Quiz

In the diagram, the circumference of the external large circle is
1) longer, or
2) shorter, or
3) equal to,
the sum of the circumferences of all inner circles centered on the common diameter, tangent to each other.

Facebook & Ranking Elo Formula

Eduardo Saverin (now a Singaporean billionaire investor) gave the wrong Elo formula to his Facebook co-founder Mark Zuckerburg, both of them became 'accidental' billionaire. Watch the video clip in the movie "Social Network":

http://m.youtube.com/#/watch?v=BzZRr4KV59I

The Elo formula is based on the theory of Normal Distribution with Logarithm function, from base of exponential e to base of 10.
The correct Elo Formula should be :
$Latex \boxed
{
E_a =\frac{1}
{1+ \frac{1}{400}.\Huge 10^{(R_b - R_a)}
}
}$

$Latex \boxed
{
E_b =\frac{1}
{1+ \frac{1}{400}.\Huge 10^{(R_a - R_b)}
}
}$

Eduardo had missed the power ^ below:

白马非马

韓非子是战国法家, 荀子的高徒, 秦始皇宰相李斯的同学。他说"白马非马", 即白马不是马, 可以用集合論(Set Theory) 证明:

Let 马 = H = {w, b, r, y ...}
w : 白马
b : 黑马
r ：红马
y：黄马

Let 白马 = W = {w}

To prove:
H = W
We must prove:
H ⊂ W and H ⊃ W

From definition we know:
$latex w \in H \supset W $
$latex H \nsubseteq W $
$latex \implies H \neq W $

白马≠马
白马非马
[QED]

其他例子:
木魚非鱼

Landau's Beautiful Proofs

Landau's Beautiful Proofs.

Solution 3 (Modelling): Monkeys & Coconuts

More on Linguistic "Half Life"

Proto Indo-European and Chinese in the Late Neolithic Age
后新石器时代的原欧-印语与汉语

Tsung-tung Chang[張聰東] 1988:
“Indo-European vocabulary in Old Chinese: A new thesis on the emergence of Chinese language and civilization in the Late Neolithic Age”, Sino-Platonic Papers 7, Philadelphia.

This Chinese scholar wrote the 1988 paper on the Chinese language origin with the proto-Indo-European (proto IE).

Interestingly very similar 'coincidence' occurs in 1500 words between Chinese and proto IE:

Take -> 得 tek (ancient Chinese sound as in Fujian dialect today)
Mort -> 殁 mo
See -> 视 see
Cow -> 牛 gu
...

http://www.sino-platonic.org/complete/spp007_old_chinese.pdf


After the Tower of Babel, God confused the human into different languages, but by the linguistic 'archaeology' 'Half Life' Theory, we can deduce ~ 4,900 years ago the Chinese and the Germanic (English, Denmark, German ...) shared the same common linguistic root.

The ancient Chinese scholar Xu Shen许慎(东汉 : 58 CE -147 CE) who compiled the first dictionary on Chinese characters and its origins "说文解字", wrote that the character :
白 (White) is "从入合二" : from the origin character 入 together with 2 (二).


For 2,000 years nobody understood what Xu meant or he had made a mistake for the character 白 (white)。

Now from the "Half-Life Theory", we trace to the Germanic common root, Chinese scholars confirm Xu was right.

The character '入' pronounced in ancient sound (still remains in southern dialect like Fujian in Singapore / Taiwan) is *njub, '入' is the ancient character for 乳头( nipple). *njub sounds like 'nipple' in Germanic languages.

白 means use 2 fingers or 2 lips (合二) to squeeze out white milk from nipple (入).

In European languages, 'White' / 'Milk' are also connected in etymology by 'b' 'mb':
English: bleach / milk
Denmark: bleg / maek
Sweden: blek / mjök
German: bleich / milch
Dutch: bleek / melk
Ancient Irish: mlicht/blicht (bleacht)
Polish: biały / mleko

Conclusion :

白 => milk => white (color)

-----------------------------------------
Language 'Half-Life' Formula

$latex \boxed{T = \frac{\log N} {2.\log 85\%}}$
(N= % of vocabulary with common root)

Modern English & German:
N=60%
=> T =1.561 (K-Yrs)
=> 1,561 yrs ago the 2 languages shared the common root @ 449 CE. [confirmed by historians]






Genesis 11:1-9 Tower of Babel

God confused Noah's descendants from their common language root 'Proto-Canaanite' to different languages:
Phoenician
Greek
Aramaic -> Arabic / Jewish
Chinese
...
1. Chinese / 日耳曼語系 Germanic (English) have common root T = 4,951 yrs ago [尧/舜]

2. Chinese / Latin (French)
T= 3,700 yrs ago [夏/商]

3. Chinese / Persian (Iran) T= 3,100 yrs ago [商/周]

Proof: [Chinese / Germanic / Latin Common Roots]

背bei -> Back

爸ba -> papa

鳳feng -> [Phoin] = Phoenix [Greek]

殁mo (=死) -> Mort (English) / Mort (French)

民min -> man (eng) / mann (German)

核he /ker -> Heart / kerd (Kernel)

酸suan -> sour

Beal's Conjecture

Polymath Project

Polymath Project initiated by Terrence Tao inviting mathematicians worldwide to improve on Yitang Zhang's 70 million bounded gap between primes.

Online Seminar by Terrence Tao on Zhang's paper:

http://terrytao.wordpress.com/2013/06/04/online-reading-seminar-for-zhangs-bounded-gaps-between-primes/

Go to this site to get regular update on progress.

http://michaelnielsen.org/polymath1/index.php?title=Bounded_gaps_between_primes

Sabbath Number 7

Sabbath Number '7':

1. Sunday: God rested on 7th day, a Holy day.

2. Professors rest on 7th year Sabbatical leave to re-charge: overseas exchange-cum-research paid holidays.

3. All farm lands must rest on 7th year before cultivation again.

4. After 6 years (Grade 1 to 6) and the stressful Primary School Leaving Exams (PSLE), relax on Secondary 1 (Grade 7); then after 4+2 years Cambridge 'O' and 'A' levels high-school pressure, 'honey-moon' in Freshman year (1st year University), or 'switch off' from study in the Army National Service.

5.  哈佛一调查报告说，人生平均只有7次决定人生走向的机会，两次机会间相隔约7年，大概25岁后开始出现，75岁以后就不会有什么机会了。这50年里的7次机会，第一次不易抓到，因为太年轻；最后一次也不用抓，因为太老。这样只剩5次，这里面又有两次会不小心错过(*)，所以实际上只有3次机会了。

人生七年之"痒" (机会)
25: 毕业寻职: 第一痒
32: 结婚成家:第二痒
39: 创业: 第三痒 (*)
46: 名利: 第四痒
53: 中年危机: 第五痒(*)
60: 金盆洗手: 第六痒
67: 夕阳无限: 第七痒

Love Math

For those who love Math ...
1x8+1=9
12x8+2=98
123x8+3=987
1234x8+4=9876
12345x8+5=98765
123456x8+6=987654
1234567x8+7=9876543
12345678x8+8=98765432
123456789x8+9=987654321

1x9+2=11
12x9+3=111
123x9+4=1111
1234x9+5=11111
12345x9+6=111111
123456x9+7=1111111
1234567x9+8=11111111
12345678x9+9=111111111
123456789x9+10=1111111111

9x9+7=88
98x9+6=888
987x9+5=8888
9876x9+4=88888
98765x9+3=888888
987654x9+2=8888888
9876543x9+1=88888888
98765432x9+0=888888888

1x1=1
11x11=121
111x111=12321
1111x1111=1234321
11111x11111=123454321
111111x111111=12345654321
1111111x1111111=1234567654321
11111111x11111111=
123456787654321
111111111x111111111=
12345678987654321

100-digit Pi

One fine day when we reach above 80 years old, if the doctor accuses us of having dementia, then prove the doctor wrong by shocking him with 100-digit Pi memory :)

With Chinese single-syllable sound for numbers, better still if can sing it as a song, memorizing 100-digit pi is easy!

Derivative Meaning

The derivative of a function can be thought of as:

(1) Infinitesimal: the ratio of the infinitesimal change in the value of a function to the infinitesimal change in a function.

(2) Symbolic: The derivative of
$Latex x^{n} = nx^{n-1} $
the derivative of sin(x) is cos(x),
the derivative of f°g is f'°g*g',
etc.

(3) Logical:
$Latex \boxed{\text{f'(x) = d}} $
$Latex \Updownarrow $
$latex \forall \varepsilon, \exists \delta, \text{ such that }$
$latex \boxed{
0 < |\Delta x| < \delta,
\implies
\Bigr|\frac{f(x+\Delta x)-f(x)}{\Delta x} - d \Bigr| < \varepsilon
}$

(4) Geometric: the derivative is the slope of a line tangent to the graph of the function, if the graph has a tangent.

(5) Rate: the instantaneous speed of f(t), when t is time.

(6) Approximation: The derivative of a function is the best linear approximation to the function near a point.

(7) Microscopic: The derivative of a function is the limit of what you get by looking at it under a microscope of higher and higher power.

(8) The derivative of a real-valued function f in a domain D is the Lagrangian section of the cotangent bundle T*(D) that gives the connection form for the unique flat connection on the trivial R-bundle ßxR for which the graph of f is parallel.

[Source]: Extract from "On Proof and Progess in Mathematics" by William Thurston.

Math Chants

Math Chants make learning Math formulas or Math properties fun and easy for memory . Some of them we learned in secondary school stay in the brain for whole life, even after leaving schools for decades.

Math chant is particularly easy in Chinese language because of its single syllable sound with 4 musical tones (like do-rei-mi-fa) - which may explain why Chinese students are good in Math, as shown in the International Math Olympiad championships frequently won by China and Singapore school students.

1. A crude example is the quadratic formula which people may remember as a little chant:
"ex equals minus bee plus or minus the square root of bee squared minus four ay see all over two ay."

