻人憎。

有關 iblog

早前四至五月相信同學們都間中看到科大裏有宣傳 blog.ust 的 poster，今天好奇試試看。嗯，觀看網頁蠻流暢的；嗯，有一些 wordpress.com 本身沒提供的 theme (???)；嗯，有 100 MB 的上傳空間。

對一個以文字為主的 blogger 來說 (不是 google 那個)，它是很容易令人着迷。既然它是 wordpress，那我嘗試把自己原有在 wordpress 的所有東西 import 看看吧。看到效果後我立刻告訢自己，blog.ust.hk 不適合我。

http://blog.ust.hk/cclee/

本來相信它和 wordpress 一樣，卻在 最重要 的地方和 wordpress 不同。另外，在 blog.ust.hk 修改自己的文章時，會有以下選項：



其中沒有 preview change 。你在修改原文章後，view this post (圖中的) 不會顯示任何改變，按 save change 後基本上整個網頁也會刷新，別妄想 undo。對我來說真的諸多不便。還是我太挑剔？

妖!

時間唔夠用!

煩左好耐

究竟我呢個 claim 岩唔岩=.=。

Let $  L$ be closed, then $  \displaystyle \bigcap_{n=1}^\infty \left(\bigcup_{x\in L}B\left(x,\frac{1}{n}\right)\right)=L$ (i.e. $  L$ is a $  G_\delta$ set).

李教授話我個 proof 睇落去冇問題，咁即係佢 expect ge 證法唔係咁喇 ... sosad。

黑仔

食飯又撞到；
拉巴又撞到；(唔好再問我 equivalence relation)
放學又撞到‥‥‥

不過傾傾下發現大家某 d 想法幾相似。

btw 聽日 sosc 162 midterm，溫得真係好輕鬆，
因為我冇 jot 過 notes ...。

父親節=.=

又係一家人食飯 ge 日子，其實對我黎講冇乜特別，基本上日日都係一家人食晚飯。不過同以往一樣，會係同埋我啊媽果家人一齊食... 發現我o係同啊媽果邊 d 人溝通開始出現問題。

細細個果陣好鍾意黏住啊媽去見 d 姨姨呀 X 父呀，因為同 d 表哥表弟一見到面都有野玩 (我果個年代興 gameboy)。不過人一大左...興趣又唔同左，變左少打機，而條路又愈行愈歪...搞到俾同學話我個腦裏面得 $  \delta$ 同 $  \epsilon$ ...。我同佢地仲有 d 咩共同興趣呢？波我又唔多睇，機又唔多打，街又少出。o係科大都話好 d，有物理系 ge 人同我傾傾 maths excalibur ge problem corner。

我都唔知點解係物理系...

最近 search 到某個師兄 ge xanga，順便又搵到 soarer 同他唔識 ge 人 ge blog，等我慢慢睇先 ccc。

(中六、七) 以下題目可視為 Matrix 特性的復習。

Problem. Given that $  A$ is an invertible square matrix and $  AA^T = A^TA = I$. Suppose that $  B=A^{-1}A^T$ and $  \det A$ is not equal to $  \det B$. Show that $  \det (A + B) =0$.

391P Homework

Problem. Let $  A,B,C$ be sets. $  f:A\to B$ is a surjection and $  g:A\to C$ is a function such that if $  f(a_1)=f(a_2)$, then $  g(a_1)=g(a_2)$. Prove that there exists a unique function $  h:B\to C$ such that $  h\circ f = g$.

用返自己對 function ge 認識去做，但做做下硬係覺得怪怪地=.=。

以下為某中學授課員的網誌：
http://johnmayhk.wordpress.com/2010/06/17/double-counting-example/
我們來看看 a(iii) 這道題目，介紹一些新的 technique。(熟悉的人可無視~)

如同網誌所說，利用``容斥原理" (inclusion-exclusion principle)，易得到分子的數目。現在額外地說說 generating function，這裏不需要 pure 課程裏額外的知識。相信有讀 241 的同學可能已接觸過一些 generating function，例如 moment generating function。在組合數學裏有類似的計算技巧。

例如 $  (1+x+x^2+x^3+\cdots+x^8)(x^2+x^3+\cdots+x^8$) 裏，$  x^8$ 的系數, 7, 是有 ``計數" 的意義的。假設這裏有 8 個蘋果及 8 個橙，從中抽 8 個水果且其中最少有 2 個橙 (蘋果數目可 0)，那麼有多少種「抽水」組合？那就是剛剛所提到的 7，即 $  x^8$ 的系數。

回到 a(iii) 的問題，看看 $  (e^x-1)^3 = e^{3x}-3e^{2x}+3e^x-1$ 裏 $  x^n/n!$ 的系數是甚麼？剛巧也是 $  3^n-3\times 2^n+3$，原因是 \[\left(x+\frac{x^2}{2!}+\cdots\right)\left(x+\frac{x^2}{2!}+\cdots\right)\left(x+\frac{x^2}{2!}+\cdots\right)\\=\sum_{n=3}^\infty\left( \sum_{\scriptstyle x_1+x_2+x_3=n\atop \scriptstyle x_1,x_2,x_3\ge 1}\frac{n!}{x_1!x_2!x_3!}\right)\frac{x^n}{n!}.\]
由此可見，$  x^n/n!$ 的系數就是把 $  \displaystyle \binom{n-1}{2}$ (這個數字與我們的運算無關) 個組合的排列加起來所得的結果！另外我們已證明\[  \sum_{\scriptstyle x_1+x_2+x_3=n\atop \scriptstyle x_1,x_2,x_3\ge 1}\frac{1}{x_1!x_2!x_3!}=\frac{3^n-3\times 2^n+3}{n!}.
\] 類似手法可證得更 General 的結果 (例如 $  (e^x-1)^k =\dots $)。

ps. 大劑，點解 wordpress 有 pingback 呢樣野。

AL 就黎放榜, LANG

雖然話就黎，不過都係兩個星期多 d 後 ge 事。最近上一 d 大專生討論區，搵搵下搵到以下鏈結。

一個考了4次al的重讀生故事+感想文:
http://www.miniforum.net/showpost.fcgi?tempid=2&MGID=2672873&page=1

帖中有 5 篇文章。
作者係我地呢屆 ge ust math 學生...，有冇人願意介紹俾我識？
我o係 202 mid term room arrangement 個表到搵唔到有瞬字 ge 人，
睇黎係網名黎 ...。

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LANG 出左 grade，我由 C range 拉返上 B-，我口技一向都平平無奇，唔叻，睇黎都係靠呢篇字數懷疑不足 ge essay。

http://ihome.ust.hk/~cclee/document/springlangessay.pdf

隨手畫畫

下圖意境是``對暑假充滿期待！"XD

筆稿，線稿，彩稿均在同一篇日記內替換)

唉...

今日印好左某一科 ge ``notes" (其實書黎) 諗住去釘裝，去到先知大件事...。有一班土木工程 ge 人一早已經霸佔左部釘裝機...。我一點去到，一點半又去一去，三點鐘再去一去，仍然係果一班人。

釘裝機使用冇時限，我可以點？就算有時限，同樣可以以 call 馬黎達到長時間使用目的。我唔係想話呢班人自私，我估佢地都係幫 department 做野。但佢地釘裝十幾份好厚 ge notes (最少 100 頁)，而我只係釘裝一份 50 頁 notes，問問我趕唔趕住用算唔算難啟齒？

Texmaker 還是 LyX？

最近在想那，有某個好數學的 (我猜) 預科生好像不太喜歡用 word 的 equation editor 和我交流。取而代之他鍾情於使用那種只要輸入 $  \ $ 指令便會自動 gen 出相應的 gif 的網頁。有些更是 GUI 的...。

有天他問我是否使用 texmaker (我想他是看到我某一頁日記了)，我立時興奮起來，有人又受我感染了？！可惜好像經過若干嘗試便放棄了的樣子。這情形亦曾經出現在我的某位同學身上 (它說想在上個 winter break 學的...)。

不得不承認 $\LaTeX$ 容易令初學者望而卻步，但相比起 C++，它絕對是容易上手的，因這套程式的使用本身不涉及任何 logic。說實話我不明白有甚麼原因會令人輕言放棄。

但事實所見「選擇性學習障礙」(很懶-.-) 的人確實存在，最近 (其實數月前在科大 math lab) 發現一款 WYSIWYG 的數學排版軟件，名叫 LyX。它能夠讓初學者``更"簡單地製造出優質的文件檔，不喜歡直接輸入指令的同學可以轉而嘗試這套軟件。

一位內地高考生的網友給我的一條問題。(答案還沒派發的家課)

Problem. Let $  a,b,c > 0$,  $  \displaystyle \left(\frac{1}{a^2} +1\right)\left(\frac{1}{b^2} +1\right)\left(\frac{1}{c^2} + 1\right) = 512$ and $  k = a+b+c$, find the minimum value of $  k$.

AL 的朋友可嘗試解答 (也就是說 AL 知識已經足夠)。

感謝 YCK 提出了一個新的解法。

不等式小記

Problem (Austrian Mathematical Olympiad 2008). Prove that the inequality $ \displaystyle \sqrt{a^{1-a}b^{1-b}c^{1-c}}\leq \frac{1}{3}$ holds for all positive real numbers $  a,b,c$ with $  a+b+c=1$.

