Quantum Physics = Math

Math Online Tom Circle

CN Yang (杨振宁) Yang-Mills Equation in Quantum Physics and ‘Fibre Bundle’ in Differential Geometryby SS Chern (陈省身) are equivalent, however both men worked independently for 30 years!

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杨澜访谈录杨振宁翁帆来做客

Chinoiseries 《汉瀚》［中/英/日/韩/法］

1957, 35岁第一个华人诺贝尔奖。
2004, 82岁娶28岁。
2015, 93岁牵39岁的夫妻。

翁帆(28)从敬仰伟人而爱杨振宁(82)。

三大科学贡献:
1. 宇恒不对称 : nuclei weakforce (弱力)不对称
2. Yang-Baxter Law (Proved by Quantum Group Mathematics)
3. Yang-Mills Conjecture (Clay Prize US $1 million, 7 Millennium Unsolved Math Problems)
如果3)被证明, 可能杨振宁得第二个诺贝尔奖。

父亲杨武之是”庚子赔款”留美的中国第一位数学博士(Chicago University), 把现代代数(Modern Algebra)介绍进中国。儿子杨振宁15岁就教儿子群论(Group Theory) — 后来证明”宇恒不对称” — 用的就是他美国恩师 Dickson的名著。

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Match Algorithm

Math Online Tom Circle

1962 two American economists David Gale & Lloyd Shipley designed The “Stable Marriage Problem” aka “The Match“.

Note: ‘Stable’ means nobody would be unhappy or breakup after the match.

Applications:
1. Matching couples
2. Matching hospitals & doctor graduates
3. Match schools to students
4. Match HDB house to families
5. …

Scenario: An island with 4 men (m1, m2, m3, m4) and 4 women (w1, w2, w3, w4). You are to match 4 couples of opposite sex.

Each man would propose to a woman. However both men and women could list down their preferences with ranking, the higher ranked person would be given the choice.

Suppose the women preferences are (Table 1):

Choices	1st	2nd	3rd	4th
w1	m1	m2	m3	m4
w2	m2	m4	m1	m3
w3	m3	m4	m1	m2
w4	m4	m3	m2	m1

Suppose the men preferences are (Table 2):

Choices

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Homotopy of Paths

For this post we will explain what is a homotopy of paths.

Source: Topology (2nd Economy Edition)

The book above is a nice introductory book on Topology, which includes a section of introductory Algebraic Topology.

Definition: If f and f’ are continuous maps of the space X into the space Y, we say that f is homotopic to f’ if there is a continuous F: X x I -> Y such that

F(x, 0)=f(x) and F(x,1) = f'(x)

for each x. The map F is called a homotopy between f and f’. If f is homotopic to f’, we write $f \simeq f'$ .

If f and f’ are two paths in X, there is a stronger relation, called path homotopy, which requires that the end points of the path remain fixed during the deformation. We write $f \simeq_p f'$ if f and f’ are path homotopic.

Next, we will prove that the relations $\simeq$ and $\simeq_p$ are equivalence relations.

If f is a path, we shall denote its path-homotopy equivalence class by [f].

Proof: We shall verify the properties of an equivalence relation, namely reflexivity, symmetry and transitivity.

Reflexivity:

Given f, it is rather easy to see that $f \simeq f$ . The map F(x,t) is the required homotopy.

F(x,0)=f(x) and F(x,1)=f(x) is clearly satisfied.

If f is a path, then F is certainly a path homotopy, since f and f itself has the same initial point and final point.

Symmetry:

Next we shall show that given $f \simeq f'$ , we have $f' \simeq f$ . Let F be a homotopy between f and f’. We can then verify that G(x,t) = F(x, 1-t) is a homotopy between f’ and f.

G(x,0) = F(x, 1)=f’ (x)

G(x,1) = F(x, 0) = f(x)

Furthermore, if F is a path homotopy, so is G.

G(0,t)=F(0, 1-t) = $x_0$

G(1,t)=F(1,1-t) = $x_1$

Transitivity:

Next, suppose that $f \simeq f'$ and $f' \simeq f''$ , we show that $f \simeq f''$ . Let F be a homotopy between f and f’, and let F’ be a homotopy between f’ and f”. This time, we need to define a slightly more complicated homotopy G: X x I -> Y by the equation

$G(x,t) = \begin{cases} F(x,2t) &\text{for }t\in [0,\frac{1}{2}],\\ F'(x, 2t-1) &\text{for } t\in [\frac{1}{2}, 1].\end{cases}$

First, we need to check if the map G is well defined at t=1/2. When t=1/2, we have F(x,2t) = F(x,1)=f'(x) = F'(x,2t-1).

Because G is continuous on the two closed subsets X x [0, 1/2] and X x [1/2, 1] of XxI, it is continuous on all of X x I, by the pasting lemma.

Thus, we may see that G is the required homotopy between f and f”.

G(x,0)=F(x,0) = f(x)

G(x,1) = F’ (x, 1) = f”(x)

We can also check that if F and F’ are path homotopies, so is G.

G(0,t) = F(0, 2t) = $x_0$

G(1, t)=F'(1, 2t-1) = $x_1$

Our Daily Story #8: The Rigorous Mathematician with epsilon-delta

Math Online Tom Circle

http://en.m.wikipedia.org/wiki/Augustin-Louis_Cauchy

We mentioned Augustin Louis Cauchy in the tragic stories of Galois and Abel. Had Cauchy been more generous and kind enough to submit the two young mathematicians’ papers to the French Academy of Sciences, their fates would have been different and they would not have died so young.

Cauchy was excellent in language. He was the 2nd most prolific writer (of Math papers) after Euler in history. When he was a math prodigy, his neighbor — the great French mathematician and scientist Pierre-Simon Laplace — advised Cauchy’s father to focus the boy on language before touching mathematics. (Teachers / Parents take note of the importance of language in Math education.)

Cauchy’s language education made him very rigorous in micro-details. This was the man who developed the most rigorous epsilon-delta Advanced Calculus (called Analysis) after Newton / Lebniz had invented the non-rigorous Calculus (why?).

Rigorous epsilon-delta…

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Real Life Applications of Algebraic Topology (Big Data)

Big Data: A Revolution That Will Transform How We Live, Work, and Think

What is Algebraic Topology:

Algebraic topology is a branch of mathematics that uses tools from abstract algebra to study topological spaces. The basic goal is to find algebraic invariants that classify topological spaces up to homeomorphism, though usually most classify up to homotopy equivalence. (Wikpedia)

What is Big Data:

Big data is a broad term for data sets so large or complex that traditional data processing applications are inadequate. Challenges include analysis, capture, data curation, search, sharing, storage, transfer, visualization, and information privacy. The term often refers simply to the use of predictive analytics or other certain advanced methods to extract value from data, and seldom to a particular size of data set. Accuracy in big data may lead to more confident decision making. And better decisions can mean greater operational efficiency, cost reductions and reduced risk. (Wikipedia)

Big Data is said to be the next biggest scientific advance since the internet. Algebraic Topology is one branch of Mathematics that is directly related to Big Data.

Topological data analysis (TDA) is a new area of study aimed at having applications in areas such as data mining and computer vision. The main problems are:

how one infers high-dimensional structure from low-dimensional representations; and
how one assembles discrete points into global structure.

The human brain can easily extract global structure from representations in a strictly lower dimension, e.g. we infer a 3D environment from a 2D image from each eye. The inference of global structure also occurs when converting discrete data into continuous images, e.g. dot-matrix printers and televisions communicate images via arrays of discrete points.