$latex \boxed{
x = \frac{-b \pm \sqrt{b^{2}-4ac}}
{2a}
}$

2. $latex \mathbb{NZQRC}$
Nine Zulu Queens Rule China

3. $latex \boxed {\cos 3A = 4\cos^{3} A - 3\cos A }$

cos = $ 1 (kö, Singapore Chinese Fujian福建話 / Taiwan 台湾闽南语dialect) [一元]

cos 3= $1.3
4cos^3 = $4.3
3cos = $3

$1.3 = $4.3 - $3

4. $latex \boxed{\pi = 3.14159\dots}$
Chant in Chinese Mandarin (Beijing):
山顶一寺一壶酒
Chinese sound means: On the mountain (3) top (.) there is one (1) temple (4) with a (1) bottle (5) of wine (9)

Note : Below is the 22 decimal memory chant:
山顶一寺一壶酒，尔乐, 我三壶把酒吃，酒杀尔，杀不死，乐而乐。

$latex \boxed{
\pi = 3.14159 \:26 \:535897 \:932 \:384 \:626 \dots
}$


5. Group Properties:
$latex \boxed{\text{CAN I ?} }$
C: Closure
A: Associative
N: Neutral (e)
I: Inverse

If only 50%, it is Semi-Group(C, A) 半群
If No Inverse, it is MoNoId (C,A,N) 幺群
(MoNoId : No I)

6. The eccentric choice of letters (a,h,g,b,f,c, skipping iand e) for the coefficients in Conic equation: $latex \boxed
{ ax^{2} + 2hxy + 2gx + by^{2}+ 2fy + c = 0
}$
Math Chant for the coefficients:
{a h g b f}
"all hairy guys big feet"

c: constant term (understood)

Proof & Progress in Math

"On Proof and Progress in Mathematics" by the late William Thurston

Great article in Mathematics Education, enjoy reading !

http://www.ams.org/journals/bull/1994-30-02/S0273-0979-1994-00502-6/S0273-0979-1994-00502-6.pdf

成语数学

中国的小学离校考试 (PSLE) 「神题」: 猜成语
1) 20 除 3
2）1 除100
3）9寸+1寸=1尺
4）12345609
5）1,3,5,7,9

答案：:
1) 20/3= 6.666 六六大顺
2）百中挑一
3）得寸進尺
4）七零八落
5）举世无双

Sequence Limit

Definition: $latex \text{Sequence } (a_n) $
has limit a

$latex \boxed{\forall \varepsilon >0, \exists N, \forall n \geq N \text { such that } |(a_n) -a| < \varepsilon}$

$latex \Updownarrow $

$latex \displaystyle \boxed{ \lim_{n\to\infty} (a_n) = a }$

What if we reverse the order of the definition like this:

∃ N such that ∀ε > 0, ∀n ≥ N,
$latex |(a_n) -a| < \varepsilon$

This means:

$latex \boxed {\forall n \geq N, (a_n) = a }$

Example:

$latex \displaystyle (a_n) = \frac{3n^{2} + 2n +1}{n^{2}-n-3}$

$latex \displaystyle\text{Prove: } (a_n) \text { convergent? If so, what is the limit ?}$

Proof:
$latex \displaystyle (a_n) = 3 + \frac{5n +10}{n^{2}-n-3}$

$latex n \to \infty, (a_n) \to 3$

Let's prove it.

$latex \text {Let } \varepsilon >0$
$latex \text{Choose N such that } \forall n \geq N, $
$latex \displaystyle |(a_n) -3| = \Bigr|\frac{5n +10}{n^{2}-n-3}\Bigr| < \varepsilon$

$latex \text{Simplify: } \displaystyle \Bigr|\frac{5n +10}{n^{2}-n-3}\Bigr|$
$latex \text{Let } n > 10 $
$latex \displaystyle \Bigr|\frac{5n +10}{n^{2}-n-3}\Bigr| < \frac{6n}{\frac{1}{2}n^{2}}= \frac{12}{n} < \varepsilon$

$Latex \text{Choose } N = \max (10, \frac{12}{\varepsilon})$
$Latex \displaystyle\forall n \geq N,
|(a_n) -3 | < \frac{12}{n} < \varepsilon$

Therefore,
$latex \displaystyle \boxed{ \lim_{n\to\infty} (a_n) = 3 }$ [QED]

[Source]: Excellent Introduction in Modern Math:
"A Concise Introduction to Pure Mathematics (3rd Edition)"
by Martin Liebeck
CRC Press @ 2011

Abel Prize 2013

Algebraic Geometry

Belgian mathematician Pierre Deligne is a 'perfect' mathematician: he won all the coveted 'Oscar' Math Prizes:
Fields Medal, Wolf Prize and in 2013 $1m Abel Prize.

http://www.nature.com/news/mathematician-wins-award-for-shaping-algebra-1.12644

Turn Sphere Inside Out

Watch this amazing video, mathematically you can turn a sphere inside out, but not a circle:

http://www.snotr.com/video/3107/How_to_turn_a_sphere_inside_out

Prime Gap by Unheralded Mathematician

http://simonsfoundation.org/features/science-news/unheralded-mathematician-bridges-the-prime-gap/

On April 17, 2013, a paper arrived in the inbox of Annals of Mathematics, one of the discipline’s preeminent journals. Written by a mathematician virtually unknown to the experts in his field — a 50-something lecturer at the University of New Hampshire named Yitang Zhang — the paper claimed to have taken a huge step forward in understanding one of mathematics’ oldest problems, the twin primes conjecture.

Editors of prominent mathematics journals are used to fielding grandiose claims from obscure authors, but this paper was different. Written with crystalline clarity and a total command of the topic’s current state of the art, it was evidently a serious piece of work, and the Annals editors decided to put it on the fast track.


Yitang Zhang (Photo: University of New Hampshire)

Just three weeks later — a blink of an eye compared to the usual pace of mathematics journals — Zhang received the referee report on his paper.

“The main results are of the first rank,” one of the referees wrote. The author had proved “a landmark theorem in the distribution of prime numbers.”

Rumors swept through the mathematics community that a great advance had been made by a researcher no one seemed to know — someone whose talents had been so overlooked after he earned his doctorate in 1991 that he had found it difficult to get an academic job, working for several years as an accountant and even in a Subway sandwich shop.

“Basically, no one knows him,” said Andrew Granville, a number theorist at the Université de Montréal. “Now, suddenly, he has proved one of the great results in the history of number theory.”

Mathematicians at Harvard University hastily arranged for Zhang to present his work to a packed audience there on May 13. As details of his work have emerged, it has become clear that Zhang achieved his result not via a radically new approach to the problem, but by applying existing methods with great perseverance.

“The big experts in the field had already tried to make this approach work,” Granville said. “He’s not a known expert, but he succeeded where all the experts had failed.”



“There are a lot of chances in your career, but the important thing is to keep thinking,” Zhang said.

Relative Entropy & WinZip

Data compression:
Low entropy: a string of 1000 repeated 0 has little information content, can be compressed as "1000 x 0".

High Entropy: a string of 1000 random '0' and '1' cannot be compressed.

Relative Entropy: the best optimized compression.

Example:
Morse code let 1 dot '.' represents the most commonly used alphabet 'e', and less commonly used alphabet 'q' is '--.-'

Applications: (use WinZip tool)
1. Analyze two articles, if they are written by same author or two different authors: the later case has higher relative entropy, requires more disk space for the file. If the compressed file is smaller, likelihood they are from the same author.

2. Analyzing 52 European languages: French and Italian have low Relative Entropy, they belong to same language family (Latin); Swedish and Croatian have high Relative Entropy, they are from different family.

3. WinZip can tell if your article after compressed is only 1/3 of the original size, most likely 2/3 of its content are redundant.

4. WinZip could be used to analyze information from data string of DNA sequences or Stock market movements.

Gödel's Proof: God's Existence

Kurt Gödel's Mathematical Proof of God's Existence

Axiom 1: (Dichotomy) A property is positive if and only if its negation is negative.

Axiom 2: (Closure) A property is positive if it necessarily contains a positive property.

Theorem 1. A positive is logically consistent (i.e., possibly it has some instance).

Definition. Something is God-like if and only if it possesses all positive properties.

Axiom 3. Being God-like is a positive property.

Axiom 4. Being a positive property is (logical, hence) necessary.

Definition. A Property P is the essence of x if and and only if x has P and is necessarily minimal.

Theorem 2. If x is God-like, then being God-like is the essence of x.

Definition. NE(x): x necessarily exists if it has an essential property.

Axiom 5. Being NE is God-like.

Theorem 3. Necessarily there is some x such that x is God-like.

Source: Wang Hao (1987) Reflections on Kurt Goedel. MIT Press: Cambridge, Mass. (Page 195).

More on 666

$latex 666 = 1^{6} - 2^{6} + 3^{6}$

$latex 666 = 6 +6 +6 +6^{3}+ 6^{3}+6^{3}$

$latex 666 = 2^{2}+ 3^{2}+ 5^{2}+ 7^{2}+ 11^{2}+ 13^{2}+ 17^{2}$

$latex \phi(666) = 666 $
where
$latex \phi(n) = \text{number of integers less than n and co-prime with n }$

Bible Code

Appeal to my Blog's international readers (currently at 41 countries):
As of today I have verified Bible Code works for 6 languages : English, Chinese, French, Irish, Japanese and Spanish.

Please help to comment below if the "Bible Code" works in your native language (Russian, Spanish, Italian, Korean, Indonesian...) It works best with Bible in King-James version.

Bible Code:

Step 1: From Genesis 1:1 chose any 1 word from the 10 words. (Eg. 'the')

Step 2: Count the length (n1) of the selected word ('the' = n1 = 3)

Step 3: Jump n1 (3) words to the next word (eg. 'created')

Step 4: count the length (n2) of the selected word ('created' = n2 = 7)

Step 5: Jump n2 (7) words to the next word.

...
Repeat till the selected word first appears in Genesis 1:3 verse.

You will always end with the word "God".

Note1: It also works for Chinese Bible with strokes笔划:
( )神 A blank to respect God's name ：can be counted as 1 or 2 words(上帝).

起10-＞ 空8->神9->1:3 神
初7-＞神...是9 -> 神9 ->1:3 ( )神
神9-＞混12-＞水4->1:3( )神
创6-＞空8-＞神9-＞1:3( )神
造10 ->面9->水4 ->1:3( )神
天4-＞空8-＞神9-＞1:3( )神
地6-＞沌7-＞灵7-＞1:3( )神

Note 2: it works for French Bible too!
1.1 Au 2 ->Dieu 4 -> la2 ->Or 2 ->terre5 -> les 3 ->l'abîme 6 ->sur 3
->1.3 Dieu

Note 3: It works for Japanese Bible too!
1:1 さ3->地6 ->な4->み3->の2-> も3
->あ4->の2->が5->て2->お4->い2
->1:3神

Note 4: It also works for Irish Bible !
1:1 báire 5 -> talamh 6 ->an 2-> agus 4 -> aghaidh 7-> ag 2 -> os 2 ->na 2 ->
1:3 (Dúirt : said) Dia (noun God)

Note 5: It works for Spanish too !
(à 1 mot près : +/- 1 word,
God said = Said God
Dios dijo = dijo Dios)

Mozart Pieces

Mozart Piece P

$latex \boxed{P= 0.027465 + 0.157692K + 0.000159446K^{2}}$
where
K = Köchel number (sequenced by time)

Note: This formula is 85% accurate, error not exceeding 2.

Example:
Mozart No. 40 (G Minor) Symphony
is K.550

Background: Wolfgang Amadeus Mozart (1756-1791) was one of the most prolific composers of all time. In 1862, the German musicologist Ludwig von Köchel made a chronological list of Mozart's musical work. This list is the source of the Köchel numbers, or "K numbers", that now accompany the titles of Mozart's pieces (e.g., Sinfonia Concertante in E-flat major, K. 364). The table below gives the Köchel numbers and composition dates of 17 of Mozart's works.

Top 10 Tough Math

These are the top 10 tough Mathematics:
1. Motivic cohomology or cohomology Theory 上同调理论
2. Langlands Functoriality Conjecture
3. Advanced Number Theory (eg. Fermat's Last Theorem) 高等数论
4. Quantum Group 量子群
5. Infinite Dimensional Banach Space 无穷维度巴拿哈空间
6. Local and Micro-local Analysis of Large Finite Group 大有限群之局部与微局分析
7. Large and Inaccessible Cardinals 大与不可达基数
8. Algebraic Topology 代数拓扑学
9. Super-String Theory 超弦论
10. Langlands Theory 非阿贝尔互反性，自守性表现和模数变化

The Eye of God

This NASA picture is "The Eye of God" with a majestic look over the Earth and the humans He created. He speaks to us in "The Language of God" - Mathematics!

"The great book of nature," said Galileo, "can be read only by those who know the language in which it was written. And this language is mathematics."

Note: There are three universal things (3M) which transcend races, cultures and languages, that God never changed after the Tower of Babel:
M: Music (formed by 7 musical notes)
M: Money (based on Gold)
M: Mathematics (derived from basic structures $latex \mathbb{NZQRC}$)

It was also observed by Leibniz that Music and Math are connected: "Human mind appreciates music through calculation without knowing."

