Mindset: The New Psychology of Success

In my spare time, I am intending to read this psychology book: Mindset: The New Psychology of Success. This book has rocketed to one of the top books on Amazon, and it must have a good reason. Parents and educators should read this groundbreaking book. The book asserts that a subtle change in mindset can have a great difference.

Dweck explains why it’s not just our abilities and talent that bring us success—but whether we approach them with a fixed or growth mindset. She makes clear why praising intelligence and ability doesn’t foster self-esteem and lead to accomplishment, but may actually jeopardize success. With the right mindset, we can motivate our kids and help them to raise their grades, as well as reach our own goals—personal and professional. Dweck reveals what all great parents, teachers, CEOs, and athletes already know: how a simple idea about the brain can create a love of learning and a resilience that is the basis of great accomplishment in every area.

Wedderburn’s Structure Theorem

In abstract algebra, the Artin–Wedderburn theorem is a classification theorem for semisimple rings and semisimple algebras. The theorem states that an (Artinian) ^[1] semisimple ring R is isomorphic to a product of finitely many n_i-by-n_i matrix rings over division rings D_i, for some integers n_i, both of which are uniquely determined up to permutation of the index i. In particular, any simple left or right Artinian ring is isomorphic to an n-by-nmatrix ring over a division ring D, where both n and D are uniquely determined. (Wikipedia)

This is quite a powerful theorem, as it allows semisimple rings/algebras to be “represented” by a finite direct sum of matrix rings over division rings. This is in the spirit of Representation theory, which tries to convert algebraic objects into objects in linear algebra, which is relatively well understood.

Wedderburn’s Structure Theorem

Let $A$ be a semisimple $R$ -algebra.

(i) $A\cong M_{n_1}(D_1)\oplus\dots\oplus M_{n_r}(D_r)$ for some natural numbers $n_1,\dots,n_r$ and $R$ -division algebras $D_1,\dots,D_r$ .

(ii) The pair $(n_i,D_i)$ is unique up to isomorphism and order of arrangement.

(iii) Conversely, suppose $A=M_{n_1}(D_1)\oplus\dots\oplus M_{n_r}(D_r)$ , then $A$ is a right (and left) semisimple $R$ -algebra.

Some Notes

The definition of a semisimple $R$ -algebra is: An $R$ -algebra $A$ is semisimple if $A$ is semisimple as a right $A$ -module.

Example: $R=\mathbb{R}$ and $A=M_2(\mathbb{R})\oplus M_2(\mathbb{R})$ . Then $A$ is semisimple by Wedderburn’s theorem.

Super All-rounder

Source: http://qz.com/603267/an-nfl-player-was-just-accepted-to-the-math-phd-program-at-mit/

This guy is the ultimate definition of “All rounded”. Good at both sports and studies. His motivation to play football is a little creepy though: “I play because I love the game. I love hitting people.” Note: American football is the “Upsized” version of the rugby in Singapore schools. Protective helmet and gear are needed.

The National Football League offseason is supposed to be a time for players to relax, recover from the brutality of the prior season, and prepare for the next one.

Not for Baltimore Ravens player John Urschel. The 6’3”, 305-pound offensive lineman will begin a PhD in mathematics at the Massachusetts Institute of Technology this year. The Hulk-like math geek, who graduated from Penn State with a 4.0 grade point average,will study spectral graph theory, numerical linear algebra, and machine learning.

JC Cut Off Points

Source: http://www.straitstimes.com/singapore/education/little-change-in-junior-college-entry-scores-this-year

Despite the latest O-level results being the best in decades, there was little change in the minimum entry requirements for most junior colleges this year.

Students (future batches) thinking of which JC to enter should read this book by Malcolm Gladwell: David and Goliath: Underdogs, Misfits, and the Art of Battling Giants

In it he discusses whether it is better to be a big fish in a small pond or small fish in big pond at early stage of life. Also read: http://news.bitofnews.com/malcom-gladwells-mindblowing-theory-on-why-its-better-to-be-a-big-fish-in-a-small-pond/. This is very true, as entering an elite JC can be quite demoralizing, not to mention not a good fit as the lectures progress too fast, leading to students requiring either tuition or very intensive self-revision at home. The final result may be that the student may do better in ‘A’ levels in a mid-tier JC than in the elite JC’s like RI or HCI.

Hence, students and parents who are undecided and asking the question “Should I go to X JC” should read the above book. If you decide to go to the elite JCs, do not be demoralised if you are not at the top of the cohort. In fact, even if you are in the bottom half of the cohort, you do still have a good chance in doing well in the ‘A’ levels. It is all about the mindset, which is discussed in the above excellent book. The bottomline is that it may not be the best idea to enter the JC with the “lowest” cut-off point, some decision may be required.

st_20160129_amcutoff29_2027059

USA “Common Core” Math

Math Online Tom Circle

To the outsiders of the USA, we can understand the “Common Core Math Program” is good for American children education, which is supported by Bill Gates with generous $6-billion donation.

The American parents who resist and oppose the ‘Common Core’ Math were like the Singaporean parents who opposed the “Singapore Math Program” in the 1980s. The Singapore parents had to go through remedial classes before they could coach their kids in the ‘new way’. Result proved that after 20 years, Singapore is ranked 2nd in the PISA Test World Ranking in Math (after Shanghai of China).

All good medicines are bitter to swallow. 良药苦口 !