The main method used by topological data analysis is:

Replace a set of data points with a family of simplicial complexes, indexed by a proximity parameter.
Analyse these topological complexes via algebraic topology — specifically, via the theory of persistent homology.^[1]
Encode the persistent homology of a data set in the form of a parameterized version of a Betti number which is called a persistence diagram or barcode.^[1]

Source: Wikipedia

Very interesting!

Measurability of product fg

In the previous chapters, Bartle showed that that if f is in M(X,X), then the functions $cf, f^2, |f|, f^+, f^-$ are also in M(X,X).

The case of the measurability of the product fg when f, g belong to M(X,X) is a little bit more tricky. If $n\in\mathbb{N}$ , let $f_n$ be the “truncation of f” defined by $f_n (x)=\begin{cases}f(x), &\text{if }|f(x)|\leq n, \\ n, &\text{if } f(x)>n,\\ -n, &\text{if }f(x)<-n\end{cases}$

Let $g_m$ be defined similarly. We will work out the proof that $f_n$ and $g_m$ are measurable (Bartle left it as Exercise 2.K).

Proof:

Each $f_n$ is a function on $X$ to $\mathbb{R}$ .

$\{x\in X:f_n (x) >\alpha\}=\begin{cases}\{x \in X: f(x)>\alpha\}, &\text{if }-n<\alpha <n,\\ \emptyset, &\text{if }\alpha\geq n,\\X, &\text{if }\alpha\leq -n \end{cases}$

All of the above sets are in X.

Thus, we may use an earlier Lemma 2.6 to show that the product $f_n g_m$ is measurable.

We also have $f(x)g_m (x)=\lim_n f_n (x)g_m (x)$ , and using an earlier corollary that says that if a sequence $(f_n)$ is in M(X,X) converges to f on X, then f is also in M(X,X), we have that $f(x)g_m (x)$ belongs to M(X,X).

Finally, (fg)(x)=f(x)g(x)= $\lim_m f(x)g_m (x)$ , and hence fg also belongs to M(X,X).

This is a very powerful result of Lebesgue integration, since we can see that the theory includes extended real-valued functions, and prepares us to integrate functions that can reach infinite values!

Source: The Elements of Integration and Lebesgue Measure

Borel Measurable

This is a continuation of the study of the book The Elements of Integration and Lebesgue Measure by Bartle, listing a few examples of functions that are measurable. Bartle is a very good author, he tries his very best to make this difficult subject accessible to undergraduates.

Example:

If X is the set R of real numbers, and X is the Borel algebra B, then any monotone function is Borel measurable.

Proof:

Suppose that f is monotone increasing, i.e. $x\leq x'$ implies $f(x)\leq f(x')$ .

Then, $\{x\in\mathbb{R}:f(x)>\alpha\}$ consists of a half-line which is either of the form $\{x\in\mathbb{R}:x>a\}$ or the form $\{x\in\mathbb{R}:x\geq a\}$ . (We will show later that both cases can occur.) Thus, the set will belong to the Borel algebra B which is the $\sigma$ -algebra generated by all open intervals (a,b) in R.

Both cases can indeed occur. For example, if f(x)=x, then the set will be of the form $\{x\in\mathbb{R}:x>a\}$ . More interestingly, if the set is the step function $f(x)=\begin{cases}-1, &\text{if }x<0\\1, &\text{if }x\geq 0\end{cases}$ , then when $\alpha=0$ , the set will be $\{x\in\mathbb{R}:x\geq 0\}$ .

Lemma: An extended real-valued function f is measurable if and only if the sets $A=\{x\in X:f(x)=+\infty\}$ , $B=\{x\in X:f(x)=-\infty\}$ belong to X and the real-valued function $f_1$ defined by $f_1 (x)= \begin{cases} f(x), &\text{if }x\notin A\cup B,\\ 0, &\text{if }x\in A\cup B,\end{cases}$ is measurable.

This lemma is often useful when dealing with extended real-valued functions.

Proof: If f is in M(X,X), it is proven earlier in the book by Bartle that A and B belong to X. Let $\alpha\in\mathbb{R}$ and $\alpha\geq 0$ , then we have that $\{ x\in X:f_1 (x)>\alpha\}=\{ x\in X:f(x)>\alpha\}\setminus A$ which is in X since it is the complement of the union of A and $X\setminus \{x\in X:f(x)>\alpha\}$ .

If $\alpha<0$ , then $\{ x\in X:f_1 (x)>\alpha \}=\{ x\in X:f(x)>\alpha \}\cup B$ , which is a union of two sets in X and hence also in X.

Hence, $f_1$ is measurable.

Conversely, if $A, B\in \mathbf{X}$ and $f_1$ is measurable, then $\{x\in X:f(x)>\alpha\}=\{ x\in X: f_1 (x) >\alpha \}\cup A$ when $\alpha \geq 0$ , and $\{x\in X:f(x)>\alpha\}=\{x \in X:f_1 (x)>\alpha\}\setminus B$ when $\alpha <0$ , due to a similar reason as above. Therefore f is measurable!

‘Beautiful Mind’ mathematician John Nash killed in US car crash

Very sad news…. Rest in peace, Professor John Nash.

Source: https://sg.news.yahoo.com/beautiful-mind-mathematician-john-nash-killed-us-police-143603056.html

Nobel Prize-winning US mathematician John Nash, who inspired the film “A Beautiful Mind,” was killed with his wife in a New Jersey car crash.

Nash, 86, and his 82-year-old wife Alicia were riding in a taxi on Saturday when the accident took place, State Police Sergeant Gregory Williams told AFP.

“The taxi passengers were ejected,” Williams said, adding that they were both killed.

The Princeton University and Massachusetts Institute of Technology (MIT) mathematician is best known for his contribution to game theory — the study of decision-making — which won him the Nobel economics prize in 1994.

His life story formed the basis of the Oscar-winning 2001 film “A Beautiful Mind” in which actor Russell Crowe played the genius, who struggled with mental illness.

“Stunned… my heart goes out to John & Alicia & family. An amazing partnership. Beautiful minds, beautiful hearts,” Crowe said on Twitter.

A Beautiful Mind

Synopsis: “HOW COULD YOU, A MATHEMATICIAN, BELIEVE THAT EXTRATERRESTRIALS WERE SENDING YOU MESSAGES?” the visitor from Harvard asked the West Virginian with the movie-star looks and Olympian manner. “Because the ideas I had about supernatural beings came to me the same way my mathematical ideas did,” came the answer. “So I took them seriously.”

Thus begins the true story of John Nash, the mathematical genius who was a legend by age thirty when he slipped into madness, and who—thanks to the selflessness of a beautiful woman and the loyalty of the mathematics community—emerged after decades of ghostlike existence to win a Nobel Prize for triggering the game theory revolution. The inspiration for an Academy Award–winning movie, Sylvia Nasar’s now-classic biography is a drama about the mystery of the human mind, triumph over adversity, and the healing power of love.

Measure and Integration Recommended Book

I have added a new addition to the Recommended Books for Undergraduate Math, which is one of my most popular posts!

The new book is The Elements of Integration and Lebesgue Measure, an advanced text on the theory of integration. At the high school level, students are exposed to integration, but merely the rules of integration. At university, students learn the Riemann theory of integration (Riemann sums), which is a good theory, but not the best. There are some functions which we would like to integrate, but do not fit nicely into the theory of Riemann Integration.

I am personally reading this book as well, as I didn’t manage to study it in university, but it is a key component for graduate level analysis. Students interested in advanced Probability (see this post on Coursera Probability course) would be needing Lebesgue theory too!

Time Management Tips for Students (What to do if fail JC Test / Promo Exam?)

Do you wish there is a method to improve your grades? How do you improve your grades after failing a Common Test for Secondary School or JC?

The Four Quadrant Method is an ideal method for students (especially higher level students like O Level or A Level students) to plan their study schedule and revision time table.