Self-study Advanced Math

I came across this review at Amazon in 2007 on how to study Advanced Math on your own. Wonderful advice !

Give yourself 10-15 years, with passion, interest, dedicated commitment, disciplined, you could self-study Math to be a next Fermat, or Hua Luogen - both learned Math by themselves through self-learning from books.

http://www.amazon.com/gp/richpub/syltguides/fullview/R1GE1P236K3YSV/ref=cm_syt_dtpa_f_1_rdssss1/102-4263436-5550568?pf_rd_m=ATVPDKIKX0DER&pf_rd_s=sylt-center&pf_rd_r=00JK8KDA3S1T2JRNBCVC&pf_rd_t=201&pf_rd_p=253457301&pf_rd_i=0821839675

Origamics

Origami + Mathematics = Origamics

The Origamics was invented and coined by its inventor a biologist Prof Kazuo Haga (Japan) in 1994.

Haga's First Theorem

Fold a square paper of size 1 unit x 1 unit: join the right-bottom vertex to the mid-point of the top edge.

This one fold creates respective edge points which mark out various ratios:

$latex \frac{1}{2}, \frac{1}{6}, \frac{1}{8}, \frac{3}{8}, \frac{\sqrt{5}}{2}, \frac{5}{24}, \dots $


Without ruler, this is the most accurate way to obtain length of $latex \frac{1}{3}, \sqrt{5} \dots $
As summarized in the diagram:

Google Linear Algebra

Google Search Engine & Linear Algebra:
1) Let M(nxn) matrix of size n (say 1 billion) web pages:

$latex
\begin{pmatrix}
m_{11} & m_{12} & \ldots & m_{1n}\\
m_{21} & m_{22} & \ldots & m_{2n}\\
\vdots & \vdots & m_{jk} & \vdots\\
m_{n1} & m_{n2} &\ldots & m_{nn}
\end{pmatrix}
$

$latex m_{jk} = \begin{cases} 1, & \text{if Page }j \text{ linked to Page k} \\
0, & \text{if } \text{ not}
\end{cases}
$

Note: This PageRank of 0 & 1 is over-simplied. The actual PageRank is a fuzzy number between 0 and 1, based on Larry Page's patented PageRank formula, taking into accounts of the importance of the pages linked from and to, plus many other factors.

2) Let v(n) eigenvector of n webpages' PageRank ak:
$latex \begin{pmatrix}
a_1 \\
a_2 \\
\vdots\\
a_k \\
\vdots\\
a_n
\end{pmatrix}
$ $latex \displaystyle \implies a_k= \sum_{j}m_{jk} $
(all Page j pageRanks)
The page k pointed to by all pages j.

3) Let λ eigenvalue: M.v =λ.v

4) Iterate n times: $latex \boxed{(M^{n}).v = \lambda{^n}.v}$

=> page k is ranked more important if many important pages j point to it;
& pages j themselves pointed by other important pages, ...(iterate n times).
=> highest ranked pages appear first in Search list.

Cut a cake 1/5

Visually cut a cake 1/5 portions of equal size:

1) divide into half:




2) divide 1/5 of the right half:




3) divide half, obtain 1/5 = right of (3)

$latex \frac{1}{5}= \frac{1}{2} (\frac{1}{2}(1- \frac{1}{5}))= \frac{1}{2} (\frac{1}{2} (\frac{4}{5}))=\frac{1}{2}(\frac{2}{5})$





4) By symmetry another 1/5 at (2)=(4)




5) divide left into 3 portions, each 1/5

$latex \frac{1}{5}= \frac{1}{3}(\frac{1}{2}+ \frac{1}{2}.\frac{1}{5}) = \frac{1}{3}.\frac{6}{10}$

Mathew Effect of e^x

"For unto every one that hath shall be given, and he shall have abundance: but from him that hath not shall be taken even that which he hath."

—Matthew 25:29, King James Version

Or, "the rich gets richer, the poor gets poorer."

Mathematically, this is $latex e^{x} \text { increases much faster than x increases} $

$latex \frac{d}{dx} e^{x} = e^{x}$

Pigeonhole Principle

$latex \pi = 3.14159265358979323846264 $
$latex \text{Let } a_1, a_2,\dots a_{24} \text{ represent the first 24 digits of } \pi$
Prove:
$latex (a_1 - a_2)(a_3 - a_4) \dots (a_{23} - a_{24}) \text{ is even}$

Proof:

13 Odd digits = {3.14159265358979323846264 }

11 Even digits

$latex \text {12 brackets :}(a_1 - a_2)(a_3 - a_4) \dots (a_{23} - a_{24})$

Put 13 odds into 12 brackets, by Pigeonhole Principle, there is certainly one bracket where
$latex (a_j - a_k) \text{ is a difference of 2 odds, which is an even = 2n}$

2n multiplies with any number will always give even.
The product of 2n with the other 11 brackets will always be even.

Therefore
$latex (a_1 - a_2)(a_3 - a_4) \dots (a_{23} - a_{24}) \text { is even}$

Love + Hate Math

Book on Loving and Hating Mathematics:

ISBN: 978-0-691-142470
Reuben Hershey and Vera John-Steiner
Princeton Press

Φ and 666

We still don't understand these mysterious numbers appearing everywhere in the Nature.
Mathematicians call them 'Transcendental number' (超函数) - basically they are irrational numbers (like $latex \sqrt {2}$ ), in additional, they are NOT solution of any polynomial equations (otherwise, they are called Algebraic number).

Top 3 mysterious numbers are:
$latex \pi, e, \phi$
connected by this formula:
$latex \boxed{e^{i.\pi} + 2\phi = \sqrt{5}}$
$latex \pi \text { involves anything in circle}$

$latex e \text{ anything with exponential growth:}$ $latex \text{ e.g. epidemic, bank interest...}$

$latex \phi \text{ the golden ratio involves symmetry and beauty. }$


Golden Ratio Φ

$latex \boxed{ \phi = 1.61803 \dots = -2 \sin 666 ^\circ}$

Recognize the Satanic number 666?

No wonder we praise in Chinese a gorgeous pretty lady having 魔鬼身材 ('satanic' body shape). It has mathematical truth with the golden ratio of beauty Φ and 666. :)

See also my next blog "More on 666 "

Note:
$latex \boxed{\frac{6}{5} \phi^{2}=\pi }$

Harvard Abstract Algebra Video

The excellent lecture videos of "Introduction to Abstract Algebra", taught by Prof Benedict Gross at Harvard, can be downloaded here:
http://www.extension.harvard.edu/openlearning/math222/

I met Prof Benedict Gross in Singapore last year at the NUS Public Lecture. I thanked him for these Abstract Algebra videos recorded many years earlier which helped me to understand and follow this advanced public lecture on "Elliptic Curve". He was thrilled that remotely he could influence an unknown student through his Internet lectures.

Analysis by Timothy Gowers

Why easy analysis problems are easy
by Timothy Gowers (UK, Fields Medal 1998)

Timothy Gowers is teaching in Cambridge, he wrote the thick volume of "Princeton Math Encyclopedia."

He is a very good mathematician, who likes to explain simple fundamental Math questions (like why 2+2=4, multiplication is commutative,...), in the process making abstract math simple to understand.

"If you have recently met epsilons and deltas for the first time, then you may find the problems you are asked to solve on examples sheets very hard. On the other hand, you will notice that your lecturers, supervisors etc. do not find them hard at all. Why is this? " Read on ...

https://www.dpmms.cam.ac.uk/~wtg10/autoanalysis.html

Below is my attempt to rewrite the Example 1 with Latex epsilon-delta notation for easy reading.

Example 1.

I wish to prove that the sequence (1,0,1,0,1,0,...) does not converge.

$latex \text{Let me set the sequence }
\{a_n\} \text{ to be:} $

$latex
\{a_n\}=
\begin{cases}
1, & \text{if }n \text{ is odd} \\
0, & \text{if }n\text{ is even}
\end{cases}
$

$latex \Large\text{ Then the statement that }
\{a_n\} \Large\text{ converges to } a \Large\text{ can be written: }$

$latex \exists a, \forall \varepsilon >0 ,\:\:\exists N ,\:\:\forall n > N , \:\:|a_n - a| < \varepsilon $

For divergence, we want to write the negation of the above as:

$latex \boxed{\forall a,\: \exists \varepsilon >0, \:\:\forall N, \:\:\exists n > N, \:\:|a_n-a| \geq \varepsilon}$

Take arbitrary a as below:

$latex a_n = 1 \text{ if n is odd, choose }a < 1/2$
$latex a_n = 0 \text{ if n is even, choose }a \geq 1/2$


$latex \text {Let } \varepsilon = \frac {1}{2}$
For either case whether n is even or odd,
$latex \forall N, \:\:\exists n > N, \:\: |a_n- a| \geq \frac{1}{2}$

$latex \iff \{a_n\} \:\: diverges$

Exercise:
Prove:
1-1+1-1+1...
=1, or
=0, or
= 1/2 (Leibniz said 50% -1 50% 0) ?

Sub-group Test

2-Step Test subgroups:
H subset of group G is subgroup if:
1. H is non-empty
(check: identity of G ∈ H)
2. $Latex a.b^{-1} \in H$

Prove Subset not a subgroup:

1. For infinite Group: sufficient to prove subset doesn't contain e (identity).

2. For finite group: sufficient to prove subset not closed.
H is subgroup of G
$latex \iff a*b^{-1} \in H, \forall a, b \in H$

Proof Set Technique

Proof Set Technique:
Let sets A, B
1. Prove A ⊆ B:
∀x ∈ A
(show x ∈ B )
=> A ⊆ B
2. Prove B ⊆ A:
∀x ∈ B
(show x ∈ A )
=> B ⊆ A
3. Prove A = B:
(A ⊆ B) & (B ⊆ A)
=> A = B

Mind, Brain And Education

Mind, Brain And Education
1. Spaced Repetition
2. Retrieval Practice

Tool: Test
Not to assess what students know, but to reinforce it.
Memory is like a storage tank, a test as a kind of dipstick that measure how much information we've put in there.

But that's not how the brain works.

Every time we pull up a memory, we make it stronger and more lasting, so that testing doesn't just measure, it changes learning.

Simply reading over materials to be learnt, or even taking notes and making outlines, as many homework assignments require, doesn't have this effect.

Language learner: 80% retained.
Science: 50% retained.

Self-quizzing (focus less on input of knowledge by passive reading, focus more on output by calling out that same information from brain.)

Cognitive disfluency:
Tough topic, recall better.
Interleaved assignment: mix up different kinds of problems instead of grouping by type.

Group Problem

Let G be a group.
∀x,y ∈ G
Prove that xy and yx have the same order n ?
ie. $latex (xy)^{n}= (yx)^{n} = e$

Proof:
Let
$latex (xy)^{n}=e$
(xy)(xy).....(xy)(xy) = e
[(xy) n times]
By associativity,
x.(yx)(yx)...(yx).y=e
$latex (yx)(yx)...(yx).y = x^{-1}$
$latex (yx)^{n-1}= x^{-1}.y{^-1}$
$latex (yx)^{n-1}= (y.x)^{-1}$
$latex (yx) ^{n-1}.(yx) = e$
$latex (yx)^{n} = e$
ie yx has order n.
[QED]

4 Pillars of Mathematics

4 Pillars of Mathematics:

1. Fundamental Theorem of Arithmetics (Prime)
2. Fundamental Theorem of Algebra (Gauss)
3. Fundamental Theorem of Calculus (Leibniz)
4.Fundamental Theorem of Finite Group (Galois)

Beautiful Trigonometry

Beautiful Trigonometry
tan α +tan β +tan γ = tan α.tan β.tan γ

Descartes 6th Meditation

Descartes 6th Meditation

Body is by nature divisible.