Those who always discuss math with me will know how to solve it, that's not hard. Now it can be easily generalized to \[ \sqrt{a_1^{1-a_1}a_2^{1-a_2}\cdots a_n^{1-a_n}}\leq \frac{1}{n^{(n-1)/2}}\] with $  a_1+a_2+\cdots +a_n=1$. From this we get for any $  a_i>0$,\[\frac{a_1+a_2+\cdots+a_n}{n}\ge \left(\frac{a_1a_2\cdots a_n}{(a_1^{a_1}a_2^{a_2}\cdots a_n^{a_n})^{1/\sum_{i=1}^na_i}}\right)^{1/(n-1)}.\]
We have established an extremely ugly lower bound. This inequality is useful if we are given a condition on $  \prod a_i^{a_i}$, for example from now on I can show that if \[\boxed{x^xy^yz^z=1, x,y,z>0},\] then \[ \sum_{cyc}\sqrt{\frac{x}{yz}}\ge 3.\]
Take $n =2$, set $  (a_1,a_2)=(x,y)$. The inequality reduces to \[\tfrac{1}{2}(x+y)\ge (x^yy^x)^{1/(x+y)}\iff \left(\tfrac{1}{2}(x+y)\right)^{x+y}\ge x^yy^x
\] (at least I think this inequality is somehow useful), by a little bit argument the following holds for all positive $ x,y$, \[
\sqrt[m]{\frac{x^m+y^m}{2}}\mathop{\ge}\limits_{m\ge n}\sqrt[n]{\frac{x^n+y^n}{2}}\mathop{\ge}\limits_{n\in\mathbb{N}}\frac{x+y}{2}\ge \sqrt{xy}\ge \frac{2}{\frac{1}{x}+\frac{1}{y}}\ge (x^yy^x)^{1/(x+y)}.\]

網友 (未來 ust phy 人?) 的朋友問了一條中(山)大(學)的入學試試題。

Problem 1. Let $  \displaystyle f(x)=\left|1-\frac{1}{x}\right|, x>0$, if $  0<a<b$ and $  f(a)=f(b)$, then prove that we must have $  ab>1$.

不難的，因最近都沒甚麼墨水...，放條問題吧。

學期尾

前日係最後一日上課日。諗返起呢個 sem 真係好充實。

話說因為 math190，mathdb 呢個網陪左我半年。
但 mathdb 只有給中學生玩的奧數 material 嗎?

錯了！

大家記住 year 2 take math 301 時黎呢到再睇睇：
http://www.mathdb.org/notes_download/analysis_ad.htm

讓自己在學校看的

http://civilliberty.about.com/od/abortion/f/abortion_legal.htmhttp://civilliberty.about.com/od/abortion/p/fetus_rights.htm

最近迷上這套漫畫：
http://acg.178.com/mh/b/bhgxbohe.shtml
http://acg.178.com/mh/l/liferenshen.shtml

真係咩口味都有 -_-：
http://acg.178.com/dh/s/weiniang.html

正體的積分號

在網上找到一個 package，它能夠把原本的 integral 換成直立的────國內出版的數學書籍常用的一種。

詳細安裝可在網上找到，package 為 mathabx。

它所修改的符號不是十全十美的，我們需要自己另外再作定義，我的是如此：

\usepackage{mathabx}
\DeclareMathSymbol{\sum}{\mathop}{largesymbols}{"50}
\DeclareMathSymbol{\supseteq}{\mathrel}{symbols}{"13}
\DeclareMathSymbol{\subseteq}{\mathrel}{symbols}{"12}
\DeclareMathSymbol{\subset}{\mathrel}{symbols}{"1A}
\DeclareMathSymbol{\supset}{\mathrel}{symbols}{"1B}
\DeclareMathSymbol{\cap}{\mathbin}{symbols}{"5C}
\DeclareMathSymbol{\cup}{\mathbin}{symbols}{"5B}
\DeclareMathSymbol{\complement} {\mathord}{AMSa}{"7B}
\DeclareMathDelimiter{\langle}{\mathopen}{symbols}{"68}{largesymbols}{"0A}
\DeclareMathDelimiter{\rangle}{\mathclose}{symbols}{"69}{largesymbols}{"0B}
\DeclareMathSymbol{\infty}{\mathord}{symbols}{"31}

\makeatletter
\DeclareRobustCommand\sqrt{\@ifnextchar[\@sqrt\sqrtsign}
\def\@sqrt[#1]{\root #1\of}
\makeatother

 

另一種方法，直接在 preamble 內加上

% upright integrals

%% \int will be upright (display mode)
%% \smallint will be upright (text mode)

%% \oldint will be slanted (display/text mode)
%% (\oldsmallint available, but not necessary)

\let\oldintop\intop
\def\oldint{\oldintop\nolimits}

\let\oldsmallint\smallint

\DeclareSymbolFont{EUEX}{U}{euex}{m}{n}

\DeclareSymbolFont{euexlargesymbols}{U}{euex}{m}{n}
\DeclareMathSymbol{\intop}{\mathop}{euexlargesymbols}{"52}
\def\int{\intop\nolimits}

\DeclareSymbolFont{euexsymbols} {U}{euex}{m}{n}
\DeclareMathSymbol{\smallint}{\mathop}{euexsymbols}{"52}

即能得到垂直的積分號。

---

好一段時間沒有用 $latex \LaTeX$ 打 project (語文功課)，原來我總是忘記在 \setmainfont 後加上 [Mapping=tex-text]，在 xeCJK package 下 \setCJKmainfont 同樣可使用這個 option。

終於唔駛補習 (住)

次次備課真係攰死人，所謂備課其實係準備 material。
有冇住 hall ge 人想接我呢單補習？非常之近科大，中文好更佳 (家長要求)。

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最近功課真係忙死，MATH 190 做一次功課好似玩一次比賽咁...。

自大得濟上 kin li 190 lecture 從來都唔會帶 notes，唔會記低重點。
每次 quiz、功課殺到黎都唔駛點睇，諗下諗下就有野出。
今次功課殺到上黎先意識到──``敝"！
未落手落腳用過 complex number 同 vector 去解決 geometry ge problem，
望住條問題真係會呆左，搞到呢幾日都博哂老命係咁溫 geometry，
可能我唔想o係呢科到臨尾香之故。

係唔係我錯覺呢？ geometry 呢課好似特別長。

為左科 geometry 搞到我呢幾日都冇時間溫 measure theory，
好彩第 9 份功課係星期 4 先交，不過星期 2 又有好多新野教 ...。

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題外話，唔係好鍾意而家 mathlinks 個介面...。

不等不等不等式, math190 quiz, math202 (extra), math370

Problem. Let $  a,b,c > 0$ be such that $  \displaystyle a^2+b^2+c^2+ab+bc+ca=\frac{3}{2}$, prove the following:

• $\displaystyle \frac{3}{4}\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}-1\right)\ge \frac{a^3c}{b}+\frac{b^3a}{c}+\frac{c^3b}{a}$.
• $ \displaystyle\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\frac{12}{(ab+bc+ca)\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)-1}$.

Math190: 20/20
Math202 (extra): 5.5/5.5
(醜不外傳，外傳不醜)

攞返 math190 份卷果陣我見到一個得意現像，李教授改卷改到進入左另一個竟界，幫同學搵卷唔係用名，而係用分！

我地終於開始 measure theory，見到嚴民教授份 lecture notes 汗都滴多幾滴‥‥‥。
急急腳去印返李教授 math301 裏面關於 Lxxxx measure 及 integration 的 notes，``user friendly" 得多。
本來以為 kin li notes 會由 25 版一直去到尾都係 Lxxxx 野 (師兄講過重點 focus o係佢到= =)。
最後印多左好多唔相關 ge 野 (implicit, inverse function theorem, etc...)，不過錯有錯着，
到就黎考試之前可以當 review，話哂 ``user friendly" 好多。

提一提大家，math 370 有得 reg 喇！

math190 quiz

quiz 已過，是時候收拾心情準備其他科。
原諒我記憶力有限，以下題目大體題意和原問題相同。

Problem 1. Show that there exist 2010 consecutive positive integers which are divisible by a sq number (not necessaily the same), where sq number is of the form $  m^2$, with $  m$ being an integer.

Problem 2. Let $  x,y\in\mathbb{N}$, find all the ordered pair(s) $  (x,y)$ satisfying
$  y^2-(x+1)2^x=1$

with proof. (Hint given: There is only one solution)

drawing in latex

作品：

http://ihome.ust.hk/~cclee/document/proofwithoutwords.pdf

Record some useful command for the environment tikzpicture.