Many students do ok in primary school, but start to falter and fail in secondary school or JC. This may be due to many factors, some of which can be remedied using effective time management.

According to this model, which comes from the book First Things First by Stephen Covey (Highly recommended to read), there are four types of activities:

Quadrant 1) Important and Urgent (crises, deadline-driven projects)
Quadrant 2) Important, Not Urgent (preparation, prevention, planning, relationships)
Quadrant 3) Urgent, Not Important (interruptions, many pressing matters)
Quadrant 4) Not Urgent, Not Important (trivia, time wasters)

The key to doing well in school and exams is actually Quadrant 2! It is highly related to human psychology. Most people would think Quadrant 1 is more important, but actually Quadrant 2 is the most important type of activity for students.

Quadrant 1 activities (in the Singapore context) are activities like assignment due next day, test next day, exam the next day, and so on. They are important and also urgent. The thing is, these things are usually done by most people since there is a time pressure factor to it. Most students will actually do and complete Quadrant 1 activities. However, as you would know by now, just doing the homework the teacher assigns is not enough to do well for the test / exam under the Singapore syllabus. Firstly, the work that the teacher assigns may be basic material, while in Singapore, the school tests and exams all contain advanced and challenging material.

Quadrant 2 activities are long-ranged planning and strategies, like preparing for a test that is 3 months later, preparing for the Promo Exam that is half a year later. Since these activities are not urgent, most people skip them altogether. However, it is highly important to do Quadrant 2 activities everyday. Stephen R. Covey is a genius for discovering that Quadrant 2 is the secret to time management. Students should set aside some time everyday to do long-ranged preparation, e.g. preparing for a test that is a few months into the future.

Quadrant 3 activities are things that are urgent but not important. Examples are checking Email, checking Whatsapp for class group notifications. Yes, checking email and Whatsapp is compulsory nowadays, but it is not considered an important activity in the grand scheme of things. One should set a minimum amount of them for these activities. CCA may also be classified under this category. This Quadrant is highly deceptive, and a huge time sink, but in the end the activities in Quadrant 3 rank very low in importance.

Quadrant 4 activities are things that are not urgent and not important. Examples are checking Facebook, playing computer games, and so on. These activities should be kept to a bare minimum, and only during scheduled breaks for destressing.

The Four Quadrant technique can be coupled with the Pomodoro Technique which is another good technique for time management.

Hope it helps! This method is for parents to teach their child about Time Management, provided their child is motivated and wishes to improve. For children that are not motivated to study / not interested in learning, parents should check out these Motivational books to motivate students instead.

Tuition Agency / Chinese Tuition

Recommended Tuition Agency:

Startutor is Singapore’s most popular online agency, providing tutors to your home. There are no extra costs for making a request. The tutors’ certificates are carefully checked by Startutor.

(Website: http://startutor.sg/request,wwcsmt)

There are many excellent tutors from RI, Hwa Chong, etc. at Startutor, teaching various subjects at all levels.
High calibre scholars from NUS/overseas universities are also tutoring at Startutor.

(Website: http://startutor.sg/request,wwcsmt)
(Please use the full link above directly, thanks!)

Chinese Tuition: http://chinesetuition88.com/

Math Resources / Short, Summarized Math Notes for Sale:
https://mathtuition88.com/math-notes-worksheets-sale/
10% Discount on all products
(Free Exam Papers from Top Schools to accompany each Math Resource.)

Singapore Math Books: https://mathtuition88.com/buy-singapore-math-books/

让孩子读书更厉害的书籍

现在的家长都很注重孩子的学习，但是有时候孩子不专注，或者对学习没兴趣怎么办？

俗话说：你可以把马牵到水边，但你无法强迫它饮水（意指有的事情必需本人自愿，强迫无济于事）；老牛不喝水，不能强按头。

可见，强逼孩子读书是没有用的，反而会造成孩子厌倦学习。最重要是培养孩子读书的兴趣，这样往往事半功倍，孩子的学习成绩突飞猛进。

在此，让我介绍一些帮忙孩子读书厉害的书：

1）我的第一本专注力训练书（专注的孩子更聪明）

现在很多孩子都有多动症，就算没有多动症也很难静下来读书，这是21世纪普遍的问题。因为现在太多引诱，比如电脑，手机，电视。毋庸置疑，专注的孩子更聪明，学习也肯定比较好。《我的第一本专注力训练书》为《看到找不到》系列之精彩合集。精选本系列中最经典和最受欢迎的形象页面，按综合难度由易到难编排，增加了专注能量级的划分和“目标锁定”等小细节，提升孩子寻找之后的成就感，逐步提升专注力、记忆力、观察力三大能力。《我的第一本专注力训练书》足足128页，让孩子一次玩得过瘾、找得开心。

2）学会提问(原书第10版)

学会提问是一门很大的学问。批判性思维领域“圣经”之作！权威大师30年畅销不衰的经典！史上最有内涵的思维训练书！亚马逊思维科学领域no.1！俞敏洪高度推荐美国大学生人手一本！打开心智，提早具备未来创新人才的核心竞争力！

3) 棚车少年(套装共8册)(中英双语)（当孩子遇到挫折，这本书能让他们笑着面对人生）

天下没有100%顺利的事，孩子总有一天会遇到挫折。遇到挫折该怎么办，怎么面对？读了这本书能让孩子笑着面对人生，不如买给孩子看看。这本书是中英双语，还能帮助孩子练习语文。《棚车少年》：亨利、杰西、维莉、班尼四兄妹从小就是孤儿，他们知道自己有一个爷爷在绿野镇，但是他们不喜欢他。为了躲避爷爷，孩子们在一个破旧的棚车里安了家，开始相依为命的生活。他们积极向上，阳光开朗，不惧生活的挫折。不仅如此，他们还相互帮助，一起寻找生活的乐趣，在树林里安家、收留小狗望望、探宝、自由赛……最后爷爷找到了他们，原来爷爷很年轻、慈祥、很爱他们。最后他们跟爷爷一起回家，过上幸福快乐的日子。

WordPress to Sina Weibo 微博 Automatic Posting

I recently discovered a way to post (automatically) from WordPress to Sina Weibo 微博(China’s version of twitter, which has more than half a billion users!)

The trick is to use IFTTT.com (If this then that).

Steps:

1) Setup up publicize for WordPress to Twitter. (WordPress.com can do this automatically).

2) Go to IFTTT.com, and set up a recipe from Twitter to Weibo. (There is a premade template for that, takes less than 5 minutes to sign up)

Done!

There may be a way for WordPress –> Weibo direct posting, I am still researching on that. (Update: Yes, there is a recipe for direct WordPress –> Weibo too!) It depends on whether you want a short summary, in which case WordPress –> Twitter –> Weibo may suit you better. If you want a full text, then WordPress –> Weibo is great. Or, you can use both!

Hope it is helpful!

无微不至:微博营销实战指南

《无微不至:微博营销实战指南》内容简介：企业如何利用微博进行营销？如何了解消费者的购买心理？如何把握微博的传播机制，发现用户的行为模式，找到有价值的客户？如何挖掘数据价值，制定营销方案，实现营销的最佳效果？《无微不至:微博营销实战指南》从如何搭建企业微博营销平台、构建微博体系、塑造企业微形象、选择微博营销模式，以及微博营销的技能、微博写作技巧等方面详尽地讲述微博营销的方法、技巧，具有实操性强，案例经典，拿来就能用的特点。在《无微不至:微博营销实战指南》中，读者还会学到以下经典内容：微博营销已不是简单开个账户，发发帖子。微博传播永远是内容为王，无论是重口味，还是小清新，一定要与草根文化血脉相通。写微博和说相声是一样的，要善于抖包袱，要在140字中写出跌宕起伏。10%的人影响了90%的人的购买行为，微博是影响他人购买决策的一个有效工具。社交广告即将或者已经成为最主流的社会化营销解决方案。高质量的内容和互动永远是提高粉丝转发率、留住粉丝的不二法宝。中国移动、中国电信应该如何做微博营销？《独唱团》爆单，快书包如何转危机为商机？如何打造企业官微？1000个真实的粉丝意味着什么？如何用微博编织人脉？微博内容写作十大技巧是什么……

Chinese Math and Science Books

Just to introduce a few books that can simultaneously improve your child’s Math and Science knowledge, and Chinese at the same time!