If so and if Mind and Body are one and the same,
then Mind is also divisible.

However, the Mind is entirely indivisible.
It follows that the Mind and Body are not the same.

庖丁解牛数学方法

"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.

IMO Technique

(a+b)³ = a³ + 3a²b+ 3ab² + b³

Different equivalent forms:
(1)：(a+b)³ = a³ + b³+3ab(a+b)
(2)：a³ + b³ = (a+b)³ - 3ab(a+b)
(3): a³ + b³ = (a+b)(a² -ab + b²)
(4)：(a+b)³ - ( a³ + b³ ) = 3ab(a+b)

1997 USAMO Q5:
Prove:
$latex \frac{1}{a^{3}+b^{3}+abc} +
\frac{1}{b^{3}+c^{3}+abc} +
\frac{1}{c^{3}+a^{3}+abc} \leq
\frac{1}{abc}$

Proof:
Apply (3):
a³ + b³ = (a+b)(a² -ab + b²) ≥ (a+b)ab

Note:
a² -ab + b²= (a-b)² + ab ≥ ab
since (a-b)² ≥ 0

$latex \frac{abc}{a^{3}+b^{3}+abc}
\leq \frac{abc}{(a+b)ab + abc}
= \frac{c}{a+b+c}$


Symmetrically,
$latex \frac{abc}{b^{3}+c^{3}+abc}
\leq \frac{a}{a+b+c}$

$latex \frac{abc}{c^{3}+a^{3}+abc}
\leq \frac{b}{a+b+c}$

Add 3 RHS:
$latex \frac{a+b+c}{a+b+c} = 1$

$latex \frac{abc}{a^{3}+b^{3}+abc} +
\frac{abc}{b^{3}+c^{3}+abc} +
\frac{abc}{c^{3}+a^{3}+abc} \leq 1$

$latex \frac{1}{a^{3}+b^{3}+abc} +
\frac{1}{b^{3}+c^{3}+abc} +
\frac{1}{c^{3}+a^{3}+abc} \leq
\frac{1}{abc}$

[QED]

Multi-variable Limit

Analysis is the study of Functions, using the main tool "Limit".

Limit problems appear in:
1. Continuity
2. Derivative
3. Integral
4. Sequence

Multi-variable Functions have different approach of Limit compared to Single-value functions.
Eg. L'Hôpital Rule is not applicable to Multi-variable Functions.

Case 1: Find the Limit (L) of
$latex \displaystyle f(x,y)= \frac{xy}{x^{2}+y^{2}} \text{ at point (0,0)} $

Solution:
Consider the point P(x,y) on f(x,y)
$latex \displaystyle \lim_{x\to 0} f(x,y)=f(0,y)= 0 \text{...(1)}$
$latex \displaystyle\lim_{y\to 0} f(x,y)=f(x,0)=0 \text{...(2)}$

but when P moves along y=x straight line approaching (0,0),
ie. x->0, y=x->0,
$latex \displaystyle\lim_{x\to 0} \frac{xy}{x^{2}+y^{2}}= \frac{x^2}{2.x^2} = \frac{1}{2} \text{...(3)}$

From (1),(2),(3) there are 3 limits {0, 0, 1/2}, hence the Limit L does not exist.

Case 2:
Find limit L of
$latex \displaystyle f(x,y)= \frac{x^{2}+y^{2}}{x^{4}+y^{4}}
\text { at point} (\infty, \infty)$

Solution:
Let y= kx
$latex \displaystyle f(x,y)= \frac{x^{2}+k^{2}y^{2}}{x^{4}+k^{4}y^{4}} = \frac{1+k^{2}}{(1+k^{4})x^2}$
When x -> $latex \infty$,
f(x,y) independent of k => possible limit of 0

Prove: $latex \displaystyle\lim_{(x,y)\to \infty} f(x,y)= L = 0$

$latex \displaystyle \left| \frac{x^{2}+y^{2}}{x^{4}+y^{4}} - 0 \right| = \frac{x^{2}+y^{2}}{x^{4}+y^{4}}
\leq \frac{x^{2}+y^{2}}{2x^{2}y^{2}} = \frac{1}{2x^{2}} + \frac{1}{2y^{2}}$

Note:
$latex (x^{2} - y^{2})^2 \geq 0$
$latex x^{4}-2x^{2}y^{2}+y^{4}\geq 0 $
$latex x^{4}+y^{4} \geq 2x^{2}y^{2}$

$latex \displaystyle\forall \epsilon > 0 \text{, take } \delta \geq \frac{1}{\sqrt{\epsilon}} \text{ such that}$
$latex |x|>\delta, |y|>\delta$
$latex \implies \displaystyle \left| \frac{x^{2}+y^{2}}{x^{4}+y^{4}} - 0 \right| \leq \frac{1}{2x^{2}} + \frac{1}{2y^{2}}
< \frac{1}{2{\delta}^{2}}+\frac{1}{2{\delta}^{2}}$

$latex \implies \displaystyle \left| \frac{x^{2}+y^{2}}{x^{4}+y^{4}} - 0 \right| \leq \frac{1}{2}\left({\frac{1}{\sqrt{\epsilon}}}\right)^{-2}+\frac{1}{2}\left({\frac{1}{\sqrt{\epsilon}}}\right)^{-2} = \frac{\epsilon}{2}+ \frac{\epsilon} {2}$
$latex \implies \displaystyle \left| \frac{x^{2}+y^{2}}{x^{4}+y^{4}} - 0 \right| \leq \epsilon$
$latex \implies \displaystyle\lim_{(x,y)\to \infty} f(x,y)= L =0 $
[QED]

Source: Prof Zhang ShiZao 章士藻(1940-) "Collected Works of Mathematics Education" 数学教育文集
for Ecole Normale Supérieure in China 高等师范学院

Generalized Analytic Geometry

Generalized Analytic Geometry

Find the equation of the circle which cuts the tangent 2x-y=0 at M(1,4), passing thru point A(4,-1).

Solution:

1st generalization:
Let the point circle be:
(x-1)² + (y-4)² =0

2nd generalization:
It cuts the tangent 2x-y=0
(x-1)² + (y-4)² +k(2x-y) =0 ...(C)

Pass thru A(4,-1)
x=4, y= -1
=> k= -2
(C): (x-3)² + (y-1)² = 0
[QED]

Galois Theory Simplified

Galois discovered Quintic Equation has no radical (expressed with +,-,*,/, square root) solutions, but his new Math "Group Theory" also explains:
$latex x^{5} - 1 = 0 \text { has radical solution}$
but
$latex x^{5} -x -1 = 0 \text{ has no radical solution}$

Why ?

$latex x^{5} - 1 = 0 \text { has 5 solutions: } \displaystyle x = e^{\frac{ik\pi}{5}}$
$latex \text{where k } \in \{0,1,2,3,4\}$
which can be expressed in
$latex x= cos \frac{k\pi}{5} + i.sin \frac{k\pi}{5} $
hence in {+,-,*,/, √ }
ie
$latex x_0 = e^{\frac{i.0\pi}{5}}=1$
$latex x_1 = e^{\frac{i\pi}{5}}$
$latex x_2 = e^{\frac{2i\pi}{5}}$
$latex x_3 = e^{\frac{3i\pi}{5}}$
$latex x_4 = e^{\frac{4i\pi}{5}}$
$latex x_5 = e^{\frac{5i\pi}{5}}=1=x_0$

=>
$latex \text {Permutation of solutions }{x_j} \text { forms a Cyclic Group: }
\{x_0,x_1,x_2,x_3,x_4\} $

Theorem: All Cyclic Groups are Solvable
=>
$latex x^{5} -1 = 0 \text { has radical solutions.}$

However,
$latex x^{5} -x -1 = 0 \text{ has no radical solution } $
because the permutation of solutions is A5 (Alternating Group) which is Simple
ie no Normal Subgroup
=> no Symmetry!

Life Changing Book

The book which changed their life:
1. GH Hardy: by Carmille Jordan's Cours d'Analyse:
"I shall never forget the astonishment with which I read the remarkable work ... and I learnt for the first time as I read it what mathematics really meant."

2. Ramanujan : George Carr's
"A Synopsis of Elementary Results in Pure & Applied Mathematics"
(4,400 results without proofs)

3. Riemann : Legendre's book

4. Hardy/Littlewood:
Landau 2-volume "Handbuch der Lehre von Der Verteilung der Primzahlen"
(Handbook of the Theory of the Distribution of Prime Numbers)

5. Atle Selverg (Norway): Ramanujan's "Collected Papers"

Note: This blogger's mathematics 'fire' is rekindled by John Derbyshire's "Unknown Quantity"

French Taupe: 3/2 & 5/2

French elite Grandes Écoles (Engineering College), established since Napoleon with the first Military College (1794) École Polytechnique (nickname X because the College logo shows two crossed swords like X), entry only through very competitive 'Concours' Entrance Exams - to gauge its difficulty, Évariste Galois failed in two consecutive years.

Before taking Concours, there are two years of Prépas, or Classe Préparatoire (Preparatory class) housed in a Lycée (High school) to prepare the top Math / Science post-Baccalaureat students. These two undergraduate years are so torturous that French call these students Taupes (Moles) - they don't see sunlight because most of the time they are studying 24x7, minus sleeping and meal time.

Most students take 2 years to prepare (Year 1: Mathématiques Supérieures, Year 2: Mathématiques Spéciales) for the Concours in order to enter X. These students are nicknamed 3/2 (Trois-Demi), so called playfully by the integration of X:

$latex \displaystyle\int_{1}^{2} xdx= \frac{1}{2}x^{2}\Bigr|_{1}^{2}=\frac{3}{2}$

If by the end of second year some students fail the Concours, they can repeat the second year, then these repeat students are called 5/2 (Cinq-Demi) - integrating X from Year 2 to Year 3:

$latex \displaystyle\int_{2}^{3} x dx=\frac{1}{2}x^{2}\Bigr\vert_{2}^{3}=\frac{5}{2}$


Évariste Galois was 5/2 yet he still failed X, not because of his intelligence but the incompetent X Examiner at whom the angry Galois threw the chalk duster. (Well done !)

Another famous 5/2 is René Thom (Fields medal 1958) who discovered 'Chaos Theory'.

There are few rare cases of 7/2 (Sept-Demi):
$latex \displaystyle\int_{3}^{4} x dx=\frac{1}{2}x^{2}\Bigr\vert_{3}^{4}=\frac{7}{2}$
for those who insist on attempting 3 times to enter X or other elite Grandes Écoles. Equally good - if not better - is École Normale Supérieure (ENS) where Galois finally entered after having failed X twice. The tragic Galois was expelled by ENS for his involvement in the Revolution.

Note: Only 200 years later that ENS officially apologized in recent year, during the Évariste Galois Anniversary ceremony, for wrongfully expelled the greatest Math genius of France and mankind.




One of the top Classe Préparatoire "Lycée Pierre de Fermat" named after the 17th century great Mathematician of the "Last Theorem of Fermat", in his hometown Toulouse, Southern France.