• \begin{tikzpicture}[>=stealth,

  execute at end picture=%

  {

  \begin{pgfonlayer}{background}

  \path[fill=yellow,rounded corners]

  (current bounding box.south west) rectangle

  (current bounding box.north east);

  \end{pgfonlayer}

  }]

  %drawing command here

  \end{tikzpicture}

• \begin{tikzpicture}[remember picture,overlay]

  \node [scale=10,text opacity=0.2]

  at (current page.center) {Sample};

  \end{tikzpicture}

• \begin{tikzpicture}[remember picture,overlay]

  \node [xshift=1cm,yshift=1cm] at (current page.south west)

  [text width=7cm,fill=red!20,rounded corners,above right]

  {

  text

  };

  \end{tikzpicture}

• \usepackage{tikz}

  \usetikzlibrary{calc,through,backgrounds}

  \usetikzlibrary{decorations.pathmorphing}

  \usetikzlibrary{decorations.text}

  \usetikzlibrary{decorations.shapes}

• Introduction to 3D-plotting.
  http://www.fauskes.net/nb/introduction-to-sketch/

\usepackage{tikz}

\usetikzlibrary{calc,through,backgrounds}

\usetikzlibrary{decorations.pathmorphing}

\usetikzlibrary{decorations.text}

\usetikzlibrary{decorations.shapes}

202, Check the convergence

Problem. Let $  [x]$ be the biggest integer $  \leq x$. Let $  a>0$. Determine the convergence of the improper integral \[
\int_0^1\left(\left[\frac{a}{x}\right]-a\left[\frac{1}{x}\right]\right)\,d x.
\]

招人報 Summer course

o係黎緊個 summer kin li 話可以搞 math370，但條件為我要搵到 5 個人。當然如果人夠多的話，呢個 course 可以以 lecture 形式進行 (唔係就要變成 reading course 了...)。有興趣的話請留低 ITSC account，姓名。

List (ITSC): Ph_ccn, cclee, chhui, ypliaa

心情好差

MATH102 大炒，唔好問我咩分喇...。
只可以說一句，「A 的炒不了，炒的 A 不了。」唉!

從前提及過 $  \prod_{k=1}^{n-1}\sin \frac{k\pi}{n}=\frac{n}{2^{n-1}}$ 這樣的一條等式。
最近看到一個例子，取某些特定的整數 $  k$，再建立一個新的乘積，便能夠建立一條美麗而簡單的等式。設 $  n$ 為大於 1 的奇數，$  \varphi(n)$ 為 Euler-$  \varphi$ 函數，即 $  \varphi(n) =$ 小於 $  n$ 及與其互質的正整數的個數。設 $  a_1,a_2,\dots,a_{\varphi(n)}$ 為與 $  n$ 互質的正整數，那麼它們成為了模 $  n$ 的一個簡系 (reduced residue system)，又因為 $  (2,n)=1$，從而 $  2a_1,2a_2,\dots,2a_{\varphi(n)}$ 也是一個模 $  n$ 的簡系，因此我們有\[
\left|\prod_{k=1}^{\varphi(n)}\cos \frac{a_k\pi}{n}\right|=\frac{1}{2^{\varphi(n)}}.
\]
取 $  n$ 為質數 $  p$，則有 \[
\left|\prod_{k=1}^{p-1}\cos \frac{k\pi}{p}\right|=\frac{1}{2^{p-1}}，
\] 初等的東西果然很容易讓人``萌" 起來！繼代數不等式後，偶被初等數論萌倒了。
再在本文最開頭所說的等式中取 $  n = p$(質數)，兩式相乘，得到 \[
\left|\prod_{k=1}^{p-1}\sin \frac{2k\pi}{p}\right|=\frac{p}{2^{p-1}}=\prod_{k=1}^{p-1}\sin \frac{k\pi}{p},\] 看到這等式後，很自然會問：「這是``偶然"嗎？」

最近有同學問我，設 $  a,b$ 為正整數，證明：若 $  4ab-1|(4a^2-1)^2$，則 $  a=b$。
唔知點解果時``發左癲"，觀測唔到某 d 明顯到冇得再明顯 ge 野，搞到唔識做。
訓訓下覺呢個問題又彈返出黎，搞到訓唔着，順手寫寫下，``下?!"。

Exercises on Fubini's principle

There are two pieces of integrals (in my collection of past solved problem) that require you to use Fubini's principle to tackle. The previous one is not hard, and similar idea can be applied to the second one, enjoy them.

Integral 1. $\displaystyle \int_0^\infty\frac{\cos x-1}{xe^x}\,d x$.
Numerical answer: -0.34657

Integral 2. $\displaystyle \int_0^1 \frac{\tan^{-1}x}{x\sqrt{1-x^2}}\,d x$.
Numerical answer: 1.3845
(my solution to this one is quite long, and I forgot to extract the solution on the Internet)

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最近心情唔好，求求大家唔好再用 202 黎抽我水了= =。
上次金仔堂 totorial 已經俾人抽足成堂。

Some problems (not all of them challenging, you may try)

Computation 1. Use simple trick taught in Math190 (something call convolution?) to show that $  \displaystyle \left(\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{2^{2n+1}(2n+1)!}\right)^{2}=-\frac{1}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n}}{(2n)!}x^{2n}.$ I don't know if there is any trigonometric function that proves this formula instantly.

Problem 2. Determine if the series $  \displaystyle \sum_{n=2}^\infty \frac{\log n}{n(n-1)}$ (base $  e$) converges.

Problem 3. Find all $  p\in\mathbb{R}$ such that $  \displaystyle \sum_{k=2}^\infty \frac{1}{\big(\log (\log k)\big)^{p\log k}}$ converges.

Problem 4. Let $  0<x_1<1$ and Define $  x_{n+1}=x_n(1-x_n)$. Show that the series $  \sum x_n$ diverges.

關於 LaTeX

有同學表示想學 $\LaTeX$。
要寫 C++ 同 complie 返個 program 出黎，我地要有 visualstudio。
要寫 $\LaTeX$ 同 compile 返個 pdf 出黎，我地要好多種 compiler 中其中一種。

我以前係用 CTeX (內置 WinEdt)，可惜呢個軟件對繁體中文支持十分之弱。
取而代之我係用 TexMaker，詳細安裝可參見下面連結：
http://gtchen.pixnet.net/blog/post/4072739

跟足哂步驟，係時候學點樣打。網上面有好多好多 latex 教材，其中非常適合入門 ge 算係呢個：
http://yenlung.math.nccu.edu.tw/~site/drupal-5.1/idisk/quicklatex.pdf

理論上睇完上面呢個 pdf 已經可以得到好靚 ge 排版效果，有唔明可以再問。
而下都係一個幾有用 ge 網，有問題時亦可以呢到搵搵先。
http://edt1023.sayya.org/tex/latex123/node1.html

最後，關於 latex 的中文討論網站：
http://bbs.ctex.org/index.php

我某 d 文章可以作為參考：
http://ihome.ust.hk/~cclee/document/190present111.tex

New 202 presentation assignment

今日去搵 kin li 講講 dirichlet test 條題目 (最尾果條 ge part a)，因為題目本身冇俾 $g$ 係 differentiable，令到題目難度大大提高 (但結論依然正確)。

今日俾佢睇左我呢個寫法：(Click me)
冇乜點細心睇過，話方向係可行的，同學仔得閒 ge 話幫手睇睇。

102 炒左

第 3 條 interate 返 5 個 integral 錯哂 (一開始已經畫錯圖)，
第 5 條個 Jacobian 唔識證 ~ 好野。

最近的 MIDTERM

MATH204 已炒，MATH202 出奇地考得不錯。
比較了兩份202 試卷 (我考的是 makeup 版)，僥倖！ 我的比較簡單。

總結來說較多人考的那一份難的只有第 3 第 4 條。(廢話)
暫時我只想到第 4 條的解，希望和 kin li 的不同。

Problem. Let $  f:\mathbb{R}\to \mathbb{R}$ be twice differentiable and for all $  x\in [0,1]$, $  |f''(x)|\leq 2010$. If there exists $  c\in (0,1)$ such that $  f(c)>f(0)$ and $  f(c)>f(1)$, then prove that
$  |f'(0)|+|f'(1)|\leq 2010$.

灰...

點解 limit of function 可以撚埋 equivalence of norm 黎考...。

Cloud Seeding

http://en.wikipedia.org/wiki/Cloud_seeding

名句

「呢個世界有三種數學家 一種識數數 另一種唔識數數」

一遇到 combinatorics ge 難題我就會用呢句安慰自己。

Problem on function + simple inequality from AL discussion.

Recently solved a problem that is given from my classmate. (the solution is confirmed to be right by Kin Li, at least at the moment I post the question here) You can try the following one.

Problem 1. Let $  f:[0,1]\to[0,1],g:[0,1]\to[0,1]$ be continuous and satisfy $  f\circ g(x)=g\circ f(x)$. Prove that there is a $  w\in [0,1]$ such that $  f(w)=g(w)$.

I think it is not that easy. :)

Problem 2. Suppose $  a,b,c >0$ and $  ab+bc+ca=\frac{1}{3}$, show that \[\frac{a}{a^2-bc+1}+\frac{b}{b^2-ca+1}+\frac{c}{c^2-ab+1}\ge \frac{1}{a+b+c}.\]

Once again, AL knowledge is enough.

出左分, 一些可介紹給中四的不等式。

MATH190 20/20
MATH151 100/100

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在中四的數學科 (DSE 微分應該必修？不清楚) 裏，談到不等式，最多只可以問一些 $ |\sin x|, |\cos x|\leq 1$ 這類型 fact 或 $ f(x)=ax^2+bx+c$ 可以有最大/最少值的問題。

現在問，若要求你找 (例如) $\displaystyle \frac{\sin \phi}{2-\cos\phi}$ 在 $ 0\leq \phi\leq \frac{\pi}{2}$ 的最大值，作為一個中四沒有 a.maths 的學生，怎樣找？這是一個可以考考學生的問題。

開始憎 multivariable integration

唉！星期 2 仲要頂兩堂 204 lecture，聽日內要明哂第一堂 lecture 講的野，有排受。

Simple revision on differentiation

Problem 1. Suppose $  f(1) = f(2) = 0$, $  f(3) = 1$ and $  f$ is twice differentiable on $  [0,3]$. Show that $  f''(c)>\frac{1}{2},\exists c\in(0,3)$.