The latest news is that China is building the Kra Canal, a news that would mark the beginning of the increased dominance of China in Southeast Asia, hence having a good grasp of Mandarin is no longer optional, but 100% compulsory if you want to have a slice of the pie of the jobs and benefits generated by China.

华文数学/科学书本

1）
可怕的科学•经典数学系列(套装共12册)(三度荣获国际科普图书最高奖)

This series of “Horrible Science” Books is translated into Chinese, and is an award winning series of books. Highly recommended!

2)
小学奥数700题详解:3、4、5、6年级

$math olympiad$

700 Practice Problems for Math Olympiad! These books are very useful for GAT / DSA / GEP Preparation.

3)
走进奇妙的数学世界1-3(套装共3册)

$math olympiad$ $world of math$

Into the world of Mathematics! 《走进奇妙的数学世界1-3(套装共3册)》内容简介：数学最让人困惑的是为什么这样和有什么用，很多人即使大学毕业也不明白，这套书完美地阐释了数学的本质，把数学和生活紧密联系在一起。13种基本数学思想，层层深入，完美阐释数学的本质。以两个小矮人贯穿全文，图文并茂，讲故事、出谜题、做游戏，游戏背后蕴藏数学概念让孩子以最简单、最科学的方式走近数学，爱上数学！不仅仅讲算术，更重在启发从不同角度看待事物、解决问题的思考方式，培养孩子的逻辑思维能力，提高综合素质。

Finally, for readers of my blog who are new to Chinese, and wish to learn this 5000 year old language, I would recommend some books to learn Chinese for beginners here：

轻松学中文1(课本)(附CD光盘1张)

Learn Chinese in an easy manner! Easy steps to Chinese. (With CD)

3 is everywhere

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Why 381,654,729 is awesome

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Pandigital numbers

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The Problem in Good Will Hunting

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Movie (Maths Problem Scene):

Homeomorphically Irreducible Trees with 10 dots.

Who was the real Good Will Hunting?

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神童背诵圆周率秀心算

This 5-year-old boy can remember Pi upto 600 decimal points.

Math Online Tom Circle

5岁儿童背 Pi 600小数位:

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Coursera Probability Course and Recommended Probability Book

Just completed the Coursera Probability Course by UPenn (University of Pennsylvania), lectured by Professor Santosh S. Venkatesh who is the author of the highly recommended book: The Theory of Probability: Explorations and Applications.

Coursera Review

The course isn’t very hard, it is very suitable for undergraduates and even high school students should be able to understand majority of the content. It actually overlaps with the A level syllabus in Singapore, and hence I would say that a 17-18 year old student would be able to grasp most of the concepts in this course.

The lecturer is very good at words, and his lectures are full of imagery and vivid descriptions. The homework is a little tricky, and hence would require some thought, even though the concepts tested are elementary (elementary in the sense that it doesn’t require calculus).

A sample of a tricky question is the “Six Saucer Question”: Six cups and saucers come in pairs: there are two cups and saucers that are red, white, and blue. If the cups are placed randomly onto the saucers (one each), find the probability that no cup is upon a saucer of the same color.

It is very tricky and to get it correct on the first try is a major accomplishment.

Overall, this Coursera Course is highly recommended, and students should try to take it the next time it comes out!

WW2 Enigma Machine

How to break 158,962,555,217,826,360,000 codes ?

Math Online Tom Circle

There are 158,962,555,217,826,360,000 possibilities of codes in the German Enigma Machine:

Flaw cracked by the genius Mathematician Alan Turing (Father of Artificial Intelligence) :

“A key can never be itself” — this is ‘the straw that breaks the camel’s back’, a critical clue to break the 158,962,555,217,826,360,000 possible codes !

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Math is Best Paid Job in 2015

Math Online Tom Circle

These 3 jobs all require Math: Actuary, Statistician and Mathematician.

Others good to have Math : Software Engineer, Systems Analyst, Data Scientist.

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Chinese Remainder Theorem History (韩信点兵)

I have written a guest post on https://chinesetuition88.wordpress.com on the very fascinating Chinese Remainder Theorem and its History (韩信点兵). Do check it out, you will be amazed at the genius of Chinese General Han Xin.

Students who are interested in Chinese Tuition may check out https://chinesetuition88.wordpress.com for more details.

Chinese Tuition Singapore

淮安民间传说着一则故事——“韩信点兵”，其次有成语“韩信点兵，多多益善”。韩信带1500名兵士打仗，战死四五百人，站3人一排，多出2人；站5人一排，多出4人；站7人一排，多出6人。韩信马上说出人数：1049。

Translation:

In Ancient China, there was a General named Han Xin, who led an army of 1500 soldiers in a battle. An estimated 400-500 soldiers died in the battle. When the soldiers stood 3 in a row, there were 2 soldiers left over. When they lined up 5 in a row, there were 4 soldiers left over. When they lined up 7 in a row, there were 6 soldiers left over. Han Xin immediately said, “There are 1049 soldiers.”

Amazing! How did Han Xin do that?

Han Xin was not only a brilliant mathematician and general, he was also a very magnanimous guy full of wisdom.

Once, when he was suffering from hunger, he met a woman who provided him with food. He promised to repay her for her kindness after he had made great achievements in life, but it was rebuffed by her…

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Think Out of The Box

Math Online Tom Circle

Answer (scroll down)

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Terry Tao, Ph.D. Small and Large Gaps Between the Primes

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Quiz: Can You Solve This Sum ?

For more logic puzzles, check out:

Puzzle Baron’s Logic Puzzles

Math Online Tom Circle

[Hint]: Think out of the box…

Answer below (scroll down)
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Answer:
1 + 13 +…

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Terence Tao, genius mathematician

费马大定理 Fermat’s Last Theorem

Intriguing review (Chinese) of FLT by a non-mathematician. He aims to convey the beauty of Mathematics to students, who unfortunately treat Math as a tool to pass exams from PSLE, O and A level, university math course, then ditch Math upon graduation. Math is the beauty of the universe.

Math Online Tom Circle

费马大定理 Fermat’s Last Theorem (FLT): 17世纪业余数学家法国大法官费马开的一个”玩笑”, 推动350年来现代数学突飞猛进。

FLT 数学长征英雄人物:

1. Fermat (费马 1601@ Toulouse, France)
2. Galois (伽罗瓦): Group Theory (群论)
3. Gauss (高斯)
4. Cauchy (柯西) Lamé (拉梅) Kummer (库马)
5. Solphie Germain
6. Euler (欧拉)
7. Taniyama (谷山丰), Shimura (志村五郎)

集大成:
8. Andrew Wiles (怀尔斯) (证明@1993 -1995)

1.Elliptic Curve (椭圆曲线)
2. Modular Form (模形式)
3. Fermat Last Theorem (费马大定理)

(1) = (2) = (3)

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Music and math: The genius of Beethoven

Math Online Tom Circle

Music and math: The genius of Beethoven – Natalya St. Clair:

Music = Math

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Math is Forever (Spanish)

With humor and charm, mathematician Eduardo Sáenz de Cabezón answers a question that’s wracked the brains of bored students the world over: What is math for? He shows the beauty of math as the backbone of science — and shows that theorems, not diamonds, are forever. In Spanish, with English subtitles.