Proofs of Isomorphism

Two ways to prove f is Isomorphism:

1) By definition:
f is Homomorphism + f bijective (= surjective + injective)

2) f is homomorphism + f has inverse map $latex f^{-1}$

Note: The kernel of a map (homomorphism) is the Ideal of a ring.

Two ways to construct an Ideal:
1) use Kernel of the map
2) by the generators of the map.

Two ways to prove Injective:

1) By definition of Injective Map:
f(x) = f(y)
prove x= y

2) By Kernel of homomorphism:
If f is homomorphism
$latex f: A \mapsto B$
Prove Ker f = {0}

Note: Lemma:
$Latex \text{f is injective} \iff Ker f = {0}$

Proof Isomorphism 4 Steps:
1. Define function f：S -> T
Dom(f) = S
2. Show f is 1 to 1(injective)
3. Show f is onto (surjective)
4. Show f(a*b) = f(a). f(b)

Example:
Let T = even Z
Prove (Z,+) and (T,+) isomorphic
Proof:
1. Define f: Z -> T by
f(a) = 2a
2. f(a)=f(b)
2a=2b
a=b
=> f injective

3. Suppose b is any even Z
then a= b/2 ∈ Z and
f(a)=f(b/2)=2(b/2)=b
=> f onto

4. f(a+b) =2(a+b) =2a+2b =f(a)+f(b)
Hence (Z,+) ≌ (T,+)

Isomorphism (≌)
1. To prove G not isomorphic to H:
=> Prove |G| ≠ |H|

2. Isomorphism class = Equivalence Class
B’cos “≌” is an Equivalence relation.

3. Properties of Isomorphism (≌):
i) match e
$latex \phi(e) = e'$
ii) match inverse
$latex \phi (g^{-1} )= (\phi (g))^{-1}$

iii) match power
$latex \phi(g^{k})= (\phi(g))^{k}$

Visualize Isomorphism

This is the Chinese graphical way to visualize Abstract Algebra:

Third Theorem of Isomorphism





Second Theorem of Isomorphism

Group G with normal subgroup N,
H subgroup of G (H ≤ G ).
=>
1.(H ∩ N) normal subgroup of H;
2. HN/N ≌ H /(H ∩ N)

Moonshine Monster Group & Fourier

Moonshine 196,883
What object can exist in 196,883 dimension ?

Simon Norton: "I can explain to you what Moonshine (Monster Group) is in one sentence.
"It is the voice of God."

While all the 5 mathematicians have been fully exhausted after 15 years of effort to categorize all Simple Groups in the Universe, only Simon Norton remains lonely in the search of the largest Monster Group (Moonshine).

Conway, who migrated from Cambridge to Princeton, does not want to touch anything on Group now, said, "Simon is the only person on earth who knows Moonshine".

Monster Group M, order |M|=
$latex 2^{46}. 3^{20}. 5^{9}. 11^{2}. 13^{3}. 17.19.23.29.31.41.47.59.71$

Moonshine Monster Group dimensions (dj) & relationship with Fourier expansion of coefficients (cj) in Modular Function:
$latex x^{-1} + 744+196,884x + 21,493,760 x^{2} + 864,229,970x^{3} +\dots $
$latex c_n= c_1+c_2+...c_{n-1} + d_{n}$
where
$latex d_1 = 196,883$
$latex d_2 = 21,296,876$
$latex d_3 = 842,609,326$
and
$latex c_1 = 1+ d_1 = 196,884$
$latex c_2 = c_1+d_2 = 21,493,760$
$latex c_3 = c_1 + c_2 + d_3 = 864,229,970$

What a coincidence! no wonder Conway said this discovery was the most exciting event in his life.

French Curve

The French method of drawing curves is very systematic:

"Pratique de l'etude d'une fonction"

Let f be the function represented by the curve C

Steps:

1. Simplify f(x). Determine the Domain of definition (D) of f;
2. Determine the sub-domain E of D, taking into account of the periodicity (eg. cos, sin, etc) and symmetry of f;
3. Study the Continuity of f;
4. Study the derivative of f and determine f'(x);
5. Find the limits of f within the boundary of the intervals in E;
6. Construct the Table of Variation;
7. Study the infinite branches;
8. Study the remarkable points: point of inflection, intersection points with the X and Y axes;
9. Draw the representative curve C.

Example:

$latex \displaystyle\text{f: } x \mapsto \frac{2x^{3}+27}{2x^2}$
Step 1: Determine the Domain of Definition D
D = R* = R - {0}

Step 2: There is no Periodicity and Symmetry of f
E = D = R*

[See Note below for Periodic and Symmetric example]

Step 3: Continuity of f
The function f is the quotient of 2 polynomial functions, therefore f is differentiable
=> f is continuous in $latex ]-\infty,0[ \cup ]0,+\infty[ $
[See previous post CID Relation]

Step 4: Determine f'
$latex \displaystyle\forall x \in R^{\star}, f'(x) = \frac{6x^{2}.2x^{2} - 4x (2x^{3}+27)}{4x^{4}} = \frac{4x^{4}-4.27x}{4x^{4}} = \frac{4x(x^{3}-27)}{4x^{4}}$
$latex \forall x \in R^{\star}, (x^{3} - 27 >0) \iff (x>3)$
Therefore f' has the same sign as $latex x \mapsto x(x-3)$

$latex \begin{cases} \forall x \in ]-\infty,0[ \cup ]3,+\infty[, & f'(x)>0 \\
\forall x \in ]0,3[ , & f'(x)<0
\end{cases}$

Step 5a: Limit at x=0

$latex \displaystyle\lim_{x\to 0}(2x^{3}+27) = 27$
$latex \displaystyle\lim_{x\to 0} 2x^{2} = 0 , (\forall x \in R^{\star}, x^{2} >0)$
Therefore, $latex \displaystyle\lim_{x\to 0}f(x) = + \infty$

Step 5b: Limit at $latex x= + \infty$
$latex \displaystyle\lim_{x\to +\infty} f(x) =\lim_{x\to +\infty} \frac{2x^{3}+27}{2x^{2}}=\lim_{x\to +\infty} \frac{2x^{3}}{2x^{2}} = \lim_{x\to +\infty} x = +\infty$

Step 5c: Limit at $latex x= - \infty$
Similarly,
$latex \displaystyle\lim_{x\to -\infty} f(x) = \lim_{x\to -\infty} x = -\infty$

Step 6: Construct the Table of Variation

$latex \begin{array}{|l|l|l|}
\hline
x & - \infty \rightarrow \: \: \: \: 0 & 0 \:\:\:\:\: \rightarrow \:\:3 \rightarrow \:\:\: +\infty \\
\hline
f'(x) & \:\: \: \: \:\: \: + & \:\:\:\: - \:\:\:\:\:\:\:\:\: 0 \:\:\:\:\:\:\: + \\
\hline
f(x) & -\infty \nearrow +\infty & +\infty \searrow \: \frac{9}{2} \nearrow +\infty\\
\hline
\end{array}$

Step 7: Study the infinite branches

7a) $latex \displaystyle\lim_{x\to 0}f(x) = + \infty$
=> y-axis is the asymptote

7b) $latex \displaystyle\forall x \in R^{\star}, f(x) = \frac{2x^{3}+27}{2x^{2}}= x+\frac{27}{2x^{2}}$
$latex \displaystyle\lim_{x\to +\infty}\frac{27}{2x^{2}} = 0$ , $latex \displaystyle\lim_{x\to -\infty}\frac{27}{2x^{2}} = 0$
=>
$latex \displaystyle\lim_{x\to +\infty}f(x) = x$ , $latex \displaystyle\lim_{x\to -\infty}f(x) = x$
=> y= x is another asymptote
$latex \forall x \in R^{\star}, \frac{27}{2x^{2}} >0$
=> The curve C is above the asymptote y=x

Step 8: Study the remarkable points: intersection points with x-axis
$latex \forall x \in R^{\star},(2x^{3}+27 =0)
\iff (x^{3}=-\frac{27}{2})
\iff (x=-\frac{3}{\sqrt[3]{2}}) = -2.38$

Step 9: Draw the representative curve C of f.

[caption id="attachment_2564" align="alignnone" width="500"] french curve[/caption]

Note:
$latex \displaystyle\text{Let g: } x \mapsto \frac{sin x}{2- cos^{2}x}$
D = R
g(x) is periodic of 2π => E = [0 , 2π]
$latex \displaystyle\forall x \in R, g(-x)= \frac{sin (-x)}{2-cos^{2}(-x)}=-\frac{sin x}{2-cos^{2}x}=-g(x) $
=> g(x) is symmetric with respect to the origin point O

We can restrict our study of g(x) in E = [0,π]

$latex \displaystyle\forall x \in R, g(\pi-x)= \frac{sin (\pi-x)}{2-cos^{2}(\pi-x)}=\frac{sin x}{2-cos^{2}x}=g(x) $
=> g(x) is symmetric w.r.t. to the equation x= π/2

Finally, we can further restrict our study of g(x) in E = [0, π/2]

TeX Math Editor

1989 Stanford Professor Donald Knuth published the first version of TeX mathematical software.

Subsequent versions follow π:
3.14,
3.141,
3.1415,
....
3.1415926 (current version)

La Ligne Directe du Dieu

Cédric Villani (Médaille Fields 2010) "Théorème Vivant":

"La fameuse ligne directe, quand vous recevez un coup de fil du dieu de la mathématique, et qu'une voix résonne dans votre tête. C'est très rare, il faut l'avouer!"

"The famous direct line, when you receive a 'telephone call' from the God of the Mathematic, and that a voice resonates in your head. It is very rare, one has to admit."

QuYuan 屈原 Symmetry

屈原 QuYuan (343–278 BCE) Symmetry:
http://en.wikipedia.org/wiki/Qu_Yuan

离騷《天问》
1. "九天之际, 安放安属,
隅隈多有, 谁知其数 ?"
=> 天 (Sky) 和 地 (Earth) must be 2 symmetric spheres.

If 地 (Earth) were flat, then there would be (隅隈) edges and angles at the 天 (Sky) & 地 (Earth) boundary (九天之际).

2. "东西南北, 其修孰多,
南北顺橢, 其衍几何。"
=> 南北顺橢 = The Earth is ellipse (橢), with north-south (南北) slightly flatten.

几何 = Geometry

3. How did QuYuan know this advanced astronomy & geometry in ~ 300 BCE?

[caption id="attachment_2393" align="alignnone" width="201"] 屈原[/caption]

墨子 Mozi & Force

[caption id="attachment_2380" align="alignnone" width="166"] Mozi[/caption]

墨子 Mozi (468 BCE~ 376 BCE), 2000 years earlier than Newton

"墨子" :  “力, 行之所以奋也。”

行: Moving
奋: Acceleration
力: Force is due to acceleration by the moving object.

F ∝ a
F = m.a

εδ Confusion in Limit & Continuity

1. Basic:
|y|= 0 or > 0 for all y

2. Limit: $latex \displaystyle\lim_{x\to a}f(x) = L$ ; x≠a
|x-a|≠0 and always >0
hence
$latex \displaystyle\lim_{x\to a}f(x) = L$
$latex \iff $
For all ε >0, there exists δ >0 such that
$latex \boxed{0<|x-a|<\delta}$
$latex \implies |f(x)-L|< \epsilon$

3. Continuity: f(x) continuous at x=a
Case x=a: |x-a|=0
=> |f(a)-f(a)|= 0 <ε (automatically)
So by default we can remove (x=a) case.