Problem 2. Suppose $  f(0) = 0, f(1) = 1$, $  f$ is differentiable on $  [0,1]$. Show that $  \displaystyle \frac{1}{f'(a)}+\frac{1}{f'(b)}=2$, for some distinct $  a,b\in (0,1)$.

兩條都是從某中學教師的 BLOG 中抽出來，後一條加上 distinct 的原因是為了把難度增加 (從 BLOG 中抽出來時沒有加上 distinct)。若把 distinct 移走，那很明顯存在 $  f'(c)=1$, 取 $  a=b=c$ 便完成證明。

重新再想 Lagrange's multiplier

學了多元微分學後重新再想及證明 Lagrange's multipler，有錯處的話請提省我！
http://ihome.ust.hk/~cclee/document/analysiswaytothink.pdf

利用相同的證法裏面的思想可推廣至任何一般情形，基本上我是閱讀了那個一般情形的證明再自行嘗試證明這個較常用的 case。

讀者有 math301 背景比較容易利解。
需要知識：Multivariable differentiation, implicit function theorem.

Just a review of Lagrange's Multiplier (to be corrected)

We try to create a concrete computation to find the local extreme of $  f:\mathbb{R}^n\supset U\to\mathbb{R}$ subject to the constraint $  g(\vec{x})=c$. Here $  c$ is a regular point, namely, a point such that $  g(\vec{x})=c\implies \nabla g(\vec x)\neq \vec{0}$.

Moreover, we call the surface specified by $  g(\vec x) = c$ to be $  S$, hence we are finding the local extreme of $  f$ restricted to the surface $  S$, namely, we are finding the local extreme of $  \big.f\big|_S$. Suppose now $  \big.f\big|_S$ attains the local extreme at $  \vec x = \vec x_0$, then for any $  \vec v\in\{\vec u\in\mathbb{R}^n:\nabla g(\vec x_0)\cdot \vec u = \vec 0\}:=T_{\vec x_0}S$, i.e. any tangent vector moving on $  S$ at $  \vec x_0$, we have $  D_{\vec v}f(\vec x_0)=\nabla f(\vec x_0)\cdot \vec v=0$ (for otherwise it is not an extrema).

Let's summarize the implication, $  \forall \vec v\in T_{\vec x_0}S\implies \nabla f(\vec x_0)\cdot \vec v=0$.

Since $  \nabla g(\vec x_0)$ spans the normal space of $  S$ at $  \vec x_0$, it follows that $  \nabla f(\vec x_0)=\lambda \nabla g(\vec x_0)$.

ODE

有陣時 integrator factor 真係 ``observe" 出黎，冇既定 procedure =.=。

$  \boxed{\text{\textbf{\sffamily Problem}}}$
Find an appropriate integrating factor and solve $  (x^3y^2-y)\,\mathrm dx+(x^2y^4-x)\,\mathrm dy=0$

为什么要用 (1-5)

屋企真係一個寶庫

無喇喇搵到一本關於 ODE ge 書 (學校 textbook 太貴買唔起)，內有 500 多條 solved problem = =，又有 brief review of 相關 material，正好補返我冇上堂聽 ge 不足，睇 powerpoint 真係睇到好攰 (我一直都係電腦睇)。

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今日返學校去拎返份 math 190 功課，TA 語重心長咁俾左一個 comment 我。

在 latex 裏打 $\sum$ 會有 $\sum$，打 $\displaystyle \sum$ 會有 $\displaystyle \sum$，這就是 displaystyle。

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Problem. Let  $ f:[0,\infty)\to\mathbb{R}$ with $ f(0)=-1$ be a differentiable function so that $ |f(x)-f'(x)|<1,\forall x\ge 0$.

a) Prove that $ f$ does have a limit that is infinite.
b) Give an example of such a function.

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Problem. Evaluate $\displaystyle\lim_{n\to\infty }n\left(\frac{1^{\alpha }+2^{\alpha }+...+n^{\alpha }}{n^{\alpha+1 }}-\frac{1}{\alpha+1 }\right)$, $ \alpha > 1 $.

陳志雲陳水扁有「真的假不了，假的真不了」

我有「好的說不了，說的好不了。」
我的嘴蠻臭。

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最近在高登看到不錯的不等式題目，當然你知道 Jensen's inequality 的話可以把這三條秒殺。
現在限制自己只知道 Cauchy-Schwarz inequality，考考腦筯，試證明：
對於正數 $  x_1,x_2,\dots,x_n$， $  m,n\in\mathbb{N}$，有

(a) $  \displaystyle \frac{\sum_{i=1}^nx_i}{n}\leq\left(\frac{\sum_{i=1}^nx_i^2}{n}\right)^{1/2}\leq \left(\frac{\sum_{i=1}^nx_i^3}{n}\right)^{1/3}$
更一般地，證明
(b) $  \displaystyle \frac{\sum_{i=1}^nx_i}{n}\leq\left(\frac{\sum_{i=1}^nx_i^m}{n}\right)^{1/m}$.

這裏所謂的一般跟你所期望的可能有點不同，事實上，一般地對於正整數 $  m$ 以下的不等式依然成立，\[
\left(\frac{\sum_{i=1}^nx_i^m}{n}\right)^{1/m}\leq \left(\frac{\sum_{i=1}^nx_i^{m+1}}{n}\right)^{1/{(m+1)}}.
\]

容易由 Jensen's inequality 得到結論，well... 我不確定 Cauchy 是否也能夠成功用在它的證明上 (我相信它不是萬能吧.....)，有興趣的可以試試，做到的話記得通知我，我會放到我的 problem book 裏。
(c) 對於 $  \beta_1,\beta_2,\dots,\beta_n\in\mathbb{Q}^+$，$  \beta = \beta_1+\beta_2+\cdots+\beta_n$，有 \[
\frac{\sum_{i=1}^n\beta_ix_i}{\beta}\leq \left(\frac{\sum_{i=1}^n\beta_ix_i^m}{\beta}\right)^{1/m}.
\]
現在 MATH202 正在教一些有關「測度」的知識。它令我解決了 4 個多月來的疑惑。
從前嚴民教授提到：「An integrable function must be continuous somewhere」，嚴民教授只是順便講講這個 fact，我弄不明白，便走去問了 LCM。但最後是不了了之 (應該是考慮到我不懂 Lebesgue's theorem 吧)。有了這個定理要證明它真係十分簡單，假設那個 function 處處不連續，那麼它的所有斷點所集成的一個集並不是一個零測集 (of measure zero)，也就是說那 function 本身不是 integrable 的，矛盾。因此，一個 integrable function 必定在某處連續。有了這個 fact，又可以證其他 fact 了。

煩

Problem. Verify that $  z=z(x,y)$ which is implicitly defined by $  \displaystyle x^2+y^2+z^2=yf\left(\frac{z}{y}\right)$ satisfies the partial diiferential equation
$  \displaystyle (x^2-y^2-z^2)\frac{\partial z}{\partial x}+2xy\frac{\partial z}{\partial y}=4xz$.

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要證 2 個 statement 等價，只須證明 $  (1)\implies (2)\implies (1)$, 但去到 4, 5 或更多個 statement，順序證明有可能令我們讚進死胡同。打個比方，若要證明 4 個 statement 等價，先證 $  (1)\implies (2)\implies (3)$ 卻發現自己 $  (3)\implies (4)$ 怎樣也想不到，可是 $  (2)\implies (4)$ 卻十分簡單，不妨先完成 $  (1)\implies (2)\implies (3)\implies (1)$，再從中把 $  (2)$ 抽出來證明 $  (2)\iff (4)$。這裏我們需要多做一步，但相對地可把問題變得簡單。

Math202 那份關於 integrability 的 notes，最後的第二條，有關證明四個命題等價的問題，若發現由第 3 到第 4 出現困難的話，不妨蹺一條長一點，但較平坦的路。

最後，當大家大至上認為自己對 integrability 有一定的認識，可嘗試 09 spring math203 final 有關 integrability 的題目：

Problem. Suppose $  f(x)$ and $  g(x)$ are integrable on $  [a,b]$. Prove that for any $  \epsilon > 0$, there is $  \delta>0$, such that for any partition $  P$ satisfying $  \|P\|<\delta$ and choices $  x_i^*,x_i^{**}\in [x_{i-1},x_i]$, we have \[
\left|\sum f(x_i^*)g(x_i^{**})\Delta x_i-\int_a^bf(x)g(x)\,\mathrm{d} x\right|<\epsilon.
\] 這大至上證明了，就算 $  x_i^*$ 和 $  x_i^{**}$ 所取的值不同，同樣有和相同選擇時的結果。

痾清嘔清一身輕

過度進食科大酸辣粉，嘔返出黎。

科大獸醫

一位我們學校某學生的經歷 = =。

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###!!!@@@ceev 說: (18小時前)
我比科大獸醫玩死

###!!!@@@@ceev 說: (18小時前)
上個星期去看獸醫 諗住傷風 easy job啦
開左三日藥比我 點知食完之後無好之餘 仲要大獲左
今日頂唔順去看醫生 點知話個鼻發炎
真係吹漲 獸醫可以完全唔知我個鼻發炎 我一早d鼻涕就green

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把 pdf 合併

同學以前問我懂不懂把數個 pdf 合併，嗯... 我不懂。今天心血來潮在網上找找，比較易找的都是 freeware，要錢的。用 ``Merge pdf" 找找看，發現一個不錯的網頁 (link)。

不用下載任何工具，上傳數個想要合併的 pdf，按 merge，完成！最後下載回來，方便快捷簡單易用。

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Problem. 設函數 $f$ 在 $latex x=0$ 處連續，如果 $ \displaystyle \lim_{x\to 0}\frac{f(2x)-f(x)}{x}=m$，求證 $ f'(0)=m$.