Here are 4 methods of subtraction:

1st : Traditional (aka Abacus) “Carry” Method
2nd: Mental Calculation
3rd : “Singapore Math” (no Carry)
4th: Common Core: “Count Up” (French Method), with which French supermart cashiers usually perform changes at counter — much…

View original post 108 more words

Geometrical Meaning of Matching Faces

Let $X$ be a simplicial set. The elements $x_0,\dots,x_{i-1},x_{i+1},\dots,x_n\in X_{n-1}$ are said to be matching faces with respect to $i$ if $d_jx_k=d_kx_{j+1}$ for $j\geq k$ and $k,j+1\neq i$ .

Geometrically, matching faces are faces that “match” along lower-dimensional faces. In other words, they are “adjacent”.

In the 2-simplex, let $x_0=v_1v_2$ , $x_1=v_0v_2$ , $x_2=v_0v_1$ . Then $x_0, x_2$ are matching faces with respect to 1, since $d_1x_0=v_1=d_0x_2$ .

In the 3-simplex, let $x_0=v_1v_2v_3$ , $x_1=v_0v_1v_2$ , $x_2=v_0v_1v_3$ , $x_3=v_0v_1v_2$ . Then $x_0$ , $x_2$ , $x_3$ are matching faces with respect to 1, since the following hold:
$\begin{aligned} d_1x_0&=v_1v_3=d_0x_2\\ d_2x_0&=v_1v_2=d_0x_3\\ d_2x_2&=v_0v_1=d_2x_3 \end{aligned}$

Thallium Poisoning Mystery: Zhu Ling

This is a very old case, but the mystery of Thallium poisoning endures till today. The murderer is still walking scot-free to this day! Do spread this news, as the name “Zhu Ling” is in danger of being forgotten after more than 20 years. Zhu Ling was a very talented student in chemistry, music (she plays the ancient instrument Guqin) and literature.

To learn more or to donate, visit: http://www.helpzhuling.org/english.aspx

From late 1994 to early 1995, 21-year old Zhu Ling 朱令 was poisoned at Tsinghua, Beijing, one of the most prestigious universities in China. Globally, physicians reviewed her symptoms, which were sent to the Internet through a Usenet group after no correct diagnostics availed themselves for months. This world first ever large scale tele-medicine trial suggested thallium poisoning, which was subsequently proven, and its treatment. Her life was ultimately saved, but she suffered serious neurological damage and permanent physical impairment.

Who poisoned this promising and multi-talented young college girl? One of her dormitory roommates with strong political connections was the prime suspect. Yet the police closed the case in 1998 without a definitive conclusion, claimed due to evidences being stolen.

After 19 years, the murderer is still at large. And it may be someone around you, with a changed identity.

Fibrant Simplicial Set

Let $X$ be a simplicial set. Then $X$ is fibrant if and only if every simplicial map $f:\Lambda^i[n]\to X$ has an extension for each $i$ .

Assume that $X$ is fibrant. Let $f:\Lambda^i[n]\to X$ . The elements $f(d_0\sigma_n),f(d_1\sigma_n),\dots,f(d_{i-1}\sigma_n),f(d_{i+1}\sigma_n),\dots,f(d_n\sigma_n)$ are matching faces with respect to $i$ . This is because for $j\geq k$ and $k,j+1\neq i$ ,

$\begin{aligned} d_jf(d_k\sigma_n)&=f(d_jd_k\sigma_n)\\ &=f(d_kd_{j+1}\sigma_n)\\ &=d_kf(d_{j+1}\sigma_n) \end{aligned}$
Thus, since $X$ is fibrant, there exists an element $w\in X_n$ such that $d_jw=f(d_j\sigma_n)$ for $j\neq i$ . Then, the representing map $g=f_w:\Delta[n]\to X$ , $f_w(\sigma_n)=w$ , is an extension of $f$ .

Conversely let $x_0,\dots,x_{i-1},x_{i+1},\dots,x_n\in X_{n-1}$ be any elements that are matching faces with respect to $i$ . Then the representing maps $f_{x_j}:\Delta[n-1]\to X$ for $j\neq i$ defines a simplicial map $f:\Lambda^i [n]\to X$ such that the diagram

Screen Shot 2016-01-26 at 11.21.11 PM
commutes for each $j$ .

By the assumption, there exists an extension $g:\Delta[n]\to X$ such that $g|_{\Lambda^i[n]}=f$ . Let $w=g(\sigma_n)$ . Then $d_jw=d_jg(\sigma_n)=g(d_j\sigma_n)=f(d_j\sigma_n)=x_j$ for $j\neq i$ . Thus $X$ is fibrant.

Motivation of Simplicial Sets

A simplicial set is a purely algebraic model representing topological spaces that can be built up from simplices and their incidence relations. This is similar to the method of CW complexes to modeling topological spaces, with the critical difference that simplicial sets are purely algebraic and do not carry any actual topology.

To return back to topological spaces, there is a geometric realization functor which turns simplicial sets into compactly generated Hausdorff spaces. A topological space $X$ is said to be compactly generated if it satisfies the condition: A subspace $A$ is closed in $X$ if and only if $A\cap K$ is closed in $K$ for all compact subspaces $K\subseteq X$ . A compactly generated Hausdorff space is a compactly generated space which is also Hausdorff.

Tips on giving (Math) Talks

Here are some tips on how to give (Math) Talks. Talks on other scientific topics should be similar.

Source 1: http://www.math.wisc.edu/~ellenber/mntcg/TalkTipSheet.pdf

One tip I found very useful is this: For long talks (1 hour) and above, it is better to use the whiteboard / blackboard. As this will give you “the flexibility to add or omit material as you see fit, and it forces you not to go to fast.”