Yes, indeed, 1000 years from now, students will still be learning Pythagoras’ Theorem, while other fragments of human knowledge would have faded away.

Check out also this book: Arithmetic and Algebra Again: Leaving Math Anxiety Behind Forever, suitable for students who really need some encouragement and motivation to overcome fear of math! Albert Einstein once said, “You never fail until you stop trying.” Hence, even if you have not done well in math for the past years, there is still hope, don’t give up!

April Fools Video Prank in Math Class

Check out this really funny video on a April Fools Prank during a Math Class!

The teacher played a trick on his math class for April Fool’s Day. In this one, he’s showing a “homework help” video that gets some trigonometry wrong.

Looking for more Math Jokes? Check out the book below!

Math Jokes 4 Mathy Folks

Excellent MITOpenCourseware

Math Online Tom Circle

Strongly recommended free excellent MIT Math for high school, undergrads/grads and any self-study learners.

Thanks Prof. Gilbert Strang for the unselfish sharing.

http://ocw.mit.edu/faculty/gilbert-strang/

I find extremely pleasure when I discovered his brilliant lecture notes in “Generating Function” – a Discrete Math technique for computing sequencing using function, and the application in complex Combinatorics. Download here:

Example:

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The Math of Shuffling Cards

Previously, the first YouTube video wasn’t working. I have added a new link to the interesting “Looking at Perfect Shuffles” video. 🙂

Mathtuition88

A magic trick based on the “Perfect Shuffle”. Featuring Professor Federico Ardila. I watched his videos on Hopf Algebras while learning the background material for my honours project on Quantum Groups.

Mathemagician Persi Diaconis discusses which is the best way to shuffle: Overhand shuffle, Riffle Shuffle, or “Smoosh” Shuffle? Watch the video to find out!

Magical Mathematics: The Mathematical Ideas That Animate Great Magic Tricks is an interesting book by Professor Diaconis, featuring Magic Tricks that have a mathematical background! This book is a great idea for a gift for students, teachers, or friends!

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Mathematicians have prevented a world disaster, behind the scenes

Recently, after taking the Coursera course on Cryptography, I had a better appreciation of mathematics and the role of cryptography in our modern society.

I was pleased to read this article Quantum compute this: Mathematicians build code to take on toughest of cyber attacks, and Washington State University mathematicians have designed an encryption code capable of fending off the phenomenal hacking power of a quantum computer.

The quantum computer, though not yet invented, is widely believed to be available soon in the next few years. In the hands of hackers, the quantum computer would be a formidable weapon as current cryptographic methods are extremely vulnerable to the quantum computer as it can factor numbers extremely quickly, leading to number theoretic codes being broken.

What would happen if a Quantum Computer is built

Quantum computers are near

Quantum computers operate on the subatomic level and theoretically provide processing power that is millions, if not billions of times faster than silicon-based computers. Several companies are in the race to develop quantum computers including Google.

Internet security is no match for a quantum computer, said Nathan Hamlin, instructor and director of the WSU Math Learning Center. That could spell future trouble for online transactions ranging from buying a book on Amazon to simply sending an email.

Hamlin said quantum computers would have no trouble breaking present security codes, which rely on public key encryption to protect the exchanges.

In a nutshell, public key code uses one public “key” for encryption and a second private “key” for decoding. The system is based on the factoring of impossibly large numbers and, so far, has done a good job keeping computers safe from hackers.

Quantum computers, however, can factor these large numbers very quickly, Hamlin said. But problems like the knapsack code slow them down.

Fortunately, many of the large data breaches in recent years are the result of employee carelessness or bribes and not of cracking the public key encryption code, he said.

Hence, when many people say mathematics is useless, they are actually extremely wrong, as mathematics permeates every aspect of life! Even though maths like calculus is not directly used in everyday life, it is part of our phone, computer, and every part of the modern lifestyle.

Kudos to the mathematicians who have averted a world disaster, before quantum computers are even invented!

If you are interested in what a quantum computer is, and what it can do (it is so powerful that whoever has one would hold the keys to the entire internet), check out this book Schrödinger’s Killer App: Race to Build the World’s First Quantum Computer.

Written by a renowned quantum physicist closely involved in the U.S. government’s development of quantum information science, Schrödinger’s Killer App: Race to Build the World’s First Quantum Computer presents an inside look at the government’s quest to build a quantum computer capable of solving complex mathematical problems and hacking the public-key encryption codes used to secure the Internet. The “killer application” refers to Shor’s quantum factoring algorithm, which would unveil the encrypted communications of the entire Internet if a quantum computer could be built to run the algorithm. Schrödinger’s notion of quantum entanglement—and his infamous cat—is at the heart of it all.

What is a Tensor?

Most people don’t encounter Tensors (the higher level advanced version of Matrices) until they reach senior undergraduate, or even graduate level.

What is a Tensor?

The best explanation I have ever seen, comes from this video by the author of A Student’s Guide to Vectors and Tensors, Daniel A. Fleisch. Using children’s blocks and laymen language, he explains what is a tensor clearly and succinctly in a way that is unbelievably crystal clear.

This YouTube video is watched over 200,000 times, a very commendable achievement for a math video!

Official Definition by Wikipedia

Tensors are geometric objects that describe linear relations between vectors, scalars, and other tensors. Elementary examples of such relations include the dot product, the cross product, and linear maps. Vectors and scalars themselves are also tensors. A tensor can be represented as a multi-dimensional array of numerical values. The order (also degree) of a tensor is the dimensionality of the array needed to represent it, or equivalently, the number of indices needed to label a component of that array. For example, a linear map can be represented by a matrix (a 2-dimensional array) and therefore is a 2nd-order tensor. A vector can be represented as a 1-dimensional array and is a 1st-order tensor. Scalars are single numbers and are thus 0th-order tensors. The dimensionality of the array should not be confused with the dimension of the underlying vector space.

If you have some programming knowledge, you may view tensors as a type of multidimensional array. A more mathematical abstract way can be achieved by defining tensors in terms of elements of tensor products of vector spaces, which in turn are defined through a universal property.

Cool? The word “tensor” really strikes me as a word that is really sophisticated and complicated!

How to find the distance of a plane to the origin

Given the equation of a plane: ax+by+cz=D, or in vector notation $\mathbf{r}\cdot \left( \begin{array}{c} a\\ b\\ c\\ \end{array}\right)=D$ , how do we find the (shortest) distance of a plane to the origin?

(When a question asks for the distance of a plane to the origin, by definition it means the shortest distance.)

One way to derive the formula is this:

Derivation

Let X be the point on the plane nearest to the origin.

$\overrightarrow{OX}$ must be perpendicular to the plane, i.e. parallel to the normal vector $\mathbf{n}=\left(\begin{array}{c}a\\b\\c\\\end{array}\right)$ .

Furthermore, X lies on the plane, hence we have $\boxed{\overrightarrow{OX}\cdot\mathbf{n}=D}$

Using the formula for dot product, we can get $|\overrightarrow{OX}\cdot\mathbf{n}|=|\overrightarrow{OX}||\mathbf{n}|\cos \theta=D$

Since $\overrightarrow{OX}$ is parallel to $\mathbf{n}$ , $\theta$ is either 0 or 180 degrees, hence $\cos \theta$ is either 1 or -1.

Thus, we have $|\overrightarrow{OX}||\mathbf{n}|=|D|$ .