做這題時可能還須要用到以下結論。

Prerequisite. 設 $ \lim_{x\to 0}f(x)=0$，且 $ \displaystyle f(x)-f\left(\frac{x}{2}\right)=o(x)$ ($x\to 0$)，求證：$\displaystyle f(x)=o(x)$ ($latex x\to 0$)。

Well...

睇黎我要背鬼左呢兩條 inequality 佢=.=，起碼要由右手邊 sense 到左手邊舊野。

• $  a^3+b^3+c^3+3abc \geq a^2 b+a^2 c+b^2 a+b^2 c+c^2 a+c^2 b$.
• $  a^4+b^4+c^4+abc(a+b+c) \geq a^3 b+a^3 c+b^3 a+b^3 c+c^3 a+c^3 b$.

熱死人

呢幾日都好翳焗，熱死人了……。

看着嚴民教授那 inverse function theorem 的三頁證明，目前還抱着監賞的目光。

Problems of MATH202 # 17 of the last week

Since I am busy doing my work in another course, this weekly post is delayed til today :(.
Of the problems, only 5, 6, 7 are of our interest.

Problem 5. Let $  f:\mathbb{R}\to\mathbb{R}$, be a three times differentiable function. If $  f(x)$ and $  f'''(x)$ are bounded functions on $  \mathbb{R}$, show that $  f'$ and $  f''$ are also bounded functions on $  \mathbb{R}$.

My way.
Just make good use of the expansion $  f(x+h)=f(x)+f'(x)h+\frac{1}{2}f''(x)h^2+\frac{1}{3!}f'''(x+\theta h)h^3$, for some $  \theta \in (0,1)$.

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Problem 6. If $  f(x)$ and $  g(x)$ are $  n$ times differentiable and $  f^{(n-1)}(x),g^{(n-1)}(x)$ are both continuous in $  [a,b$. Then there exists a number $  c \in(a,b)$ such that \[ \frac{\displaystyle f(b)-f(a)-\sum_{k=1}^{n-1}\frac{(b-a)^k}{k!}f^{(k)}(a)}{\displaystyle g(b)-g(a)-\sum_{k=1}^{m-1}\frac{(b-a)^k}{k!}g^{(k)}(a)}=\frac{(m-1)!}{(n-1)!}(b-c)^{n-m}\left(\frac{f^{(n)}(c)}{g^{(m)}(c)}\right).\]

NO IDEA, I am just able to prove the case when $  m=n$.

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Problem 7. Let $  f$ be $  p$ times differentiable on $  \mathbb{R}$ and let $  M_k=\sup\{|f^{(k)}(x)|:x\in\mathbb{R}\}<\infty$, $  k=0,1,2,\dots,p$ and $  p\ge 2$. Prove that $  M_1\leq \sqrt{2M_0M_2}$ and \[M_k\leq 2^{\frac{k(p-k)}{2}}M_0^{1-\frac{k}{p}}M_p^{\frac{k}{p}}$, for $  k=1,2,\dots,p-1.\]
My way.
(1) Same as the case of problem 5, replace $  h$ by $  -h$ to construct another equation (remember to choose different $  \theta$), subtract two equation, observe that $  0\leq 2M_0 +2M_1h+M_2h^2$ for any $  h$, while discriminant $  \leq 0$, we are done.
(2) In exactly the same manner as (1), we conclude that $  M_{j+1}\leq \sqrt{2M_jM_{j+2}}$ for all $  j\leq p-2$. Now we take product $  \prod_{j=m}^{n}$ on both sides, having \[ \sqrt{M_{n+1}M_{m+1}}\leq (\sqrt{2})^{n-m+1}\sqrt{M_{n+2}M_m}.
\] Before we proceed, we first consider two cases. If $  M_k=0$, then the inequality we are asked to prove obviously holds since right hand side is always non-negative. In case if $  M_k>0$, then we take the product $  \prod_{m=0}^{k-1}$ on both sides of $  \sqrt{M_{n+1}M_{m+1}}\leq (\sqrt{2})^{n-m+1}\sqrt{M_{n+2}M_m}$, it results in \[ M_n^k\leq \left(\frac{M_0}{M_k}2^{k(2n-k+1)/2}\right)M_{n+1}^k.\]
(I remember I have replaced $  n$ by $  n-1$ to make the inequality seem better), we are interested in this because it is a beautiful (in the sense of solving the problem) reccurence relation, we have a direct consequence \begin{align*}
 M_n^k&\leq \left(\frac{M_0}{M_k}\right)^{p-k}2^{\sum_{j=n}^{n+p-k-1}k(2j-k+1)/2}M_{n+p-k}^k\\
&=\left(\frac{M_0}{M_k}\right)^{p-k}2^{k(p-k)(2n-2k+p)/2}M_{n+p-k}^k.
\end{align*} Finally, we take $  n=k$, $  M_k^p\leq M_0^{p-k}2^{\frac{k(p-k)p}{2}}M_p^{k}$.

Math202 presentation

這是我預備的 Presentation soft copy (Click)
TA 說：「又屈條 inequality 出黎....」= =。

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今天有人再一次問我，甘仔 (定金?) 先前提出的問題的解法，已知 $g$ 是 differentiable (其餘 condition 略)

他問：「為甚麼一定存在 $c$ 令到 $ g'(c) = 0$？」
(這個問題沒甚麼，我再一次解釋一番)

我道：「因為由 slope 看到 local extrema 必定在那開區間裏發生」
(再明確一點 $ g'(a) < 0\implies$ $ g(a)$ 不是 minima，$ g'(b)>0\implies$ $ g(b)$ 也不是 minima，又由於一個 continuous function 必定在閉區間裏達到最大最小值，因此最小值必定在開區間 $ (a,b)$ 裏發生。)

他又問道：「那 $g$ 起角怎麼辦？」
我：「……」

事後又給我撞正他說我甚麼解釋得不清楚，諸如此類的，我快心臟病發。有時腦內補完也是蠻重要的吧?

MATH190 presentation next week

I am here to post my answer (numerical and method only) for my teammate, you can use it to check. If our answers are different,  please discuss with me.

Problem (1990 Austrian-Polish Math Competition)
Let $  n>1$ be an integer and let $  f_1,f_2,\dots, f_{n!}$ be the $  n!$ permutations of $  1,2,\dots,n$ (each $  f_i$ is a bijective function from $  \{1,2,\dots,n\}$ to itself).  For each permutation $  f_i$, let us define $  \displaystyle S(f_i)=\sum_{k=1}^n|f_i(k)-k|$. Find $  \displaystyle \frac{1}{n!}\sum_{i=1}^{n!}S(f_i)$.

My numerical answer. $  \displaystyle \frac{n^2-1}{3}$.

Next question I just use Jensen's inequality once.

Lazy to choose what to dress

惡補 linear algebra

.... bilinear ... multilinear .... 甚麼鬼東西...。
最近上 MATH204 愈來愈一頭霧水。

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Lift 3 Room 5003

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Problem (1990 IMO Shortlisted Problem).
For any $  x,y,z >0$, $  xyz=1$, prove that \[ \frac{x^3}{(1+y)(1+z)}+\frac{y^3}{(1+z)(1+x)}+\frac{z^3}{(1+x)(1+y)} \ge \frac{3}{4}.\]

我第一次看這條題目時它給的 condition 是 $  x+y+z\ge 3$，代替了 $  xyz=1$，不知道其用意。重提這問題的原因是那時因年少無知，提出了這樣的答法 (按我)。這很明顯是錯的，我們從來沒有學過考慮三組數字的排序不等式。所以以前看過我這做法的人，對不起，我錯了。

首次看到它的原因是在瀏灠 Mathdb 時看到這頁(按我)，看到「頗有難度的不等式題目」的 pdf，正巧那時我正研究着不等式，便嘗試着解決裏面的問題。

李教授提出這個例子的目的是要表現出 Muirhead 有多暴力。有學過 H$  \ddot{\text{o}}$lder's 不等式的話，我們能給出一個較``雅"的做法。證法是很明顯的，在這裏不花口水了。

內地的高考

不等式是內地高考一個重要的課題。懷着好奇的心去做做看，不懂的還蠻多的...。以下都是些解了的題目，在此不提供解答了。

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Problem 1. Prove that for any $  a,b,c\in\mathbb{R}$, $  \displaystyle \sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\ge \frac{3\sqrt{2}}{2}.$

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Problem 2 (Mircea Lascu).  Let $  a,b,c$ be positive real numbers such that $  abc=1$. Prove that $  \displaystyle \frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{2}}\ge \sqrt{a}+\sqrt{b}+\sqrt{c}+3.$

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Problem 3. Let $  a,b,c,x,y,z$ be positive real numbers such that $  x+y+z=1$. Prove that $  \displaystyle ax+by+cz+2\sqrt{(xy+yz+zx)(ab+bc+ca)}\leq a+b+c.$

An integral that even wolfram integrator cannot find

Here is the ``discussion" from uwants (click me)
Another discussion from mathlinks (click me)
But I find it in a way totally different from them, that's a challenge to find the formula.