For shorter talks, it is better to use slides. “A good rule of thumb: you should allow between 30 seconds and 1 minute per slide.” So if you are preparing for a 30 minute talk, around 30-60 slides would be ideal.

Source 2: https://faculty.washington.edu/heagerty/Courses/b572/public/HalmosHowToTalk.pdf

This is advice by the legendary Paul Halmos. His first advice is to “Make it simple, and you won’t go wrong.”

Evaluating Integrals using Complex Analysis

Our next few posts on complex analysis will focus on evaluating real integrals like $\displaystyle\int_0^\infty \frac{1}{x^2+1}\,dx$ using residue theory from Complex Analysis. This is something amazing about Complex Analysis, it can be used to solve integrals in real numbers, something which is not immediately obvious.

To calculate those real integrals, the first step is to study the theory of residues and poles. This can be found in Chapter 6 of Churchill’s book Complex Variables and Applications (Brown and Churchill).

The extremely powerful theorem that one first needs to know is called Cauchy’s Residue Theorem:

Let $C$ be a simple closed positively oriented contour. If a function $f$ is analytic inside and on $C$ except for a finite number of singular points $z_k$ ( $k=1,2,\dots,n$ ) inside $C$ , then $\displaystyle \int_Cf(z)\,dz=2\pi i\sum_{k=1}^n\text{Res}_{z=z_k}\,f(z)$ .

A Summary of the 3 types of Isolated Singular Points:

Pole of order $m$ . The coefficients $b_n$ of the Laurent series contain a finite (nonzero) number of nonzero terms, i.e. $b_n$ eventually becomes zero after a certain number. i.e. $\displaystyle f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n+\frac{b_1}{z-z_0}+\frac{b_2}{(z-z_0)^n}+\dots+\frac{b_m}{(z-z_0)^m}$ .
Removable singular point. Every $b_n$ is zero.
Essential singular point. An infinite number of the coefficients $b_n$ in the principal part are nonzero.

Shortcut for calculating Residues at Poles

Theorem: An isolated singular point $z_0$ of a function $f$ is a pole of order $m$ if and only if $f(z)$ can be written in the form $f(z)=\frac{\phi(z)}{(z-z_0)^m}$ where $\phi(z)$ is analytic and nonzero at $z_0$ . Moreover $\text{Res}_{z=z_0}f(z)=\phi(z_0)$ if $m=1$ and $\text{Res}_{z=z_0}f(z)=\frac{\phi^(m-1)(z_0)}{(m-1)!}$ . if $m\geq 2$ .

This is a very short crash course on the theorems needed. The next blog post on complex analysis will go into calculating some actual integrals.

WordAds: Advertising on WordPress

There has been a drastic WordAds revenue drop recently. For those using WordPress, check out this post, which is the most comprehensive post on WordAds on the internet. Hope the situation gets better once WordAds 2.0 stabilises.

Excellent Math Books for General Audience

Reading Math “General Audience” books can be a really enlightening experience. Not as dense as Math textbooks, they tell the story behind the great discoveries of Mathematicians. It is an excellent choice for parents who want to inspire and motivate their child about the beauty of Math, which is not just about exams / doing endless exercises.

Recently two books on the Poincare conjecture on Amazon caught my eye. This is the famous conjecture solved recently by Grigori Perelman, who is the first person ever to reject the Fields Medal. Definitely looking forward to read these two books.

The Poincare Conjecture: In Search of the Shape of the Universe

Poincare’s Prize: The Hundred-Year Quest to Solve One of Math’s Greatest Puzzles

Semisimple Modules Equivalent Conditions

Proposition: For a right $A$ -module $M$ , the following are equivalent:

(i) $M$ is semisimple.

(ii) $M=\sum\{N\in S(M): N\ \text{is simple}\}$ .

(iii) $S(M)$ is a complemented lattice, that is, every submodule of $M$ has a complement in $S(M)$ .

We are following Pierce’s book’s Associative Algebras (Graduate Texts in Mathematics) notation, where $S(M)$ is the set of all submodules of $M$ .

This proposition is can be used to prove two useful Corollaries:

Corollary 1) If $M$ is semisimple and $P$ is a submodule of $M$ , then both $P$ and $M/P$ are semisimple. In words, this means that submodules and quotients of a semisimple module is again semisimple.

Proof: Since $M$ is semisimple, by (iii) we have $M\cong P\oplus P'$ , where $P'\cong M/P$ . Let $N$ be a submodule of $P\leq M$ . Then $N$ has a complement $N'$ in $S(M)$ : $M=N\oplus N'$ .

$P=P\cap(N\oplus N')=N\oplus (N'\cap P)$ with $N'\cap P\in S(P)$ . Thus, we have that $P$ is semisimple by condition (iii). Similarly, $M/P\cong P'$ is semisimple.

Corollary 2) A direct sum of semisimple modules is semisimple.

Proof: This is quite clear from the definition of semisimple modules being direct sum of simple modules. A direct sum of (direct sum of simple modules) is again a direct sum of simple modules.

Free Gift by Giant (Singapore only)

http://www.giantsingapore.com.sg/cny2016/subscribe-to-newsletter/

Just to share a good deal with readers of my blog.

Once you subscribe to their newsletter, you will receive an email something like this. I have not claimed the prize yet, but heard it is a pair of USB speakers.

*The disclaimer is “Limited Stocks only”, and free things do run out quickly in Singapore!

Thank you for subscribing with us.