The shortest distance from the point X to the origin is then $\displaystyle|\overrightarrow{OX}|=\frac{|D|}{|\mathbf{n}|}=\frac{|D|}{\sqrt{a^2+b^2+c^2}}$

Ans: Shortest distance from point to plane is $\displaystyle\boxed{\frac{|D|}{\sqrt{a^2+b^2+c^2}}}$

H2 Maths Condensed Notes and Prelim Papers

If you are looking for a short summarized H2 Maths Notes, with Prelim Papers to practice, do check out our Highly Condensed H2 Maths Notes!

GEP Questions: The Gauss Trick

We will continue our series on GEP Questions. To learn more about Recommended Books for GEP, to practice GEP Questions, visit the link here.

Today, we will discuss the quintessential GEP Question: The Gauss Trick. This GEP question illustrates the fact that giftedness can be trained to a large extent.

Question:

Find the sum of 1+3+5+7+…+95+97+99.

Solution will be below after this text, scroll down after you are ready to see the answer!

If a 9 year child can solve this on his/her first attempt (i.e. see the question for the very first time), then the child is actually at the level of Carl Friedrich Gauss, the legendary mathematician! When Gauss was a young kid, his teacher set the class a difficult question, 1+2+3+…+99+100, hoping to keep the children quiet and busy while he could have some time to relax. Little did he expect Gauss to come up with the right answer minutes later.

Most 9 year old kids would not be able to solve this on their first try. I wouldn’t be able to solve it correctly even if given an hour when I was a kid! If the children have seen it before and practiced, that is a different story, as it becomes so easy even for a 7 year old kid. Hence, practicing GEP Questions actually leads to an indirect boost of IQ in this manner! The transition from ignorance to knowledge, leading to increased intelligence, can be accomplished by practice! To practice more of these GEP Questions, check out the Math Olympiad Recommended Books page. As a tutor, I know that this Gauss Trick is a “must-know” question for students aiming high for school Math / GEP Selection Test, since almost every kid knows this nowadays, and it is highly popular in tests.

In fact, this technique (Gauss Trick) commonly tested in GEP Questions can be used all the way to JC and beyond! In JC it is covered under the topic of AP/GP (Arithmetic Progression and Geometric Progression).

Solution:

Now, for the solution. This sum is the famous arithmetic progression, where each term differs from the next by a fixed constant. In this case, the constant is 2.

Some solutions use a number pair matching, which can be problematic to explain when the number of terms is odd, but still works nevertheless. We can use another method of writing the sum backwards.

First we note that there are 50 terms in this series. We can know that either by noting that they are the odd numbers from 1 to 100, and half of the numbers are odd, hence there are 100/2=50 terms. Alternatively, we can note that 1=2(1)-1, 3=2(2)-1, 5=2(3)-1, …, 99=2(50)-1, where the number in the brackets acting like a counter.

1+3+5+7+…+95+97+99
99+97+95+…+5+3+1

Note that each pair, 1+99, 3+97, 5+95 actually add up to the same thing, i.e. 100.

Adding up the two expressions above, we get 100×50=5000.

Dividing that by two (since we double counted), we will get 5000/2=2500.

Ans: 2500

Hope you enjoyed solving this question!

Check out this amazing book on Gauss: The Prince of Mathematics: Carl Friedrich Gauss

Gauss is a true Math genius, and you can read more about his life in this interesting biography! This historical narrative will inspire young readers and even curious adults with its touching story of personal achievement.

This Is How Long Your Teen Needs to Spend on Homework to Be Better at Math and Science

So it is official, scientists have discovered that a minimum of 1 hour of homework per day is needed to be better at Math!

GEP Screening Test Question Sample: The Tap Question

The Tap Question is another one of those questions that only involve fractions and whole numbers, and hence technically within the grasp of a 9 year old kid sitting for the GEP Screening Test.

However, looks are highly deceiving, and whoever tries the Tap Question for the very first time is highly likely to get stuck. (I was one of them years ago!) The Tap Question is highly popular as a challenging question, due to its psychological nature it is a hard question to grasp. This is a question you wouldn’t want to meet for the first time in the GEP Screening Test. However, if you know how to solve it, it is easy as a piece of cake, and you will be able to solve it during the GEP Screening Test no matter how they twist the question.

GEP Screening Test Sample Math Question (The Tap Question):

A fish tank is connected to three taps.
Tap A can fill the tank in 2 hours.
Tap B can fill the tank in 3 hours.
Tap C can drain the tank in 6 hours.
If all three taps are turned on at the same time, how long would it take to fill the empty fish tank?

Do try out the question before looking at the answer below!

There is a huge difference solving a question for the first time, and solving a question that one has seen before. To familiarize yourself with GEP Screening Test questions that can come out, do check out our Recommended GEP Books. Reading one of those books would increase your child’s repertoire of questions, and hence boost your child’s IQ indirectly. Also, do check out Math Olympiad books, as it is well known that GEP Screening Test Math questions do incorporate some Math Olympiad questions.

A book like The Original Collection of Math Contest Problems: Elementary and Middle School Math Contest problems would be suitable for children training for GEP Screening Test and Math Olympiad simultaneously at the same time.

Solution:

The key insight is to find out what each tap can do in one hour.

Tap A can fill 1/2 of the tank in 1 hour.
Tap B can fill 1/3 of the tank in 1 hour.
Tap C can drain 1/6 of the tank in 1 hour.

Hence, if all of them are turned on simultaneously,
$\displaystyle\frac{1}{2}+\frac{1}{3}-\frac{1}{6}=\frac{2}{3}$ of the tank can be filled in 1 hour.

2/3 of the tank takes 1 hour to fill.
Multiplying this statement by 3,
2 tanks takes 3 hours to fill.
1 tank takes 3/2=1.5 hours to fill.

Ans: 1.5 hours

Hope you had fun solving this!

Also check out the following GEP Screening/Round 2 questions:

By the way, for my foreign readers who are curious what is GEP Screening Test, it is a test conducted in Singapore for entry to the GEP (Gifted Education Programme). The screening and selection tests are conducted at the end of Primary 3, equivalent to Grade 3 or age 9. The official website on GEP Screening Test is available at: http://www.moe.gov.sg/education/programmes/gifted-education-programme/faq/general/

The GEP Screening Math questions can be viewed as the epitome of the Singapore Math system, as it features highly challenging Math questions that are technically in syllabus but few students know how to solve!

Buy H2 Maths Notes / Buy H2 Maths Exam Papers

After a long time, the Highly Condensed H2 Maths Notes is finally ready!

Are you looking for a short, summarized notes for H2 Vectors, H2 Complex Numbers, or even H2 Statistics? Do you remember the formula for the sum of a Geometric Progression?

Purchase this Highly Condensed H2 Maths Notes to quickly review and get ready for your exam!

Free Exam Papers included:
Numerous H2 Maths (Prelim) Free Exam Papers in PDF Format, with Solutions. Schools include ACJC, AJC, HCI, NJC, all the way to YJC, and more.

Note: The set of notes is 9 pages long (2 columns per page, total of 18 columns), i.e. highly condensed and summarized short notes.

Do check it out at our page of Math Resources for Sale!

buy h2 maths exam papers — We specially drew this beautiful graph of a Hyperbola using Geogebra for this Highly Condensed H2 Maths Notes.

GEP Sample Question: The Worker Question

Here is a type of a typical GEP Exam Question that can come out. Technically, this question is in syllabus since it only involves Whole Numbers. However, in practice, this is an extremely tough GEP Exam question for students who have not seen the Worker Question before.

GEP Test Question Sample (Worker Question):

6 men working 8 hours a day can paint a house in 2 days. In how many more days will 4 men, working 3 hours a day at the same rate, complete the same job?