Problem. Find the explicit formula of the integral $  \displaystyle \int \frac{x^2-1}{(x^2+1)\sqrt{1+x^4}}\,dx$.

My way is as follows

局部調整法

Kin Li 教授 d material 真係教左種好方便的 skill，由其是個 Local Refinement Method 感覺由複雜變簡單。同學學左個技巧就可以解決到呢題。
 
Problem (IMO-1984-1). Let $  x,y,z$ be non-negative integers satisfying $  x+y+z=1$, prove that \[xy+yz+zx -2xyz\leq \frac{7}{27}.\]

一開始 fix 數切忌不要 fix 得太多。Smoothing method 到而家都仲未領會到。

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soarer TA 已經發放了下星期 4 將會用的 tutorial notes...，然後我突然看到這一題。

Problem (USA-1997). Let $  a, b, c > 0$. Prove that $\displaystyle \frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{c^3+a^3+abc}\leq \frac{1}{abc}$.

只是借題發揮一下，這是一條需要很少不等式知識的題目。觀察到 $  (a^2-b^2)(a-b)\ge 0 \iff a^3+b^3\ge ab(a+b)$，我們便完成了這條問題。

My way.
$  \displaystyle\sum_{cyc} \frac{1}{a^3+b^3+abc}\leq\sum_{cyc}\frac{1}{ab(a+b) + ab(c)}=\sum_{cyc}\frac{c}{abc (a+b+c)}=\frac{1}{abc}$.

Muirhead 確實是很黃很暴力的不等式，它能夠令我們不加思索也能夠解決一條不等式的問題。在我的問庫中 (點我)，第 41 條正正是好例子。利用電腦運算它等同於要證明 \[  \displaystyle\sum_{sym}a^{12}b^3c^3 + \sum_{sym} a^9b^3c^3+\sum_{sym}a^6b^3c^3\leq \sum_{sym}a^{14}b^2c^2+\sum_{sym}a^{13}bc+\sum_{sym}a^{12}.\] 成立。比較相同次方的和，問題得證。有時不一定要齊次的時候才成功，用電腦爆它一爆可能別有洞天。

MATH190 可以開着 notebook 考試嗎?

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An exercise in MATH204 assignment that suits MATH102

Problem. Find the range of $  p$ at which $  \displaystyle \lim_{{\normalsize (x,y)\to \vec{0},y^2>x>0}} \frac{x^py}{x^2+y^2}$ converges.

Problem of today's MATH202 tutorial lesson

Too many hints are present in this note.

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In example 4 we have come across a boring question. Here is a much interesting one.
$latex \boxed{\text{\textbf{Problem}}}$
Suppose $latex f''(0)$ exists and $latex f''(0)\neq 0$. Prove that in the mean value theorem $latex f(x)-f(0)=xf'(\theta x)$, we have $latex \displaystyle \lim_{x\to 0}\theta =\frac{1}{2}$.

This is a problem of my past assignment in MATH203, and the example 4 is exactly one of them.

Problem of today's MATH190 tutorial lesson

Problems typed below can also be found in this tutorial note, http://ihome.ust.hk/~masoarer/math190/week3.pdf. For fear that the web host will be out of service due to any reason, I have typed everything below. This text is prepared for future use, for example, my ``collection of problems".

Exercise 1. Let $  a,b,c>0$, show that $  (a^3+1)(b^3+1)(c^3+1)\ge (a^2b+1)(b^2c+1)(c^2a+1)$.

My way.
Note that $  (a^3+1)(a^3+1)(b^3+1)\ge (a^2b+1)^3$, we are done (you can see what happens when $  (a,b)=(b,c)$ and $  (a,b)=(c,a)$.).

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Exercise 2. If $  a,b,c>0$, prove that $  \displaystyle \frac{a+\sqrt{ab}+\sqrt[3]{abc}}{3}\leq \sqrt[3]{a\cdot \frac{a+b}{2}\cdot \frac{a+b+c}{3}}$.

My way.
We see that \begin{align*}
 a\cdot \frac{a+b}{2}\cdot \frac{a+b+c}{3}&=\left(\frac{a}{3}+\frac{a}{3}+\frac{a}{3}\right)\left(\frac{a}{3}+\frac{a+b}{6}+\frac{b}{3}\right)\left(\frac{a}{3}+\frac{b}{3}+\frac{c}{3}\right)\\
&\ge \left(\frac{a}{3}+\sqrt[3]{\frac{ab(a+b)}{2\cdot 3^3}}+\frac{\sqrt[3]{abc}}{3}\right)^3.
\end{align*} Finally, $  \displaystyle\frac{a+b}{2}\ge\sqrt{ab}$ does solve the problem.

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Exercise 5. Let $  a,b,c>0$, prove that $  \displaystyle \frac{a^6}{b^2+c^2}+\frac{b^6}{c^2+a^2}+\frac{c^6}{a^2+b^2}\ge \frac{abc(a+b+c)}{2}$.

My way.
Use exercise 4 once, we have a new lower bound $  \displaystyle \frac{a^6}{b^2+c^2}+\frac{b^6}{c^2+a^2}+\frac{c^6}{a^2+b^2}\ge \frac{1}{6}(a^2+b^2+c^2)^2$, since \[(a^2+b^2+c^2)^2 \ge (ab+bc+ca)^2\ge 3(abbc + bcca + caab)=3abc(a+b+c),
\] we are done.

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Exercise 6. Let $  f$ be convex on $  [a,b]$. If $  c,d\in[a,b]$ with $  c-a>b-d$, prove that $  \displaystyle 2f\left(\frac{c+d}{2}\right) \leq f(c)+f(d)\leq f(c+d-b)+f(b)$.

My way.
The left inequality is obviously true. To go through right inequality, we first determine the position of $  a,b,c,d$ and $  c+d-b$. Observe that the inequality will change nothing if we interchange $  c$ and $  d$, so without loss of generality, we assume that $  c\leq d$, finally $  c+d-b\leq c \iff d\leq b$ we have known that $  a\leq c+d-b\leq c\leq d\leq b$.

Since $  \displaystyle f(c)+f(d)\leq f(c+d-b)+f(b)\iff \frac{f(b)-f(c)}{b-c}\ge \frac{f(d)-f(c+d-b)}{d-(c+d-b)}$, the equivalent inequality is obvious by convexity.

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Exercise 9. Let $  A,B,C$ be angles of a triangle. Prove that $  \displaystyle \frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\sin C}{C}\leq \frac{9\sqrt{3}}{2\pi}$.

My way.
We observe the ineqality $  \displaystyle \frac{\sin x}{x}\leq \left(\frac{\frac{\pi}{3}\cdot \frac{1}{2}-\frac{\sqrt{3}}{2}}{\left(\frac{\pi}{3}\right)^2}\right)\left(x-\frac{\pi}{3}\right)+\frac{3\sqrt{3}}{2\pi}$. I haven't rigorously checked its validity by calculus, at least it is true for $  x\in (0,\pi)$ with the aid of graph ploting, hence replacing $  x$ by respectively $  a,b,c$ and adding them up, we get desired result.

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Exercise 10. Let $  f$ be a convex function on $  [a,b]$ and $  x,y,z\in [a,b]$. Prove by majorization inequality that $  \displaystyle f(x)+f(y)+f(z)+3f\left(\frac{x+y+z}{3}\right)\ge 2 \left(f\left(\frac{x+y}{2}\right)+f\left(\frac{y+z}{2}\right)+f\left(\frac{z+x}{2}\right)\right)$.

My way.
Correcting.

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Exercise 3, 4, 7, 8 are skipped (I would not include these problems in my pdf), I am trying exercise 10.

Interesting identity, deduced from a Probability problem of Ming (CU,CSC)

I can do all except (b) orz.

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Well this is a good example that why combinatorics and probability can generate interesting identity. (discussion from part (e) on) It can be easily deduced that $  \displaystyle P(B_i)=\frac{1}{2^{n-1}(2^n-1)}$. Now we turn our objective to finding the probability that $  A$ and $  B$ will meet at the $  r$-th round. Having known what $  P(B_i)$ is, our desired probability is obviously

$  p = \displaystyle 2^{n-r}\cdot P(B_i) = \frac{2^{n-r}}{2^{n-1}(2^n-1)}$.

We take a step further, the same probability can be duduced if $  A$ and $  B$ cannot meet and they win at the first $  r-1$-th rounds and occasionally meet at the $  r$-th round, this is the same as \[
p = \underbrace{\prod_{k=1}^{r-1}\left\{\frac{\displaystyle \binom{2^{n-k}}{2}\cdot 2^3}{2^{n-k+1}(2^{n-k+1}-1)}\cdot \left(\frac{1}{2}\right)^2\right\}}_{\text{don't meet and win at the first r - 1 rounds}}\cdot \overbrace{\left(\frac{2}{2^{n-r+1}}\cdot \frac{1}{2^{n-r+1}-1}\cdot 2^{n-r}\right)}^{\text{meet at the last round}}.
\] We now equate two $  p$'s, having \[
\prod_{k=1}^{r-1}\left(1-\frac{1}{2^{n-k+1}-1}\right)=\frac{2^n-2^{r-1}}{2^n-1},
\] by simple algebra we get a more convenient form \[\boxed{\dis \prod_{k=j}^n\left(1-\frac{1}{2^k-1}\right)=1-\frac{2^{n-j+1}-1}{2^n-1}.}
\] If this is an unprecedented identity, then I would call it CC's identity (政昌恆等式).