Also from 1) it is understood: |x-a|>0
Hence suffice to write only:
$latex |x-a|<\delta$

f(x) is continuous at point x = a
$latex \iff $
For all ε >0, there exists δ >0 such that
$latex \boxed{|x-a|<\delta}$
$latex \implies |f(x)-f(a)|< \epsilon$

Yin-Yang of Hopf Algebra

Prof Heinz Hopf invented this new Hopf Algebra during WW II, it remained unnoticed for years after he & his wife's tragic death by the Nazi.

The essence of Hopf Algebra is a weird 'Yin-Yang' view of every algebra concept.
Examples:
Associative co-Associative
Commutative co-Commutative
Algebra co-Algebra
Multiplication co-Multiplication
Unit co-Unit
Homomorphism co-Homomorphism
Monoïd co-Monoïd
...
the 'co-' notion represents the 'Yin' opposite of the 'Yang' notion.

[Analogy: Set & coSet 倍集， l'ensemble à gauche ou à droite]

Hopf Algebra is applied in Quantum Group for Quantum Physics, Yang-Baxtor equation, Combinatorics and Topology, etc.

If you find a structure which has a combined properties of Group, Ring and Vector Space, then it could be a Hopf Algebra.

Applied it in Graphs, we can re-discover the Euler's (E-V+F = 2 ) formula, Möbius function, and Symmetric Group (Sn).

Note: Excellent 30-lecture series by Prof Federico Ardila (San Francisco University) on Hopf Algebra:

http://math.sfsu.edu/federico/Clase/Hopf/lectures.html

French Baccalaureat Math

Below are two French Baccalaureat Math textbooks (Volume 1 & 2) for Grade 12th (or Junior College Year 2, GCE A Level, Lycée: Terminale C) students in the late 1970s.

There was a strong Bourbaki style of influence in this Math teaching, which was later criticized as being too abstract and less applied for 18-year-old students.

Nevertheless, the syllabus proved to be excellent for those Math-inclined young minds who later entered the special 2-year Classe Préparatoire aux Grandes Ecoles - Preparatory class (equivalent to Bachelor of Science) for graduate schools in crème de la crème top universities (Grandes Écoles ) such as Ecole Normale Supérieure or Ecole Polytechnique , where the solid Preparatory Math trainings (Maths Supérieures, Maths Spéciales) are extremely rigorous and high standard, incubating most of the French great mathematicians, Fields Medalists and Nobel Prize Scientists.

Voici les 2 tomes de Mathematiques Terminales C et E
Bulletin Officiel du 24 Juin 1971
par M. Monge, M.-C. audouim -Egoriff F. Lemaire-Body
LIBRAIRIE BELIN, 1974 -4 ISBN 2-7011-0224-3

Tome 1: Algèbre et Géométrie

Chapitre 1: Structures Algebriques
1. Groupes-Anneaux -Corps
2. Homomorphisms
3. Espaces vectoriels
4. Applications linéaires et matrices

Chapitre 2: Les nombres complexes

Chapitre 3: Sous espaces vectoriels d'un espace vectoriel réel

Chapitre 4: Espace affines et sous-espace affines

Chapitre 5: Application linéaires

Chapitre 6: Application affines

Chapitre 7: Transformation orthogonales
1. Orthoganalité dans un espace vectoriel
2. Endomomorphismes orthogonaux
3. Transformations orthogonales de E2
4. Transformations orthogonales de E3
5. Orientation d'un espace vectoriel

Chapitre 8: Angles. Product vectoriel

Chapitre 9: Isométrie affines

Chapitre 10: Similitudes

Chapitre 11: Coniques

Chapitre 12: Geometrie descriptive

TOME 2: Arithmétique, Analyse et Probabilités
Programs (B.O. du 24.6.1971 et du 19.7.1973)

Chapitre 1: Entiers naturels. Numération.
1. Ensembles ordonnés
2. Ensemble N des Entiers naturels
3. Numeration

Chapitre 2: Entiers rationnels. Entiers modulo n.
1. Construction de l'ensemble Z
2. Sous-groupes de (Z,+)
3. Congruences dans Z

Chapitre 3: PGCD PPCM Nombres premiers.

Chapitre 4: Nombres réels. Suites

Chapitre 5: Continuité. Limites
1. Continuité en un point
2. Continuité sur un intervalle
3. Limites

Chapitre 6: Fonction réciproque d'une fonction continue strictement monotone

Chapitre 7: Dérivation

Chapitre 8: Pratique de l'étude d'une fonction

Chapitre 9: Fonction vectorielle d'une variable réelle

Chapitre 10: Calcul intégral

Chapitre 11: Fonction logarithmiques. Fonction exponentielles

Chapitre 12: Probabilités sur un ensemble fini

Annexé: Équations différentielles
1. Équations: y'= ay
2. Équations: y'' + ω²y = 0

Note: The 2012 French Baccalaureat (Terminale S) Syllabus is modified simpler to suit the majority of science students with more applied math and lesser abstract math:
http://www.maths-france.fr/Terminale/TerminaleS/ProgrammeOfficiel_TerminaleS_2012.pdf

Old Mathematicians Live Young

J.J. Sylvester, who coined the term Matrix, pointed out that Leibniz, Newton, Euler, Lagrange, Laplace, Gauss, Plato, Archimedes, and Pythagoras all were productive until their 70s or 80s.

"The mathematician lives long and lives young," he wrote.

"The wings of the soul do not early drop off, nor do its pores become clogged with the earthly particles blown from the dusty highways of vulgar life."

Sylvester himself was his 82nd year, in 1896, when he "found a new enthusiasm and blazed up again over the theory of compound partitions and Goldbach's conjecture."

Another mathematician Harold S. M. Coxeter (9/2/1907 - 31/3/2003) attributed his longevity to love of mathematics.

Hilbert's Problem Solving

David Hilbert was a most concrete, intuitive mathematician who invented, and very consciously used, a principle: namely, if you want to solve a problem first strip the problem of everything that is not essential. Simplify it, specialize it as much as you can without sacrificing its core. Thus it becomes simple, as simple as it can be made, without losing any of its punch, and then you solve it. The generalization is a triviality which you don't have to pay too much attention to.

Contemporary Math

Bourbaki started as rebels against the established modes of thought in French mathematics in the early 20th century.

But with new discoveries and the increasing important interaction between physics and math, their exclusionist (from Physics etc) approach lost its effectiveness.

Contemporary math is more multifaceted, it includes more varied theoretical and applied approaches.

Algorithm For Love

Each heart represents a person.

The numbers indicate how well they answered questions based on the other person's wants.

The result is a mathematical expression of how happy you could be.



- By Sam Yagan (co-founder of online dating site OkCupid)

Differentiating under integral

Prove: (Euler Gamma Γ Function)
$latex \displaystyle n! = \int_{0}^{\infty}{x^{n}.e^{-x}dx}$

Proof:
∀ a>0
Integrate by parts:

$latex \displaystyle\int_{0}^{\infty}{e^{-ax}dx}=-\frac{1}{a}e^{-ax}\Bigr|_{0}^{\infty}=\frac{1}{a}$

∀ a>0
$latex \displaystyle\int_{0}^{\infty}{e^{-ax}dx}=\frac{1}{a}$ ...[1]

Feynman trick: differentiating under integral => d/da left side of [1]

$latex \displaystyle\frac{d}{da}\displaystyle\int_{0}^{\infty}e^{-ax}dx= \int_{0}^{\infty}\frac{d}{da}(e^{-ax})dx=\int_{0}^{\infty} -xe^{-ax}dx$

Differentiate the right side of [1]:
$latex \displaystyle\frac{d}{da}(\frac{1}{a}) = -\frac{1}{a^2}$
=>
$latex a^{-2}=\int_{0}^{\infty}xe^{-ax}dx$

Continue to differentiate with respect to 'a':
$latex -2a^{-3} =\int_{0}^{\infty}-x^{2}e^{-ax}dx$
$latex 2a^{-3} =\int_{0}^{\infty}x^{2}e^{-ax}dx$
$latex \frac{d}{da} \text{ both sides}$
$latex 2.3a^{-4} =\int_{0}^{\infty}x^{3}e^{-ax}dx$
...
...
$latex 2.3.4\dots n.a^{-(n+1)} =\int_{0}^{\infty}x^{n}e^{-ax}dx$
Set a = 1
$latex \boxed{n!=\int_{0}^{\infty}x^{n}e^{-x}dx}$ [QED]

Another Example using "Feynman Integration":

$latex \displaystyle \text{Evaluate }\int_{0}^{1}\frac{x^{2}-1}{ln x} dx$

$latex \displaystyle \text{Let I(b)} = \int_{0}^{1}\frac{x^{b}-1}{ln x} dx$ ; for b > -1

$latex \displaystyle \text{I'(b)} = \frac{d}{db}\int_{0}^{1}\frac{x^{b}-1}{ln x} dx = \int_{0}^{1}\frac{d}{db}(\frac{x^{b}-1}{ln x}) dx$

$latex x^{b} = e^{ln x^{b}} = e^{b.ln x} $

$latex \frac{d}{db}(x^{b}) = \frac{d}{db}e^{b.ln x}=e^{b.ln x}.{ln x}= e^{ln x^{b}}.{ln x}=x^{b}.{ln x}$

$latex \text{I'(b)}=\int_{0}^{1} x^{b} dx=\frac{x^{b+1}}{b+1}\Bigr|_{0}^{1} = \frac{1}{b+1}$
=>
$latex \text {I(b)} = ln (b+1) + C$

Let b=0
I(0) = 0= ln (1) + C = 0+C => C=0

$latex \boxed{I(b)=ln(b+1)}$

Let b= 2
$latex \displaystyle\int_{0}^{1}\frac{x^{2}-1}{ln x} dx = I(2) = ln (3)$ [QED]

Surjective Proof

Proof Surjective (or Onto 满射 )

Let f: R² -> R²
(x,y) -> (x+1, y+2)
Prove f onto?

Proof:
By defn of f, f(R²) ⊆ R²
To be onto: f(R²) = R²
=> we must prove:
f(R²) ⊇ R²

∀ (x’, y’) ∈ R² ...(1)
We must show that
(x’, y’) ∈ f(R²),
∃ (x, y) in domian R²
s.t. f(x,y)=(x’,y’)
x’=x+1, y’=y+2
x=x’-1, y=y’-2
With these values (x,y)∈R² and f(x,y)=(x’,y’) as required,
=> (x’, y’) ∈ f(R²)...(2)
Thus R² ⊆ f(R²)

f(R²) ⊆ R² and R² ⊆ f(R²)
=> f(R²) = R²= codomain
=> f is onto
[QED]

九章算术 & Fermat Little Theorem

500 BCE Confucius Period 孔子时代，Chinese ancient Math text "Nine Chapters of Arithmetic" 中国 (九章算术) already recorded the following:
$latex p \mid (2^{p} - 2) \text { for p prime}$
$latex \iff 2^{p} \equiv 2 \mod {p} $

Fermat Little Theorem (17th century):
$Latex m^{p} \equiv m \mod {p}$

Note: Computer Cryptography is based on two math theorems: Chinese Remainder Theorem and Fermat Little Theorem.

What is Infinity

What is Infinity?
(Dedekind’s definition)

$Latex 1+ \aleph_0 = \aleph_0 $
$latex \text {where } \aleph_0 \text{ is Aleph 0} $

The Peano axiom was influenced by Dedekind.