We are giving you a gift as a token of appreciation!

You may redeem the free gift at the Customer Service Counter of these 2 Hypermarkets only:

1. Tampines Hypermarket (21 Tampines North Drive 2 #03 -01 Singapore 528765)
2. IMM Hypermarket (2 Jurong East St 21 #01 -100 Singapore 609601)

Please print out this email and present it to customer service to redeem the free gift.
Please bring your Identification Card for verification purposes.

Terms & Conditions:
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4. Giant reserves the right to change the free gift at its own discretion without prior notice.

Closure is linear subspace

Let $X$ be normed linear space, $Y$ a subspace of $X$ . The closure of $Y$ , $\bar{Y}$ , is a linear subspace of $X$ .

Proof:
We use the “sequential” equivalent definition of closure, rather than the one using open balls: $\bar Y$ is the set of all limits of all convergent sequences of points in $Y$ . Let $z_1,z_2\in \bar{Y}$ , $\alpha\in\mathbb{R}$ . There is a sequence $(a_n)$ in $Y$ such that $a_n\to z_1$ . Similarly there is a sequence $(b_n)$ in $Y$ which converges to $z_2$ .

Then $(a_n+b_n)$ is a sequence in $Y$ that converges to $z_1+z_2$ . $(\alpha a_n)$ is a sequence in $Y$ that converges to $\alpha z_1$ .

Mathematical Story Book (Flatland + Flatterland)

Parents who are looking for a book that can improve a child’s English / Math / logic skills simultaneously can consider Flatland & its sequel Flatterland. Many good reviews on Amazon, and it is considered a classic literature book.

This edition of Flatland is recommended, as it contains annotations by Ian Stewart, a famous mathematician author.

The Annotated Flatland: A Romance of Many Dimensions

Flatterland: Like Flatland, Only More So

Homotopy Theory on Simplicial Sets

Let $f,g:X\to Y$ be simplicial maps. We say that $f$ is homotopic to $g$ (denoted by $f\simeq g$ ) if there exists a simplicial map $F:X\times I\to Y$ such that $F(x,0)=f(x)$ and $F(x,1)=g(x)$ for all $x\in X$ . If $A$ is a simplicial subset of $X$ and $f,g:X\to Y$ are simplicial maps such that $f|_A=g|_A$ , we say that $f\simeq g\ \text{rel}\ A$ if there is a homotopy $F:X\times I\to Y$ such that $F(x,0)=f(x)$ , $F(x,1)=g(x)$ and $F(a,t)=f(a)$ for all $x\in X$ , $a\in A$ , $t\in I$ .

Let $X$ be a simplicial set. The elements $x_0,\dots,x_{i-1},x_{i+1},\dots,x_n\in X_{n-1}$ are said to be matching faces with respect to $i$ if $d_jx_k=d_kx_{j+1}$ for $j\geq k$ and $k,j+1\neq i$ .

A simplicial set $X$ is said to be fibrant (or Kan complex) if it satisfies the following homotopy extension condition for each $i$ :

Let $x_0,\dots,x_{i-1},x_{i+1},\dots,x_n\in X_{n-1}$ be any elements that are matching faces with respect to $i$ . Then there exists an element $w\in X_n$ such that $d_jw=x_j$ for $j\neq i$ .

We define $\Lambda^i[n]$ as the simplicial subset of $\Delta[n]$ generated by all $d_j\sigma_n$ for $j\neq i$ , where $\sigma_n=(0,1,\dots\,n)\in\Delta[n]_n$ is the nondegenerate element.

Proposition: Let $X$ be a simplicial set. Then $X$ is fibrant if and only if every simplicial map $f:\Lambda^i[n]\to X$ has an extension for each $i$ .

Life is Like a Cup of Coffee

麦片虾 Cereal Prawn

Cereal Prawn cooked by my wife 🙂

Chinese Tuition Singapore

Recently, I found a recipe of cereal prawn on the internet. I tried it.

It is not bad, even though I used cornflakes instead of cereal, and green onion leaf instead of curry leaf.
The recipe is like this:

It is an easy way to cook cereal prawn.

Ingredients: prawns, cereal, curry leaf, butter, salt.
1. Clean prawns.

2. Pour some cooking oil into the pan, and deep fry the prawns.

3. Melt the butter with low heat. Put in the curry leaf, stir fry. Then put in the cereal.

4. When the cereal is fried evenly, put in the prawns. Stir fry a little and ladle out.

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Cauchy’s Theorem

Cauchy’s Theorem:

Let $R$ be the closed region consisting of all points interior to and on the simple closed contour $C$ .

If $f$ is analytic in $R$ and $f'$ is continuous in $R$ ,

$\int_C f(z)\,dz=0$

(This is the precursor of Cauchy-Goursat Theorem, which allows us to drop the condition that $f'$ is continuous.)

Proof Using Green’s Theorem:

Let $C$ denote a positively oriented simple closed contour $z=z(t)$ , $a\leq t\leq b$ .

$\int_C f(z)\,d(z)=\int_a^b f[z(t)]z'(t)\,dt$ where $f(z)=u(x,y)+iv(x,y)$ and $z(t)=x(t)+iy(t)$ . Thus

$\begin{aligned} \int_C f(z)\,dz&=\int_a^b (u+iv)(x'+iy')\,dt\\ &=\int_a^b (ux'-vy')\,dt+i\int_a^b (vx'+uy')\,dt\\ &=\int_C u\,dx-v\,dy+i\int_C v\,dx+u\,dy \end{aligned}$

Next, we need Green’s Theorem:

$\int_C P\,dx+Q\,dy=\iint_R (Q_x-P_y)\,dA$

By assumption $f'$ is continuous in $R$ , thus the first-order partial derivatives of $u$ and $v$ are also continous. This is exactly what we need for Green’s Theorem.