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Motivated by this problem, if we don't know the above formula, can we show that \[  \sum\limits_{k=2}^\infty\displaystyle\ln\left(1-\frac{1}{2^k-1}\right)
\] converges? The answer is yes! Take a suitable $N$ such that when $  k >N$, we always have $  \displaystyle \frac{1}{2^k - 1}<1$, consider the taylor expansion $  \displaystyle \ln (1+x)=x-\frac{1}{(1+c)^2}x^2$, for some $  c$ in between $  0$ and $  x$, now consider $  |x|<1$ such that $  |c|<\delta <1$, then we get by putting $ x=x_k$, $  |x_k|<1$, \[
 \frac{x_k^2}{(1+\delta)^2}<x_k-\ln(1+x_k)<\frac{x_k^2}{(1-\delta)^2},
 \] actually $  x>\ln (1+x),\forall x>-1$ simply because $  e^x\ge 1+x$. Now let $  \displaystyle x_k=-\frac{1}{2^k-1}$, our required $  N$ is simply 1, so we know that $  \sum_{k=2}^\infty x_k^2$ converges, thus by above inequality, $  \sum_{k=2}^\infty (x_k-\ln(1+x_k))$ also converges by comparison test, since $  \sum_{k=2}^\infty x_k$ converges, $  \sum_{k=2}^\infty \ln(1+x_k)$ also converges, we are done. Moreover, $  \sum\limits_{k=2}^\infty\displaystyle\ln\left(1-\frac{1}{2^k-1}\right)=-\ln 2$.

一年多前

我原來發過這樣的題目 ...

歌…

最近迷上左遠藤正明同古泉一樹 d 高音 ... sosad。

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From Yan Min:

(12/2/2010) I will organize a hiking on Saturday, Feb 20. We will meet at noon at the university bus station. You and your friends are welcome. The hiking in Hong Kong is generally not strenuous. However, I am not reponsible for your safety.

MATH204 (Analysis II) 及 MATH323 (Topology) 的同學仔都會去，不是這兩個 course 的同學也歡迎參加！

人情朱古力

收到 d 咁 ge 野，唔知點算好 = =。

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今日陪屋企人睇錦衣衛。呢套片由頭到尾都係打打打，我覺得非常適合以下呢幾類人去睇

1. 對甄子丹情有獨鍾


2. 鍾意睇型仔 (冇靚女)


3. 鍾意睇打鬥鏡頭


4. 饑不擇食

唔係以上呢幾類買飛前三思 (極度主觀 mode)。順帶一提，套野我睇到半路就走左，我都唔知結局係點，等到屋企人出黎就冇再提過套野。

Problems of today's MATH202 tutorial notes #15

Problem 3. Let $  \{a_n\}$ and $  \{b_n\}$ be two sequences such that $  \lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n=c$, suppose $  f'(c)$ exists. Show that $  \displaystyle \lim_{n\to\infty}\frac{f(b_n)-f(a_n)}{b_n-a_n}=f'(c)$.

My way.
I have to put some restictions to ensure my deduction is correct. If $  f'(x)$ is continuous near $  c$, then we are done with the Mean Value Theorem. But it, first and foremost, requires $  f(x)$ to be differentiable.

To avoid continuity of derivative, we inevitably set Mean Value Theorem aside, then differentiability of $  f(x)$ other than $  c$ doesn't carry weight any more. But I have to suppose we always have $  \boxed{a_n<c<b_n}$ or $  \boxed{b_n<c<a_n}$ for sufficiently large $  n$, this assumption will be used later on. By the fact that the existence of first order derivative implies the existance of linear approximation near $  c$, we have for any $  x$ near $  c$, \[f(x)=f'(c)(x-c)+f(c)+o(x-c),
\] Here for any $  \epsilon >0$, there exists a $  \delta > 0$ such that $  \displaystyle 0<|x-c|<\delta \implies \left|\frac{o(x-c)}{x-c}\right|<\frac{\epsilon}{2}$.

While for any $\delta_0 >0$, there exists an $  N$ such that $  n>N\implies |b_n-c|, |a_n-c|<\delta_0$, we take $  \delta_0<\delta$, now for $  n>N$,\begin{align*}
&{\color{white}=}\left|\frac{f(b_n)-f(a_n)}{b_n-a_n}-f'(c)\right|\\
&=\left|\frac{f'(c)(b_n-c-(a_n-c))+o(b_n-c)+o(a_n-c)}{b_n-a_n}-f'(c)\right|\\
&=\left|\frac{o(b_n-c)+o(a_n-c)}{b_n-a_n}\right| \\
&\leq \left|\frac{o(b_n-c)}{b_n-c}\right|\left|\frac{b_n-c}{b_n-a_n}\right|+\left|\frac{o(a_n-c)}{a_n-c}\right|\left|\frac{a_n-c}{b_n-a_n}\right| \\
&<\left|\frac{o(b_n-c)}{b_n-c}\right|+\left|\frac{o(a_n-c)}{a_n-c}\right|\text{(by the assumption)}\\
&<\epsilon.
\end{align*}

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Redo example 10. Show that $  \displaystyle 2<65^{1/6}<2+\frac{1}{192}$.

My way.
The left hand side is obvious. For the right hand side, observe that \[
65=2\left(1+\frac{1}{2^6}\right)^{\frac{1}{6}}<2\left(1+\frac{1}{6\cdot 2^6}\right)=2+\frac{1}{192}.
\] We have used $  (1+x)^\alpha \leq 1+\alpha x,\forall x\ge 0,\forall \alpha \in[0,1]$.

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Problem 6. Let $  f:[0,\infty)\to\mathbb{R}$ be continuous and $  f(0)=0$. If $  |f'(x)|<|f(x)|$ for very $  x>0$, show that $  f(x)=0$, for every $  x\in[0,\infty)$.

My way.
Instead of considering $  [0,\frac{1}{2}]$, we can consider any interval $  [0,p]$, $  p<1$ first. It can be shown that by mean value theorem,\begin{align*}

|f(x)-f(0)|&<|f(y_n)||x|\\
&=|f(y_n)-f(0)||x|\\
&<|f(y_{n-1})||x||y_n|\\
&<\cdots \\
&<|x||y_n|\cdots |y_2||f(y_1)|\\
&<p^n|f(y_1)|,\forall x\in [0,p].
\end{align*} Letting $  n\to\infty$, we get $  f(x)=\inf\{p^n|f(y_1)|:n\in\mathbb{N}\}=0$. We can show by induction that $  f(x)=0$, for all $  x\in [(n-1)p,np],n\in\mathbb{N}$ in the same manner.

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Problem 9. Give an example of a function that is differentiable on $  (0,1)$ and $  f(0)=f(1)$ but does not satisfy the Rolle's Theorem.

My way.
Define $  f(x)=\sqrt{x},\forall x\in[0,1)$ and $  f(1) = 0$.

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Additional problem. Let $  f(x)$ be a differentiable on $  [a,b]$, $  f'(a)<f'(b),$ for any $  y_0\in (f'(a),f'(b))$, prove that there exists $  c\in(a,b)$ such that $  f'(c)=y_0$.

My way.
Since that $  f$ is differentiable cannot imply the continuity of $  f'$. Intermediate value theorem fails to work here. We on the contrary define a new continuous function, $  g(x) = f(x)-y_0 x$. It can be easily varified that $  g'(a)=f'(a)-y_0<0$ and $  g'(b)=f'(b)-y_0>0$, then extreme value cannot be attained at the end point, the minima must lie somewhere else in the interval $  (a,b)$, hence there exists $  c\in(a,b)$ such that $  g'(c)=0\implies f'(c)=y_0$.

Today's MATH190 problems in tutorial notes (the remaining is basic)

Problem 5. Let $  a,b,c>0$ be the sidelengths of a triangle. Prove that \[
a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge 0.
\] My way.
We use the usual substitution, $(a,b,c)=(x+y,x+z,y+z)$, $  x,y,z>0$, then the original inequality is equivalent to \[
 x^3y+y^3z+z^3x\ge xyz(x+y+z).
\] Finally, the last inequality is easy enough to see by noting that \[
x^3y+y^3z+z^3x = xyz\left(\frac{x^2}{z}+\frac{y^2}{x}+\frac{z^2}{y}\right),
\] we are done.

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We are learning how to solve ODE. Now for those who has learnt ODE in secondary school (took AL applied maths paper II), you must be able to do the following.

1. Evaluate $\displaystyle u(x)=1+\frac{x^3}{3!}+\frac{x^6}{6!}+\frac{x^9}{9!}+\cdots$, how do you link it with solving ODE? Of couse the substitution of cube root of unity will do in this question.


2. Let $ f (x)$ be a differentiable function on $ [0, 1], f (0) = 0 $ and $f (1) = 1$, prove that $\displaystyle \int_0^1|f(x)-f'(x)|\,dx\ge \frac{1}{e}$. Once again, technique involved in solving ODE is needed. Can you recall it?