Z Risk Ratio

Z risk ratio defined as:

Z = 0.012A + 0.014B + 0.033C+
0.006D + 0.010E

Let T= Total Asset
A= Net Current Asset / T
B= Retained Earning /T
C= Profit b4 interest before tax / T
D= Market Cap / T
E= Sales / T

Z < 1.8 => company bankrupt
Z > 3 => company healthy


Accuracy:
bankruptcy (3 years 95%) (2 yrs 70%)

Ramanujan-Hardy number

Ramanujan-Hardy number

1729 = 1.1.1 + 12.12.12
1729 = 9.9.9 + 10.10.10

Invention Formula

InVention =
(Passion + Perseverance) / (Knowledge + Experience + Inspiration)

$latex V = \frac{2P}{KEI}$

1's Association

1's Association
= x² + y² [Radius of unity circle]
= sin² X + cos² X
= log 10
= x²/a² + y²/b² [oval]
= lim sin x / x ; x-> 0
= nCn
= (-1/2 ± i√3 /2)³ [cubit root of unity : x³ = 1]

Talent vs Genius

Talent is doing easily what others find difficult, but that genius is doing easily what others find impossible.

Homomorphism History

1830 Group Homomorphism
(1831 Galois)

1870 Field Homomorphism
(1870 Camile Jordan Group Isomorphism)
(1870 Dedekind: Automorphism Groups of Field)

1920 Ring Homomorphism
(1927 Noether)

Set Theory in John 1:1

"In the beginning was the Word, and the Word was with God, and the Word was God. " (John 1:1)

太初有道。道与神同在，道就是神。

Mathematically in Set Theory:
Let W = Word, G = God

∃ W, (In the beginning was the Word)
and
W⊂ G, W⊃ G (the Word was with God)

=> W=G (the Word was God)

(John 14:11 NIV)
"Believe me when I say that I am in the Father and the Father is in me; or at least believe on the evidence of the works themselves."

Proof: by Set Theory
⊂: in, include, inside

Jesus ⊂ Father (I am in the Father )
Father ⊂ Jesus (the Father is in me)
=> Jesus = Father (= God)
[QED]

What is “sin A”

What is “sin A” concretely ?

1. Draw a circle (diameter 1)
2. Connect any 3 points on the circle to form a triangle of angles A, B, C.
3. The length of sides opposite A, B, C are sin A, sin B, sin C, respectively.

Proof:
By Sine Rule:

$latex \frac{a}{sin A} = \frac{b}{sin B} =\frac{c}{sin C} = 2R = 1$
where sides a,b,c opposite angles A, B, C respectively.
a = sin A
b = sin B
c = sin C

杨振宁欣赏数学

1938 杨振宁高二（JC1) 时，父亲杨武之（华罗庚，陈省身的老师 ）借他二本书：
Hardy's "Pure Mathematics",
E.T. Bell's "Men of Mathematics".
并讨论 Set Theory, The Continuum Hypothesis.

40年后杨振宁说：
"我的物理学界同事们大多对数学采取功利主义态度。也许受父亲的影响，我较为欣赏数学，其优美和力量：它有战术上的机巧与灵活，又有战略上的雄才远虑。而且，奇迹的奇迹，它的一些美妙概念竟是支配物理世界的基本结构。＂

陈省身: 数学=美

陈省身 数学=美

“宇宙世界万象纷呈, 都暗含着一种伟大的秩序, 這种秩序, 可以被数学家发现, 用数学公式表示出来; 而公式的演算, 应该做到由繁就简, 最后的形式最简洁, 才最漂亮。”

New Geometry 新几何

New Geometry (新几何) invented by Zhang JingZhong (張景中) derived from 2 basic theorems:

1) Triangles internal angles =180º

2) Triangle Area = ½ base * height
=> derive all geometry
=> trigonometry
=> algebra
(These 3 maths are linked, unlike current syllabus taught separately)

The powerful Area (Δ) Proof Techniques:

1) Common Height:
Line AMB, P outside line
Δ PAM / Δ PBM = AM/BM

2) Common 1 Side (PQ):
Lines AB and PQ meet at M
Δ APQ /Δ BPQ = AM/BM

3) Common 1 Angle:
∠ABC=∠XYZ (or ∠ABC+∠XYZ = ∏ )
Δ ABC /Δ XYZ= AB.BC /XY.YZ

These 3 theorems can prove Butterfly and tough IMO problems.

Coset

The powerful notion of Coset was invented by Galois (l'ensemble à gauche ou à droit), but only named as Coset after 150+ yrs later by G.A. Miller in 1910.

Prove:
Coset * Coset = Coset
=> Normal Subgroup

[Hint] Proof technique: use
1) $latex a^{-1}$
2) e

Proof:
1) For any a ∈G, H subgroup of G,

$latex (Ha)(Ha^{-1})= H.(aHa^{-1})$

2) Given $latex H.(aHa^{-1}) $ is right coset,
Choose $latex (aHa^{-1}) = e \in G$
$Latex H.(aHa^{-1})= He = H$
=> $latex aHa^{-1} \subset H$
=> H Normal subgroup

Lord of the Ring

Lord of the "Ring":
The term Ring first introduced by David Hilbert (1862-1943) for Z and Polynomial.
The fully abstract axiomatic theory of commutative rings by his student Emmy Noether in her paper "Ideal Theory in Rings" @1921.

eg. 3 Classical Rings:
1. Matrices over Field
2. Integer Z
3. Polynomial over Field.

Ring Confusions
Assume all Rings with 1 for * operation.

Ring has operation + forms an Abelian group, operation * forms a semi group (Close, Associative).

1) Ever ask why must be Abelian + group ?
Apply Distributive Axioms below:
(a+b).(1+1) = a.(1+1) + b.(1+1)
= a + a + b + b ...[1]

Or,
(a+b).(1+1) = (a+b).1 + (a+b).1
= a + b + a + b ...[2]

[1]=[2]:
a + (a + b) + b = a + (b + a) +b
=> a + b = b + a

Therefore, + must be Abelian in order for Ring's * to comply with distributive axiom wrt +.

2). Subring
Z/6Z ={0,1,2,3,4,5}
3.4=0 => 3, 4 zero divisor

has subrings: {0,2,4},{0,3}

3). Identity 1 and Units of Ring

Z/6Z has identity 1
but 2 subrings do not have 1 as identity.
subrings {0,2,4}:
0.4=0
2.4=2,
4.4=4 => identity is 4
4 is also a unit.

Units: Ring R with 1.
∀a ∈ R ∃b ∈ R s.t.
a.b=b.a = 1
=> a is unit
and b its inverse a^-1

Z/6Z: identity for * is 1
5.5 = 1
5 is Unit besides 1 which is also unit. (1.1=1)

Group Action: Orbit

Orbit demystified
Let G= {rotation around center O }
Let X= point {x,y } ∈ R²
Orbit (X) = subset of points X on the concentric circles of different radius, centered at point O.
In fact,
Orbit = Coset
Hence is equivalent relation
=> can partition set X
=> X = Uj Orbit (Xj)

French Flag Colors

National Flag of France has 3 color stripes: red,white, blue = (r,w,b)
If we permute the 3 colors, how many flags are there ?

Eg. (w,b,r) = (r,b,w) since they are same if flip over the flag.
Eg. (b,b,b),(r,w,r)... repeat color allowed.
[Ans: 18]
Hint: Apply Group action on the Set (r,w,b)


French Flag:
Let Group G = {g1,g2 }
g1={e}= {(1)(2)(3)}
g2={(1 3)(2)}
|G|=order = 2
Apply Group Action on 3 colors set X {r,w,b}:
g1.X=e.{r,w,b} => Fix 3 stripes each with 3 colors
=> |Fix(g1)|= 3³ ways

(1 3)= Fix 1st & 3rd stripe as 1 stripe, x 3 colors => 3 ways
(2) => Fix 2nd stripe x 3 colors = 3 ways

g2.X=(1 3)(2).X=> 3x3
=> |Fix(g2)|= 3²

By Counting Theorem:
Total Flags = Orbit(X)
= ∑|Fix(g) | /|G|
= (3³ + 3² ) / 2
= 18



Exercise:
Chessboard (8x8=64) colored arbitrarily in black or white.
How many different patterns are there ?

[Consider 2 patterns as the same if a rotation (90, 180, 270 degrees) takes one to another ?]

Mystery of i

$latex \sqrt {-1} = i $ took 300 Years to be accepted.

1. 17AD - Cardano: (stole it from other) for cubic equation.

2. 18AD - Lebniz: i is like "Holy Spirit" between 'there' and 'not there'.

3. 19AD - Gauss: i as Geometry
Point (a,b) in plane RxR :
z=a+ib
a on x-axis, b on i-axis

4. 20AD - Hamilton:
i as ordered-pair (a,b) with unique operations:
(a, b)+(c,d)=(a+c, b+d)
(a, b).(c,d)=(ac-bd, bc+ad)

Langlands' Program

Langlands' Program:

Langland wrote to André Weil in 1967:
Analysis & Algebra linked up by L-function, which converts algebraic data from Galois theory into Analytic functions in complex numbers.

He goes beyong Modular Form to the Automorphic Forms (complex functions whose symmetries are described by larger matrices).

Key concepts on Algebra marry up with those from Analysis.

Elliptic Curve

Elliptic Curve (E):

$latex y^{2}= x^{3} + Ax + B$

1. Every E(Q) over Q is Modular

2. Δ = -16(4A³ + 27B²)

Modular Form

Modular Form (MF):
Is a function which takes Complex numbers from the upper half-plane as inputs and gives Complex numbers as outputs.

MF are notable for their high level of symmetry, determined not by a single number (2π for sine) but by 2x2 Matrices of Complex numbers.

Uses:
1. Proof of the FLT
2. Investigation of Monster Group.
3. Elliptic curve = MF
4. L-function provides dictionary for translating between Analysis and Number Theory.

Pedigree

Mathematician Pedigree: 名师出高徒
1. Littlewood-> Peter Swinnerton-Dyer
2. GH Hardy -> Ramajunian / Hua Luogeng
3. Richard -> Evariste Galois, Charles Hermite
4. Gauss -> Dedekind, Riemann
5. Dirichlet -> Dedekind, Riemann
6. Charles Hermite -> Lindermann
7. Hua Luogeng-> Chen Jin-run
8. SS Chern -> Yau ST
9. Camile Jordan -> Felix Klein, Sophus Lie

Princeton IAS

Princeton

Fine Hall = Institute of Adanced Study

"Raffiniert ist der Herr Gott, aber boshaft ist Er nicht."

The Lord God is subtle, but malicious he is not.