Continuing from above, we get $\int_C f(z)\,dz=\iint_R (-v_x-u_y)\,dA+i\iint_R(u_x-v_y)\,dA$ which is exactly zero in view of the Cauchy-Riemann equations $u_x=v_y$ , $u_y=-v_x$ !

Schur’s Lemma

Schur’s Lemma is a useful theorem in algebra that is surprisingly easy to prove.

Schur’s Lemma: Let $M$ and $N$ be right $A$ -modules and let $\phi: M\to N$ be a nonzero $A$ -module homomorphism.

(i) If $M$ is simple, then $\phi$ is injective.

(ii) If $N$ is simple, then $\phi$ is surjective.

(iii) If both $M$ and $N$ are simple then $\phi$ is an isomorphism.

(iv) If $M$ is a simple module, then $\text{End}_A(M)$ is a division $R$ -algebra.

Proof:

(i) $\ker\phi$ is a submodule of $M$ . Since $\phi$ is nonzero, $\ker\phi\neq M$ , which means $\ker\phi=0$ .

(ii) $\text{Im}\,\phi$ is a submodule of $N$ . Since $\text{Im}\,\phi\neq 0$ , $\text{Im}\,\phi=N$ .

(iii) Combine (i) and (ii) to get $\phi$ a bijective homomorphism.

(iv) $\text{End}_A(M):=\text{Hom}_A(M,M)$ is an $R$ -algebra. Let $\phi:M\to M$ be an element in $\text{Hom}_A(M,M)$ . Since $\phi$ is an isomorphism, its inverse $\phi^{-1}$ exists. Then $\phi\circ\phi^{-1}=\text{id}_M=\phi^{-1}\circ\phi$ . Thus $\text{End}_A(M)$ is an division $R$ -algebra.

RI’s O Level Scores: Only 1 student out of 10 made it to JC

Source: http://themiddleground.sg/2016/01/18/32412/

This is indeed very surprising news. One wonders what exactly has gone wrong in the system. One can’t help but feel sorry for the sportsmen who have trained hard and won awards for the school, but were dropped out of the IP track and were inadequately prepared for the O Level track.

The most telling phrase from the article is “He wondered if the school lost out because teachers were unfamiliar with the ‘O’ level syllabus: “When we were doing the paper, we all knew that something was wrong.””

Discussion on this topic can be found at:

https://www.reddit.com/r/singapore/comments/41jalb/ris_o_level_scores_only_one_in_class_of_10/

http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=1&t=85290&p=1630906&sid=9cff42ac3d643115c79f4b426a68ff40#p1630906

N is a simple nonzero right A-module (Equivalent Conditions)

Let $N$ be a nonzero right $A$ -module. Then the following are equivalent:

(i) $N$ is simple.

(ii) $uA=N$ for all $u\in N\setminus\{0\}$

(iii) $N\cong A/M$ for some maximal right ideal $M$ of $A$ .

Proof: (i)=>(ii) Let $u\in N\setminus \{0\}$ . $uA$ is a submodule of $N$ . (Let $ua\in uA$ and $a'\in A$ , then $ua\cdot a'\in uA$ , $ua_1+ua_2=u(a_1+a_2)\in uA$ ). Since $N$ is simple, $uA\neq 0$ implies $uA=N$ .

(ii)=>(i) Condition (ii) implies that $N$ is the only nonzero submodule of $N$ , thus $N$ is simple.

(ii)=>(iii) Let $\psi:A\to N=uA$ , $\psi(a)=ua$ . $\psi$ is an $A$ -linear map that is surjective, thus $A/\ker\psi\cong N$ . $M:=\ker\psi$ is a right ideal of $A$ . Since (ii) implies (i), $N$ is a simple module. Thus by Correspondence Theorem, $M$ is a maximal right ideal.

(iii)=>(i) Follows from the Correspondence Theorem: The map $S\mapsto S/M$ is a bijection from the set of submodules of $A$ containing $M$ and the submodules of $A/M$ . Thus if $M$ is maximal, the only submodules containing $M$ are $M$ and $A$ , thus the only submodules of $N\cong A/M$ are $M/M\cong 0$ and $A/M\cong N$ , i.e. $N$ is simple.

Norm of Cartesian Product

If we have two normed linear spaces $Z$ and $U$ , their Cartesian product $Z\oplus U$ can also be normed, such as by setting $|(z,u)|=|z|+|u|$ , $|(z,u)|'=\max\{|z|,|u|\}$ , or $|(z,u)|''=(|z|^2+|u|^2)^{1/2}$ . Note that we are following Lax’s Functional Analysis, where a norm is denoted as $|\cdot |$ , rather than $\|\cdot\|$ which is clearer but more cumbersome to write.

It is routine to check that all the above 3 are norms, satisfying the positivity, subadditivity, and homogeneity axioms. Minkowski’s inequality is useful to prove the subadditivity of the last norm.

We may check that all of the above 3 norms are equivalent. This follows from the inequalities $\frac{1}{2}(|z|+|u|)\leq\max\{|z|,|u|\}\leq |z|+|u|$ , and $M=(M^2)^{1/2}\leq (|z|^2+|u|^2)^{1/2}\leq (2M^2)^{1/2}=\sqrt 2 M$ , where $M:=\max\{ |z|,|u|\}$ . In general, we have that all norms are equivalent in finite dimensional spaces.