Before you scroll down to check the answer, do give it a try! From personal experience as a tutor, even a 16 year old typical Secondary 4 student cannot solve this question if it is the first time they see it. However, once I explain the solution to them, it is extremely easy and students will get it immediately, even for primary school kids. Someone who has seen these types of questions before can solve it under a minute!

This shows the immense advantage one has if he/she has been exposed to certain types of questions. This is same for the GEP General Ability Test (GAT), a type of “IQ test”, which is basically pattern recognition. If a child has been exposed to books like Match Wits With Mensa: The Complete Quiz Book, words cannot describe the huge advantage he/she has over someone who has not seen a logic pattern puzzle before.

Solution to GEP Sample Question (Worker Question):

There are many types of solutions to this question, but my favorite is using the man-days concept. Man-days is a unit for the amount of work that is needed for something. E.g. If building a house needs 10 man-days, it can be accomplished by either 1 man x 10 days = 10 man-days, or 5 men x 2 days = 10 man-days, etc.

For this question, we will use the unit of man-hours instead.

6x8x2=96 man-hours are needed to paint the entire house.

4 men working 3 hours a day would lead to 12 man-hours a day. Hence 96/12=8 days are needed.

Warning: This is where they trap the careless students! The question asks for how many more days. Hence, the answer is 8-2=6 more days.

Ans: 6 days

Do also check out the Chicken and Rabbit GEP Math Question, which is another type of popular GEP Selection Test and Screening Test question, and can be practiced beforehand as a GEP Mock Test.

Xinmin Secondary 2010 Prelim Paper I Q24 Solution (Challenging/Difficult Probability O Level Question)

Just to reblog this earlier post on a really challenging Probability O Level Question.

Also, do check out my other related posts on Probability:

Coursera Exciting New Probability Course

List of really useful Probability Formulae

Probability is becoming a really important branch of mathematics. One of the most famous Singaporean mathematicians is Professor Louis Chen Hsiao Yun who has a theorem named after him! (Stein-Chen method of Poisson approximation) Professor Chen researches on Probability.

To begin your journey in Probabilty, Introduction to Probability, 2nd Edition is a good book to start learning from. An intuitive, yet precise introduction to probability theory, stochastic processes, and probabilistic models used in science, engineering, economics, and related fields. You may also wish to refer to our comprehensive list of Recommended Undergraduate Books.

(One of the best books to begin your journey in studying the mysterious topic of Probability)

Mathtuition88

A bag A contains 9 black balls, 6 white balls and 3 red balls. A bag B contains 6 black balls, 2 white balls and 4 green balls. Ali takes out 1 ball from each bag randomly. When Ali takes out 1 ball from one bag, he will put it into the other bag and then takes out one ball from that bag. Find the probability that

(a) the ball is black from bag A, followed by white from bag B,
(b) both the balls are white in colour,
(c) the ball is black or white from bag B, followed by red from bag A,
(d) both the balls are of different colours,
(e) both the balls are not black or white in colours.

Solution:

(a) $latex displaystylefrac{9}{18}timesfrac{2}{13}=frac{1}{13}$

(b) Probability of white ball from bag A, followed by white ball from bag B=$latex displaystyle=frac{1}{2}timesfrac{6}{18}timesfrac{3}{13}=frac{1}{26}$

Probability of white from B, followed…

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GEP Exam Paper / GEP Questions

Firstly, do check out our GEP Recommended Books if you are interested in books that will help boost your chances of entering GEP.

A typical sample GEP Exam Paper Question (Math) is the Chicken and Rabbit Question. This test is highly popular in the GEP Screening / GEP Selection Test, Round 1 or even extremely difficult Round 2.

A sample question would go like this:

A farmer has 36 chickens and rabbits in total.
He counted 124 legs altogether.
How many chickens and how many rabbits are there?

(By the way, this question is generated from my Chicken and Rabbit Question Generator!)

This is a highly typical GEP Exam paper question that may come out in the GEP test.

How do we solve it? One way is the trial and error or Guess and Check method. However, this method may not work for high numbers. What if the farmer had 10000 sheep?

The GEP Exam Paper has limited time, hence, we would need to solve it in an efficient way.

There is one method called the Assumption method, where students can remember the acronym ASSD!

Steps of ASSD to solve GEP Exam Paper “Chicken and Rabbit” Question:

Step 1 (A): Assume) Let’s assume all the 36 animals are chickens.

Then, there would be 36×2=72 legs in total.

Step 2 (S): Subtract) Clearly, 72 legs is too few.

In reality, there are 124-72=52 legs more.

Step 3 (S): Subtract) 4-2=2

Each rabbit has 2 more legs than a chicken.

Step 4 (D): Divide)

The extra 52 legs must be due to the rabbits, and each rabbit contributes 2 more legs.

Hence, there are 52/2=26 rabbits!

There must be 36-26=10 chickens then.

Check

During the GEP Exam Paper, checking is essential to avoid careless mistakes. 26×4+10×2=124, which tallies! Hence, we are right!

To practice more Chicken and Rabbit questions, which is highly likely to come out in the GEP Exam Papers, check out our Chicken and Rabbit Worksheet Generator.

Lastly, do check out our GEP Recommended Books which may be the most useful books on the market (not found in Singapore since most major bookstores like Borders have closed down).

Scientists have proven that IQ can be increased, and hence reading a book like Match Wits With Mensa: The Complete Quiz Book would increase your score in the GEP Exam Paper (Logic) section during the GEP Screening / Selection Test.

Coursera Cryptography I Review

I have just completed the Coursera Cryptography I course by Dan Boneh successfully, and received the statement of accomplishment!

Review of the Course

Difficulty: 4.9/5

This course is really difficult for those with no computer science background. Although there is a section on number theory, most of the sections are new to me as my background is mostly undergraduate mathematics. (Though I did take a course IT1002 (from NUS) called Introduction to Programming, which is mostly on Java Programming.)

Especially the programming exercises are very tough for people with limited programming knowledge! However, note that the programming assignments are entirely optional.

Course Content

This course covers the theory and practice of cryptographic systems. Topics included symmetric encryption, data integrity, public-key encryption, and key exchange. The course emphasized the correct use of these primitives.

Interesting Things about this Course

It is interesting to note how complex the field of cryptography is, and how smart hackers have become. It is possible to do a timing attack where even the time taken to respond to say a login, can be used by hackers to guess your password. Every logical operation in a computer takes time to execute, and the time can differ based on the input; with precise measurements of the time for each operation, an attacker can work backwards to the input. – Wikipedia

Needless to say, as our world becomes increasingly digital, cryptography becomes increasingly important.

I wrote two JavaScript applications to help solve some of the programming challenges in this course:

JavaScript XOR Hexadecimal App is a simple JavaScript application that can XOR Hexadecimals (Hex), 2 characters at a time. The output is in 2 digits, meaning that “0” is written as “00”.
Javascript Application to Split Text Every Few Characters

Sadly, WordPress doesn’t support JavaScript, so I have to write them on my sister blog: http://www.mathtuition88.blogspot.com

If you are interesting in programming, particularly app programming, why not check out this book Learning iOS Game Programming: A Hands-On Guide to Building Your First iPhone Game. You may be the creator of the next “Flappy Bird” which reportedly earned its creator $50,000 a day! Wow!

Read this to be the next Flappy Bird creator! Michael Daley walks you through every step as you build a killer 2D game for the iPhone.

Solution to HP A4 Printer Paper Mysterious Question

A while ago, I posted the HP A4 Paper Mysterious Question which goes like this:

Problem of the Week

Suppose $f$ is a function from positive integers to positive integers satisfying $f(1)=1$ , $f(2n)=f(n)$ , and $f(2n+1)=f(2n)+1$ , for all positive integers $n$ .