Group Representation

Group Representation
Faithful: Isomorphic (1 to 1)
Unfaithful: homomorphic (m to 1)

Group of Equilateral Triangle
Permute the 3 vertices.
Isomorphic to S3 (permute 1 2 3)
Faithful representation by matrices:
e = $latex
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
$

C₃= $latex
\begin{pmatrix}
-\frac {1}{2} & \frac{\sqrt{3}}{4} \\
-\frac{\sqrt{3}}{4} & -\frac {1}{2}
\end{pmatrix}
$

C₃² = $latex
\begin{pmatrix}
-\frac {1}{2} & -\frac{\sqrt{3}}{4} \\
\frac{\sqrt{3}}{4} & -\frac {1}{2}
\end{pmatrix}
$

σ₁ = $latex
\begin{pmatrix}
-1 & 0 \\
0 & 1
\end{pmatrix}
$

σ₂ = $latex
\begin{pmatrix}
\frac {1}{2} & \frac{\sqrt{3}}{4} \\
\frac{\sqrt{3}}{4} & -\frac {1}{2}
\end{pmatrix}
$


σ₃ = $latex
\begin{pmatrix}
\frac {1}{2} & -\frac{\sqrt{3}}{4} \\
-\frac{\sqrt{3}}{4} & -\frac {1}{2}
\end{pmatrix}
$

Group Theorems: Lagrange, Sylow, Cauchy

1. Lagrange Theorem:
Order of subgroup H divides order of Group G

Converse false:
having h | g does not imply there exists a subgroup H of order h.
Example: Z3 = {0,1,2} is not subgroup of Z6
although o(Z3)= 3 which divides o(Z6)= 6

However,
if h = p (prime number),
=>
2. Cauchy Theorem: if p | g
then G contains an element x (so a subgroup) of order p.
ie.
$latex x^{p} = e $ ∀x∈ G

3. Sylow Theorem :
for p prime,
if p^n | g
=> G has a subgroup H of order p^n:
$latex h= p^{n}$

Conclusion: h | g
Lagrange (h) => Sylow (h=p^n) => Cauchy (h= p, n=1)


Trick to Remember:

g = kh (god =kind holy)
=> h | g
g : order of group G
h : order of subgroup H of G
k : index

Note:
Prime order Group is cyclic
(Z/pZ, +) order p is cyclic & commutative.

Order 4: Z4 not isomorphic to Z2xZ2

Order 6: only Z6 isomorphic Z2xZ3.
Z6 non-commutative

S3 = {1 2 3} ≈ D3 Not Abelian
(1 2)(1 3) = (1 3 2)

(1 3)(1 2) =(1 2 3)

Lagrange: |G|=6
=> order of subgroups in G = 1,2,3,6
6= 2x3
Cauchy : 2|6, 3|6 (2,3 prime)
=> order of elements in G
= 2, 3

Win Win Problem

Yoyo lost $3, Pierre found it and returned to Yoyo.
Yoyo wanted to give all $3 to Pierre as gift, but Pierre refused.

Joe is the judge. How to solve it ?

Solution:
Joe donates $1 to make total $4,
divide between Yoyo and Pierre，

=> Yoyo and Pierre each gets $2
Also Yoyo, Pierre and Joe all lose $1.

Note: "三方各损一两" Win-Win Problem

Aristotle & Plato

Aristotle said about his teacher:
"Plato is dear to me, but dearer still is truth."

吾爱吾师，吾更愛真理！
子曰：当仁不让于师

Poor-Rich Divide: GINI

GINI
0.459 in 2012 (Singapore)
(without govt. help: 0.478)
=>
Income of household at top 10% was 9.14 times that of the bottom 10% (after tax and transfer by government, becomes 7.87x)

OECD Method 1: (0.457)
First adult in family:1 point
Each additional adult 0.5 point
Each child 0.3 point
GINI = household income / sum of points

OECD Method 2:
GINI = square root of family size = 0.435

Note: Worsening GINI
GINI for USA : 0.43 (1990) to 0.47 (2010)
GINI for China: 0.35 (1990) to 0.47 (2010)
GINI for Singapore: 0.43 (1990) to 0.46 (2010)

Einstein & Riemann

Albert Einstein owed much to Riemann's conception, saying that if he had not been acquainted with them
"I never would have been able to develop the theory of relativity."

Manifold

Manifold introduced by Riemann
Whereby the measurement of Length, angle, curvature could vary from point to point, unlike a sphere where the curvature is the same all over.

Manifold: 流形 [文天祥：正气歌]

无用之用

周培源教授： 数学是＂无用之用＂

"Useless" Mathematics is "Useful".

Axiom

Axiom (Greek): meant request. The reader is requested to accept the axioms unquestioningly as the rules of the game.

Euclid's "Element" built the whole Geometry with only 5 axioms.
The 5th axiom "Parallel line" was not challenged for 3,000 years until 19th CE Gauss & Riemann developed the Non-Euclidian Geometry.

Richard Dedekind

Julius Wilhelmina Richard Dedekind (6 Oct 1831 - 12 Feb 1916)

- Last student of Gauss at Göttingen
- Student and closed friend of Dirichlet who influenced his Mathematical education
- Introduced the word Field (Körper)
- Gave the first university course on Galois Theory
- Developed Real Number 'Dedekind Cut' in 1872
- Accomplished musician
- Never married, lived with his unmarried sister until death
- "Whatever provable should be proved."
- By 1858: still yet established ?
$latex \sqrt{2}.\sqrt{3} = \sqrt{2.3} $
- Gave strong support to Cantor on Infinite Set.

http://www-history.mcs.st-and.ac.uk/Biographies/Dedekind.html

CID Relation

C Continuous
I Integrable
D Differentiable

The CID Relation:




$latex D \to C, C \to I$
=>
$latex D \to I$

What it means a curve (function) is :
1. Continuous = not broken curve
2. Differentiable = no pointed 'V' or 'W' shape curve
3. Integrable: can compute the area under the (unbroken) curve.

Visualize e^iπ = -1

Visualize by mapping Exp e (red and pink arrows):

$latex e^{i\pi} = -1$

What is i^i

$Latex i^{i } = 0.207879576...$
$latex i = \sqrt{-1}$

If a is algebraic and b is algebraic but irrational then $latex a^b $ is transcendental. (Gelfond-Schneider Theorem)

Since i is algebraic but irrational, the theorem applies.

1. We know
$latex e^{ix}= \cos x + i \sin x$

Let $latex x = \pi/2 $

2. $latex e^{i \pi/2} = \cos \pi/2 + i \sin \pi/2 $

$latex \cos \pi/2 = \cos 90^\circ = 0 $

$latex \sin 90^\circ = 1 $
$latex i \sin 90^\circ = (i)*(1) = i $

3. Therefore
$latex e^{i\pi/2} = i$
4. Take the ith power of both sides, the right side being $latex i^i $ and the left side =
$latex (e^{i\pi/2})^{i}= e^{-\pi/2} $
5. Therefore
$latex i^{i} = e^{-\pi/2} = .207879576...$

Transcendental Number

Transcendental numbers: e, Π, L...

What about $Latex e^{e}, \pi^{\pi} ,\pi^{e} $ ?

Aleksander (Alexis) Osipovich Gelfond (1906-68):

Gelfond-Schneider Theorem

$Latex a^ b $ transcendental if
a is algebraic, not 0 or 1
b irrational algebraic number

Examples:
$latex \sqrt{6}^{\sqrt{5}}, 3^{\sqrt{7}}$
Hilbert Number: $latex 2^{\sqrt {2}}$ (Hilbert Problem proven by Gelfond}

Is log 2 transcendental ?
[log = logarithm Base 10]

Proof:
$latex 10^{log 2} = 2$

1) Sufficient to prove log 2 irrational
Assume log 2 rational
log 2= p/q, p and q integers
$latex 10^ {log 2} = 2 = 10^ {p/q}$
raise power q
$latex 2^{q} = 10^{p} = (2.5)^{p}$
$latex 2^{q} = 2^{p}.5^{p}$

Case 1: p>q
$latex 1 = 2^{p-q}.5^{p}$
=> False

Case 2: q>p
$latex 2^{q-p}= 5^{p}$
Left is even : $latex 2^{m} \text { = even} $
Right is odd: $latex 5^{n} \text {= ....5} $
=> False

Therefore p,q do not exist,
=> log 2 irrational

Reference: Top 15 Transcendental Numbers: http://sprott.physics.wisc.edu/pickover/trans.html

Chicken & Rabbits

This is an ancient Math Puzzle before Algebra was invented, so the ancient intelligent Chinese used this funny "Method of Elimination" below:

Puzzle: In a cage housing some chicken and rabbits. There are total of  35 heads, and 94 legs. Find how many chicken and rabbits ?

Solution:

Imagine letting each animal raises up 2 legs, then there would be 35x2 = 70 legs raised.

94 - 70 = 24 legs

Note that all chicken raising 2 legs would have fallen on ground (no leg how to stand ?), so the 24 remaining legs still standing on the ground MUST belong ONLY to the rabbits standing on 2 legs (the other 2 legs raised up)

24 / 2 = 12 rabbits (Ans)

35 - 12 = 23 chicken (Ans)

Prime Secret: ζ(s)

Riemann intuitively found the Zeta Function ζ(s), but couldn't prove it. Computer 'tested' it correct up to billion numbers.

$latex \zeta(s)=1+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\dots$

Or equivalently (see note *)

$latex \frac {1}{\zeta(s)} =(1-\frac{1}{2^{s}})(1-\frac{1}{3^{s}})(1-\frac{1}{5^{s}})(1-\frac{1}{p^{s}})\dots$

ζ(1) = Harmonic series (Pythagorean music notes) -> diverge to infinity
(See note #)

ζ(2) = Π²/6 [Euler]

ζ(3) = not Rational number.

1. The Riemann Hypothesis:
All non-trivial zeros of the zeta function have real part one-half.

ie ζ(s)= 0 where s= ½ + bi

Trivial zeroes are s= {- even Z}:
s(-2) = 0 =s(-4) =s(-6) =s(-8)...

You might ask why Re(s)=1/2 has to do with Prime number ?

There is another Prime Number Theorem (PNT) conjectured by Gauss and proved by Hadamard and Poussin:

π(Ν) ~ N / log N
ε = π(Ν) - N / log N
The error ε hides in the Riemann Zeta Function's non-trivial zeroes, which all lie on the Critical line = 1/2 :

All non-trivial zeroes of ζ(s) are in Complex number between ]0,1[ along real line x=1/2

2. David Hilbert:

'If I were to awaken after 500 yrs, my 1st question would be: Has Riemann been proven?'

It will be proven in future by a young man. 'uncorrupted' by today's math.

Note (*):

$latex \zeta(s)=1+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\dots = \sum \frac {1}{n^{s}}$ ...[1]

$latex \frac {1}{2^{s}}\zeta(s) =
\frac{1}{2^{s}}(1+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\dots) $

$latex \frac {1}{2^{s}}\zeta(s) =
\frac {1}{2^{s}}+ \frac{1}{4^{s}} + \frac{1}{6^{s}} + \frac{1}{8^s} +\dots$ ... [2]

[1]-[2]:

$latex (1- \frac{1}{2^{s}})\zeta(s) = 1+ \frac{1}{3^{s}} + \frac{1}{5^{s}} + \dots + \frac{1}{p^{s}} +\dots $

$latex \text {Repeat with} (1-\frac{1}{3^s}) \text { both sides:} $

$latex (1- \frac{1}{3^{s}})(1- \frac{1}{2^{s}})\zeta(s) = 1+ \frac{1}{5^{s}} + \frac{1}{7^{s}} + \dots + \frac{1}{p^{s}} +\dots $

Finally,

$latex (1- \frac{1}{p^{s}}) \dots (1- \frac{1}{5^{s}})(1- \frac{1}{3^{s}})(1- \frac{1}{2^{s}})\zeta(s) = 1$

Or

$latex \zeta(s) = \prod \frac {1}
{1- \frac{1}{p^{s}}}= \sum \frac {1}{n^{s}}$

Note #:
$latex \zeta(s) = \prod \frac {1}
{1- \frac{1}{p^{s}}}= \sum \frac {1}{n^{s}}$

Let s=1
RHS: Harmonic series diverge to infinity
LHS:
$latex \prod \frac {1}{1- \frac{1}{p}}= \prod \frac{p}{p-1}$
Diverge to infinity => there are infinitely many primes p

[caption id="" align="alignright" width="300"] English: Zero-free region for the Riemann_zeta_function (Photo credit: Wikipedia)[/caption]