{p(x)<1} is a Convex Set

This theorem can be considered a converse of a previous theorem.

Theorem: Let $p$ denote a positive homogenous, subadditive function defined on a linear space $X$ over the reals.

(i) The set of points $x$ satisfying $p(x)<1$ is a convex subset of $X$ , and 0 is an interior point of it.

(ii) The set of points $x$ satisfying $p(x)\leq 1$ is a convex subset of $X$ .

Proof: (i) Let $K=\{x\in X\mid p(x)<1\}$ . Let $x_1,x_2\in K$ . For $0\leq\alpha\leq 1$ ,

$\begin{aligned}p(\alpha x_1+(1-\alpha)x_2)&\leq \alpha p(x_1)+(1-\alpha)p(x_2)\\ &<\alpha+(1-\alpha)\\ &=1 \end{aligned}$

Therefore $K$ is convex. We also have $p(0)=0\in K$ .

The proof of (ii) is similar.

Wedge and Smash Products

Let $X$ and $Y$ be pointed simplicial sets. The wedge of $X$ and $Y$ , denoted by $X\vee Y$ , is the simplicial set obtained by identifying the basepoint of $X$ with the basepoint of $Y$ . Hence, we can define $X\vee Y$ by the push-out diagram
wedge smash
$X\vee Y$ can be described as the simplicial subset of $X\times Y$ consisting of $(x,*)$ for $x\in X$ and $(*,y)$ for $y\in Y$ .

The smash product $X\wedge Y$ is defined to be the simplicial quotient $X\times Y/(X\vee Y)$ .

Push-out

Let $f:A\to B$ and $g:A\to C$ be simplicial maps. We define $X_n$ to be the push-out in the diagram
Screen Shot 2016-01-17 at 4.46.49 PM
i.e. $X_n=B_n\coprod C_n/\sim$ , where $\sim$ is the equivalence relation such that $f(a)\sim g(a)$ for $a\in A_n$ . Then $X=\{X_n\}_{n\geq 0}$ with faces and degeneracies induced from that in $B$ and $C$ forms a simplicial set with a push-out diagram of simplicial sets
Screen Shot 2016-01-17 at 4.47.03 PM
For example, let $\partial\Delta[n]=\langle d_i(0,1,\dots,n)\mid 0\leq i\leq n\rangle\subseteq \Delta[n]$ be the simplicial subset of $\Delta[n]$ generated by all of the faces of the $n$ -simplex $\sigma_n=(0,1,\dots,n)$ . Then

is a push-out diagram, where $f$ is the inclusion map and $S^n=\Delta[n]/\partial\Delta[n]$ .

Complex Integrals

According to Churchill’s book Complex Variables and Applications,

Integrals are extremely important in the study of functions of a complex variable. The theory of integration, to be developed in this chapter, is noted for its mathematical elegance. The theorems are generally concise and powerful, and many of the proofs are short.

Basic Contour Integrals

The basic way of computing contour integrals is to use the definition. There are more advanced and very powerful methods of computing contour integrals, which we will mention in later posts.

The summarised definition is as follows: $\int_C f(z)\ dz=\int_a^b f[z(t)]z'(t)\,dt$ where $z=z(t), a\leq t\leq b$ represents a contour $C$ .

Basic Example 1: $I=\int_C \bar{z}\,dz$ , where $C$ is the contour $z=2e^{i\theta}$ , $-\pi/2\leq\theta\leq\pi/2$ .

Using the definition, we have

$\begin{aligned} I&=\int_{-\pi/2}^{\pi/2}2e^{-it}\cdot 2ie^{it}\,dt\\ &=4i\int_{-\pi/2}^{\pi/2} 1\,dt\\ &=4\pi i \end{aligned}$

Opposite Algebra

Let $(A,+,\mu)$ be an $R$ -algebra. Define an operation $\mu': A\times A\to A$ by $\mu'(a,b)=\mu(b,a)=ba$ . This algebra $(A,+,\mu')$ is called the opposite algebra to $A$ . We verify that it forms an $R$ -algebra.

Bilinearity comes from the following computations:

$\mu'(a_1+a_2,b)=b(a_1+a_2)=ba_1+ba_2=\mu'(a_1,b)+\mu'(a_2,b)$

$\mu'(a,b_1+b_2)=(b_1+b_2)a=b_1a+b_2a=\mu'(a,b_1)+\mu'(a,b_2)$

$\mu'(ar,b)=bar=\mu'(a,b)r$

$\mu'(a,br)=bra=bar=\mu'(a,b)r$

Associativity is true from $\mu'(\mu'(a,b),c)=cba=\mu'(a,\mu'(b,c))$

The unity element is the same unity element $1_A$ : $\mu'(x,1_A)=1_Ax=x$ , $\mu'(1_A,x)=x1_A=x$ .

Center of Matrix Algebra / Matrix Ring is Scalar Matrices

We will prove that the center $Z(M_n(A))=Z(A)I_n$ , where $A$ is an $R$ -algebra (or ring with unity).

One direction is pretty clear. Let $X\in Z(A)I_n, Y\in M_n(A)$ . Then $X=zI_n$ for some $z\in Z(A)$ . $XY=zI_n Y=zY$ , $YX=YzI_n=zY$ , so $X$ is in the center $Z(M_n(A))$ .

The other direction will require the use of $E_{ij}$ matrices, which is a n by n matrix with (i,j) entry 1, and rest zero.