Find the maximum of $f(n)$ when $n$ is greater than or equal to 1 and less than or equal to 1994.

So far no one seems to have solved the question on the internet yet!

I have given it a try, and will post the solution below!

If you are interested in Math Olympiad, it is a good idea to invest in a good book to learn more tips and tricks about Math Olympiad. One excellent Math Olympiad author is Titu Andreescu, trainer of the USA IMO team. His book 104 Number Theory Problems: From the Training of the USA IMO Team is highly recommended for training specifically on Number Theory Olympiad questions, one of the most arcane and mysterious fields of mathematics. He does write on other Math Olympiad subjects too, like Combinatorics, so do check it out by clicking the link above, and looking at the Amazon suggested books.

Now, to the solution of the Mysterious HP A4 Paper Question:

We will solve the problem in a few steps.

Step 1

First, we will prove that $\boxed{f(2^n-1)=n}$ . We will do this by induction. When $n=1$ , $f(2^1-1)=f(1)=1$ . Suppose $f(2^k-1)=k$ . Then,

$\begin{aligned} f(2^{k+1}-1)&=f(2(2^k)-1)\\ &=f(2(2^k-1)+1)\\ &=f(2(2^k-1))+1\\ &=f(2^k-1)+1\\ &=k+1 \end{aligned}$

Thus, we have proved that $f(2^n-1)=n$ for all integers n.

Step 2

Next, we will prove a little lemma. Let $g(x)=2x+1$ . We will prove, again by induction, that $\boxed{g^n (1)=2^{n+1}-1}$ . Note that $g^n(x)$ means the composition of the function g with itself n times.

Firstly, for the base case, $g^1(1)=2+1=3=2^2-1$ is true. Suppose $g^k (1)=2^{k+1}-1$ is true. Then, $g^{k+1}(1)=2(2^{k+1}-1)+1=2^{k+2}-1$ . Thus, the statement is true.

Step 3

Next, we will prove that if $y<2^n-1$ , then $f(y)<n$ . We will write $y=2^{\alpha_1}x_1$ , where $x_1$ is odd. We have that $x_1<2^{n-\alpha_1}$ .

$\begin{aligned} f(y)&=f(2^{\alpha_1} x_1)\\ &=f(x_1) \end{aligned}$

Since $x_1$ is odd, we have $x_1=2k_1+1$ , where $k_1<2^{n-\alpha_1-1}$ .

Continuing, we have

$\begin{aligned} f(x_1)&=f(2k_1+1)\\ &=f(2k_1)+1\\ &=f(k_1)+1 \end{aligned}$

We will write $k_1=2^{\alpha_2}x_2$ , where $x_2$ is odd. We have $x_2<2^{n-\alpha_1-\alpha_2-1}$ .

$\begin{aligned} f(k_1)+1&=f(2^{\alpha_2}x_2)+1\\ &=f(x_2)+1 \end{aligned}$

where $x_2=2k_2+1$ , and $k_2<2^{n-\alpha_1-\alpha_2-2}$ .

$\begin{aligned} f(x_2)+1&=f(2k_2)+1+1\\ &=f(k_2)+2\\ &=\cdots\\ &=f(k_j)+j \end{aligned}$

where $k_j=1$ , $1=k_j<2^{n-\alpha_1-\alpha_2-\cdots-\alpha_j-j}$ .

Case 1: All the $\alpha_i$ are 0, then $y=2(\cdots 2(k_j)+1=g^j(1)=2^{j+1}-1$ . Then, $j+1<n$ , i.e. $j<n-1$ .

Thus, $f(y)=f(k_j)+j<1+n-1=n$ .

Case 2: Not all the $\alpha_1$ are 0, then, $1=k_j<2^{n-\alpha_1-\alpha_2-\cdots-\alpha_j-j}\leq 2^{n-j-1}$ . We have $2^0=1<2^{n-j-1}$ , thus, $0<n-j-1$ , which means that $j<n-1$ . Thus, $f(y)=f(k_j)+j<1+n-1=n$ .

Step 4 (Conclusion)

Using Step 1, we have $f(1023)=f(2^{10}-1)=10$ , $f(2047)=f(2^{11}-1)=11$ . Using Step 3, we guarantee that if $y<2047$ , then $f(y)<11$ . Thus, the maximum value of f(n) is 10.

Ans: 10

ST Yao丘成桐

Yao Sheng Tong (丘成桐) is the first Chinese (China – HK) who won the Fields Medal, he later also won the Wolf Prize after his PhD-thesis mentor SS. Chern (陈省身)。He is the first Chinese to head Harvard Math Faculty.

Math Online Tom Circle

丘成桐 (ST Yao 1949~) Fields Medal @1982 [33岁] proved Calabi Conjecture

1. 读私立培正中学, 高中遇好数学老师. @香港中文大学, 发觉 Math Beauty, ‘豁然开朗’.
2. Best Math student not necessary Mathematician, only sufficient!
3. 一名数学科学家都应对文学,哲学这类学科有基本的涉猎. 好的数学使你体验到庄子讲的
“天地与我并生, 万物与我为一” 的境界
4. 成功 = 要有数学热情.
Strategy:
a. 深入思考
b. 在心中或纸上仔细研究
c. Find clues from book, till get answer.
d. 出题目给自己

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庖丁解牛数学方法

Math Online Tom Circle

“庖丁解牛”数学方法
庄子讲庖丁(butcher)解牛有三个功夫階段：
1st Level: 看见一只全牛 (Whole Cow)

2nd Level: 三年后，不见全牛，只见牛的生理结構(Anatomy) ：骨骼，肌肉，筋腱。

3rd Level: 不以目视而是神视，＂与桑林之舞合拍，与经首之会同律。＂达到了”物我”两忘的境界。Intuition.

数学的方法也如此。

1st Level: Whole Math (Primary school to High School)

2nd Level: Component Structure – (Undergraduate Math)：
Macro- structure (Algebra : Group, Ring, Field, Vector Space… );
Micro-structure (Analysis : Calculus, Topology, etc)

3rd Level: 无处不数 Ubiquitous Math – (Graduate Math)
eg. Fermat’s Last Theorem used all Math theories available today to prove.

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Math Duality

Singapore still follows the outdated UK Math pedagogy, using the old term “Advanced Calculus” (高等微积分) for the huge discipline of “Analysis” (分析) — the ‘Micro’ view of Math.

Math Online Tom Circle

Mathematics is roughly divided into 2 categories:

‘Macro’ Math: Algebra

‘Micro’ Math: Analysis (or the outdated name Calculus)

Algebra has been transformed rapidly from 19th century after Galois’s invention of Group Theory, and expanded by David Hilbert and his students E. Noether, Artin, etc in Axiomatic Algebra, takes a very macro view of Mathematical structures in abstract thinking.

Analysis, also after 19th century Cauchy and Wierestrass’s invention of ‘epsilon-delta’ micro view of Calculus, transformed the Newton Calculus into rigourous Math.

The old school of division of Pure and Applied Math is no longer valid. Take for example, the Applied Math used in Google Search Algorithm uses abstract Vector Space of Matrices in Linear Algebra (Pure Math).

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Singaporean’s Plan for the Long Weekend (Aug 7-Aug 10)

As every Singaporean should know by now, August 7 is going to be a public holiday this year, due to SG 50, i.e. Singapore’s 50th anniversary as a nation.

A quick check using the mental calculation of dates (Doomsday Algorithm), we can know that August 8 is Saturday, hence August 7 is a Friday. Thus, August 7 to August 10 is indeed a long weekend!

What are Singaporeans doing during the long weekend? Hint: It has something to do with studying. This is a really funny cartoon by Singaporean cartoonist “Chew on it”