開始忙了

今日好充實，朝早上 (訓) 完 MATH151 再去再聽 eating style (HL001)，然後食午餐再備課備到點半上 MATH204。嚴民 d 堂講得好快，又易明，可惜一個星期得兩堂，如果一星期有四堂我都會肯去。落左堂之後同 204 同學做功課同備課去到 5:45，六點去 MATH204 tutorial lesson。

但奇怪嚴民教授明明 o係 course webpage 寫住

(5/2/2010) Tutorial times set at Tuesday, 18:00 - 18:50, in room 3215.

好在當時 save 左 ust 搵房網址入電腦 (出發時先醒起 3215 房間)，點我，慳左好多功夫，但原來 3215 係一個錯 ge 地址 (3215 係 TA 房)，有 4 至 5 個都走錯左地方。搵房果陣有個同班的人搵我講野...，o係果時先知道原來成日同班內地生傾計果個人都識講廣東話... (堔圳學生)，俾佢呃左好耐 = =，搞到我一直都唔敢搵佢講野。最後大家都要返返去 3584 上堂。

今次 tutorial 堂好奇怪 (奇怪在變返正常)，真係有個 PG 黎做 tutor...，平時都係教授一手包辦埋 tutorial 堂。老實，有少少失望。

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I have come arcoss a question in uwants, prove that $  \tan 1^{\circ}$ is irrational, and here is the discussion.
I suppose that $  \tan 1^{\circ}$ is rational, then by the identity
$  \displaystyle \tan (30 \theta)=\frac{\sum_{k=1}^{15}(-1)^{k+1}\binom{30}{2k-1}\tan^{2k-1}\theta}{\sum_{k=0}^{15}(-1)^k\binom{30}{2k}\tan^{2k}\theta}$

We have by putting $  \theta = 1^{\circ}$, $  \displaystyle \frac{1}{\sqrt{3}} =\tan (30^{\circ})=\frac{\sum_{k=1}^{15}(-1)^{k+1}\binom{30}{2k-1}\tan^{2k-1}1^{\circ}}{\sum_{k=0}^{15}(-1)^k\binom{30}{2k}\tan^{2k}1^{\circ}}$, this shows that $  \displaystyle\frac{1}{\sqrt{3}}$ is rational, a contradiction. I hope this is one of the way that apook (koopa) thinks of.

余振強紀念中學校友會成立十週年既會暨老師榮休宴

日期: 二零一零年三月二十日
時間: 下午四時正 (參觀校舍)
晚上七時正 (晚宴)
地點: 學校禮堂

內容: 週年會員大會及祝賀趙錦輝老師於09年榮休，亦邀請嘉賓老師(現任及退休老師)出席，設置自助晚宴及神秘節目，歡迎攜眷出席。

入場票價: 每位參加者$100，每席10位
付款方式: 透過銀行入數到校友會的銀行戶口
(恆生銀行戶口: 024-242-4-039747)，
入數紙需連同入數人姓名、參加者人數、日期和聯絡電話一併 fax(2760 1488) 或郵寄到學校。

截止日期: 二零一零年二月二十八日
聯絡方法: 陳學明 9031 0921
蔡文遜 9251 4420
勞國樑老師 9273 8036

Email: yckmcaa@gmail.com

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Problem. Show that for any $  a,b,c>0$ (not necessarily integer), $  \displaystyle a^ab^bc^c\ge (abc)^{(a+b+c)/3}$.

這是一條例題，李教授以此來 dem 一 dem Chebyshev 不等式的用法，其後便給我們一個練習：
重證這一條不等式，但你只能使用 AM-GM inequality。

這是不難的，中七的學生可當練習。

A PuiChing Question (to someone)

Problem. Denote $  [x]$ the greastest integer not exceeding $  x$, calculate $  \displaystyle\sum\limits_{k=1}^{100}[\sqrt{k(k+4)+20}]$.

Solution.
First note that $  \sqrt{k(k+4)+20}=(k+2)\sqrt{1+\big(4/(k+2)\big)^2}$, it conveys us some messages. As $  k$ increases, we have $  \sqrt{k(k+4)+20}= k+2 + f(k)$, where $  f(k)\to 0$ as $  k\to\infty$. But only the integral part deserves our concern, so if $  k+2 <\sqrt{k(k+4)+20}<k+3$, then everything goes smooth and we have $  [\sqrt{k(k+4)+20}]=k+2$. But when does this inequality hold? Just solve the inequality we have $  k>5.5$, so the inequality is true of $  k\ge 6$ and thus the value should be $  5+5+6+7+8+8+9+10+\cdots +102 = 5256$.     $  \blacksquare$

Last problem of MATH202 tutorial note #14

Problem. Consider a continuous function $  f$ on $  [a,b]$, suppose for any $  a\leq x\leq b$, there is $  a\leq y\leq b$, such that $  |f(y)|\leq \frac{1}{2}|f(x)|$. Prove that there is a $  c\in[a,b]$ such that $  f(c)=0$.

My way.
It is often easy to prove the converse to be impossible for this kind of problems. We on the contrary suppose that $  f(x)\neq 0,\forall x\in[a,b]$, then either $  f(x)>0$ or $  f(x)<0$ for all $  x\in[a,b]$.

We assume that $  f(x)>0$, due to continuity there exists $  x_0$ such that
$  (*):\qquad f(x)\ge f(x_0)>0, \forall x\in[a,b].$

However, we have the following observation.
$  \exists y_1\in [a,b], f(y_1)\leq \frac{1}{2}f(y_0)$;
$  \exists y_2\in [a,b], f(y_2)\leq \frac{1}{2}f(y_1)$;
$  \qquad \vdots$
$  \exists y_n\in [a,b], f(y_n)\leq \frac{1}{2}f(y_{n-1})$.

thus $  \displaystyle f(y_n)\leq \frac{1}{2}f(y_{n-1})\leq \left(\frac{1}{2}\right)^2f(y_{n-2})\leq \cdots \leq \left(\frac{1}{2}\right)^nf(y_0)$. Since right hand side tends to 0 as $  n\to\infty$, there exists an $  N$ such that $  n>N\implies f(y_n)<\epsilon < f(x_0)$, a contradiction with (*). The case that $  f(x)<0$ is essentially the same, hence there exists $  c\in[a,b]$ such that $  f(c)=0$.

Alternatively, if we know that any bounded sequence have a convergent subsequence, then this problem can be solved in a much neater way. As above we show that \[
|f(y_n)|\leq \frac{1}{2}|f(y_{n-1})|\leq \left(\frac{1}{2}\right)^2|f(y_{n-2})|\leq \cdots \leq \left(\frac{1}{2}\right)^n|f(y_0)|,\forall n \in \mathbb{N}\cup\{0\},
\] then since there exists a convergent subsequence $  \{y_{n_k}\}$ with $  \lim_{k\to\infty}y_{n_k}=c\in[a,b]$, thus taking limit on the proved inequality, we get \[
0\leq \lim_{k\to\infty}|f(y_{n_k})|=\left|f\left(\lim_{k\to\infty}y_{n_k}\right)\right| =|f(c)|\leq\lim_{k\to\infty}\left(\frac{1}{2}\right)^{n_k}|f(y_0)| = 0,
\] here we used the fact that a composite of continuous function is still continuous, we are done.

Left a name in Excalibur!

There is a fairly easy question in the problem corner of the latest mathematical excalibur:
Show that $  \displaystyle \sum_{k=0}^{n-1}(-1)^k\cos^n\left(\frac{k\pi}{n}\right)=\frac{n}{2^{n-1}}$, for any positive integer $  n$.

The term  $  2^{n-1}$ is reminiscent of something, isn't it? What is it? How it works?

But I guess there must be some combinatorics/probabilistic argument, especially the use of inclusion-exclusion principle.

I am glad that Kin  Li adopted my solution of the strange inequality and quoted my name^^.
For details:
http://www.math.ust.hk/excalibur/v14_n3.pdf
(You can search LEE Ching Cheong to see where I am XD)

Another basic counting in discusshk

Problem. There are 6 distinct presents. How many ways to distribute the presents to three people if everyone at least have one present?

In the past I have answered how to distribute 20 different presents to 6 people (of course different people!). Intrinsically two problems are the same. The complexity lies here is the way we draw presents, it is not of necessity to distribute all presents to them.

Ok, now back to this question, just applied the same trick that I have developed in the past, this is an easy problem if one can set up a reccurence relation like the following:

Define $s_n = \{\text{number of way of distribution if exactly \textit{n} people have present(s).}\}$, \[ s_n = ({n+1})^6-\sum_{i=0}^{n-1}\binom{n}{i}s_i,
\] setting  $ i=1,2$ and defining $  s_0=1$, one can have that \[
\left\{\begin{array}{l}s_0=1\\s_1 = 2^6 - 1\\s_2 = 3^6 - 2\cdot 2^6 + 1\\s_3 = 4^6 - 3^7 + 3\cdot 2^6 - 1 = 2100\end{array}\right.\]
One may ask what happens when presents are identical, the solution is much trivial and I guess every statistic (especially MAEC ?) student in UST is able to count it.here is the PDF concerning 20 presents case
Click me

銘問我的問題

解 functional equation，CU CS係咪 maths major 黎- -。
已知

找 T，b 是任意的 constant。

An ugly limit

$\displaystyle \lim_{n\to\infty}\left\{\frac{1}{n^{4}}\left (\sum_{k=1}^{n}k^{2}\int_{k}^{k+1}x\ln\big( (x-k)(k+1-x)\big) \mathrm{d} x\right )\right\}$.