Let $X=(x_{ij})\in Z(M_n(A))$ . We can write $X=\sum x_{ij}E_{ij}$ . Our key step is compute $E_{pq}X=\sum x_{qj}E_{pj}=XE_{pq}=\sum x_{ip}E_{iq}$ . Thus we may conclude that

$(E_{pq}X)_{ij}=\begin{cases}x_{qj}&\text{if}\ i=p\\ 0&\text{otherwise} \end{cases}$

$(XE_{pq})_{ij}=\begin{cases}x_{ip}&\text{if}\ j=q\\ 0&\text{otherwise} \end{cases}$

Plugging in some convenient values like $i=p, j=q$ , we can conclude that $x_{qq}=x_{pp}$ for all $p,q$ , i.e. all diagonal entries are equal.

Plug in $i=p, j\neq q$ gives us $x_{qj}=0$ for all $j\neq q$ .

Thus $X$ is a scalar matrix, i.e. $X=\alpha I_n$ for some $\alpha\in A$ .

Observing that $(\beta I_n)X=\beta\alpha I_n=X(\beta I_n)=\alpha\beta I_n$ , we conclude that $\beta\alpha=\alpha\beta$ for all $\beta$ in $A$ . So $\alpha$ has to be in the center $Z(A)$ .

Gauge (Minkowski functional) of Convex Set

Theorem: Let $K$ be a convex set.

(i) $p_K(x)\leq 1$ if $x\in K$ .

(ii) $p_K(x)<1$ if and only if $x$ is an interior point of $K$ .

We need the following definition: $p_K(x)=\inf\{a\mid a>0,x/a\in K\}$ . This $p_K$ is often known as a Gauge or Minkowski functional.

Proof:

(i) If $x\in K$ , then $x/1\in K$ , so $P_K(x)\leq 1$ . This is the easy part. The converse holds here, if $x\notin K$ , then $p_K(x)>1$ .

(ii) We first prove the “only if” part. Assume $p_K(x)<1$ . Suppose $x$ is not an interior point of $K$ , i.e. there exists $y\in X$ such that for all $\epsilon>0$ , $x+ty\notin K$ for some $|t|<\epsilon$ .

We then have $p_K(x+ty)>1$ . Combining with the subadditive property of the gauge, we have $1<p_K(x+ty)\leq p_K(x)+p_K(ty)$ . Rearranging, we get $p_K(x)>1-p_K(ty)$ . By considering the various possibilities of the sign of $t$ , and using the positive homogeneity of the gauge, we can obtain a contradiction. For example, if $t>0$ , $p_K(x)>1-tp_K(y)$ . Since $t\to 0$ as $\epsilon\to 0$ , this implies $p_K(x)\geq 1$ , a contradiction.

Conversely, if $x$ is an interior point of $K$ , for all $y$ there exists $\epsilon>0$ such that $x+ty\in K$ for all $|t|<\epsilon$ .

We have $p_K(x+ty)\leq 1$ for all $y$ , for all $|t|<\epsilon$ . Since it is a “for all” quantifier, we can choose in particular $y=x$ , $t>0$ .

Then we have $p_K((1+t)x)\leq 1$ , which leads to $(1+t)p_K(x)\leq 1$ and $p_K(x)\leq \frac{1}{1+t}<1$ .

Simplicial Map and n-simplex

A simplicial map $f:X\to Y$ is a family of functions $f:X_n\to Y_n$ that commutes with $d_i$ and $s_i$ . If each $X_n$ is a subset of $Y_n$ such that the inclusions $X_n\hookrightarrow Y_n$ is a simplicial map, then $X$ is said to be a simplicial subset of $Y$ .

The $n$ -simplex $\Delta[n]$ is defined as follows:
$\Delta[n]_k:=\{(i_0,i_1,\dots,i_k)\mid 0\leq i_0\leq i_1\leq\dots\leq i_k\leq n\}$
where $k\leq n$ .
The face $d_j:\Delta[n]_k\to\Delta[n]_{k-1}$ is defined by $d_j(i_0,i_1,\dots,i_k)=(i_0,i_1,\dots\i_{j-1},i_{j+1},\dots,i_k)$ , i.e. deleting $i_j$ . The degeneracy $s_j:\Delta[n]_k\to\Delta[n]_{k+1}$ is given by $s_j(i_0,i_1,\dots,i_k)=(i_0,i_1,\dots,i_j,i_j,\dots,i_k)$ , i.e. repeating $i_j$ . Let $\sigma_n=(0,1,\dots,n)\in\Delta[n]_n$ . Any element in $\Delta[n]$ can be written as iterated compositions of faces and degeneracies of $\sigma_n$ .

Simplicial Sets

A simplicial set $X$ is a sequence of sets, $X=\{X_0, X_1,\dots, X_n,\dots\}$ , together with maps

$\begin{aligned}d_i&: X_n\to X_{n-1}\\ s_i&: X_n\to X_{n+1} \end{aligned}$

for each $0\leq i\leq n$ . These maps are required to satisfy the simplicial identities

$\begin{aligned} d_id_j&=d_{j-1}d_i\ \ \ \text{for}\ i<j\\ d_is_i&= \begin{cases}s_{j-1}d_i&\text{for}\ i<j\\ \text{id}&\text{for}\ i=j, j+1\\ s_jd_{i-1}&\text{for}\ i>j+1 \end{cases}\\ s_is_j&=s_{j+1}s_i\ \ \ \text{for}\ i\leq j \end{aligned}$

We can use deleting-doubling for remembering simplicial identities:

$\begin{aligned} d_i&: (x_0,\dots,x_n)\mapsto (x_0,\dots,x_{i-1},x_{i+1},\dots,x_n)\\ s_i&: (x_0,\dots,x_n)\mapsto (x_0,\dots,x_{i-1},x_i,x_i,x_{i+1},\dots,x_n) \end{aligned}$

议论文论据素材的积累 Where to Collect Material for Argumentative Composition

Chinese Tuition Singapore

议论文中的论据是用来证明论点的理由和根据。主要有事实论据和道理论据。

一般可以用来作为论据的有很多，如名人名言，俗语谚语，名人轶事，甚至广告词，歌词，身边的事情都可以作为论据。

写作素材的积累需要学生增加阅读量，无论是课外书（小说，诗歌，散文），或是报纸，或者杂志，甚至看一些中文电视节目，这些都是积累的途径。读书破万卷，下笔如有神，读的多了，写文章自然会变成一件容易的事情。

读报纸，可以阅读一些社会热点，并且学习记者在报道新闻和评论新闻的思维，这对口试也是有帮助的。尤其是中学的口试需要对一些社会时事发表自己的观点，多看这类报章，会帮助学生开拓思维。在口试的时候，才会知道该怎么回答，从哪些方面来回答。

看中文电视节目，对学生的作文和口试都有帮助。推荐一档中国的电视节目：

「百家讲坛」这档中国CCTV的科教节目涉及人文科学，自然科学，哲学等。通过看这档节目可以学到很多中国历史典故和国学常识，而且很多名家讲课的内容非常有趣生动。当我还是高中生的时候，课间休息时间，全班学生经常一起看这个节目。比较著名的是（王立群读史记，易中天品三国，于丹论语心得，刘心武揭秘红楼梦，孔庆东看武侠小说，纪连海说清朝二十四臣，等）这些主讲嘉宾，既有著名大学教授，也有作家，还有高中教师。如果对中国文化，历史，文学感兴趣，这是一个很好的节目。

当然如果学生不适合这类比较深的节目，可以尝试一些比较轻松综艺性节目，比如说中国最近有一档节目叫「世界青年说」，这个节目邀请了十多个国家的青年代表（他们来自美，德，英，意，澳等国家），他们每一期节目都会围绕一个议题展开讨论。有一些议题和学生的口试题目非常接近，比如养老，环保，文化差异等等。虽然这些青年代表都是外国人，可是他们的中文水平都不错，对当下中国的流行文化也非常了解。他们在讨论的过程中，会涉及到大量的不同国家的文化风俗，会给观众科普很多典故及历史，这些都是很好的写作素材。而且在他们讨论的过程中也是可以学习他们对某一问题如何发表看法，并且如何论证自己的观点。因为嘉宾都是来自不同的国家和地区，所以在讨论过程中会产生思维碰撞，也可以打开学生的思维，让他们学会从不同的角度去看待同一个问题。

很多家长对成语情有独钟。一些家长会明确要求我给他们的孩子多教成语。我比较推荐学生通过读成语故事的方式来学习成语。因为成语故事本身要么是史实典故，要么是传说，无论是哪种，都是社会普遍认同的，所以故事来当论据是很有分量的。写成语故事还有一个好处，就是凑字数。有些学生会为自己写的作文字数太少而头疼不已。成语故事肯定比只是成语单单这几个字要长的多。

很多学生不喜欢读中文读物，也不喜欢看中文电视节目怎么办？

这种情况下，课本和阅读理解以及阅读问答也可以转变成素材的来源。其实课文中就有很多成语，俗语等。而且课文的写作思路也可以成为借鉴的材料。比如中三的课本中有”老吾老以及人之老””少年莫笑白头翁，人人都有夕阳红””百善孝为先”等可以积累的名语。所以不爱读课外读物的学生，可以把课文内容掌握好，这也算一种方法。

除了课本之外，学生往往也会忽视阅读题目的另外一个作用，就是它也可以当作写作素材。很多阅读题目会选一些名人故事当作材料，这些故事都可以作为论据，而且一个故事可以多用。比如说，有一个理解问答是关于爱因斯坦的。爱因斯坦发表了”相对论”，瑞典皇家科学院很多评委凭借这个理论提名爱因斯坦诺贝尔奖。但是有一个评委反对，因为”相对论”没有经过验证。所以爱因斯坦就没能得到诺贝尔奖。之后一次又一次，爱因斯坦被提名，却被这个评委用同样的借口拒绝。后来有一个叫奥森的年轻人，用”光电效应理论”提名爱因斯坦。这次爱因斯坦终于拿到诺贝尔奖。这个故事涉及到三个人物，而这三个人物刚好是理解这个故事的三个角度。爱因斯坦是著名的科学家，而他拿诺贝尔奖之路漫长又坎坷。开始虽然一次又一次失败，但是他也没有因此沮丧或者气馁，最终他还是拿到了诺贝尔奖，虽然不是凭借”相对论”。从爱因斯坦这个角度，至少可以提炼出两个论点：1，遇到挫折，困难，不可以放弃，要坚持。2，上帝在关闭一扇门的同时，也会打开一扇窗。天无绝人之路。如果从奥森这个角度来讲，也可以提炼出一个论点：换个角度看问题，或者遇事要灵活。如果从投反对票的评委来讲可以从两个方面进行分析：1，坚持原则，刚正不阿，不会因为舆论导向而放弃自己的准则。2，墨守成规，做事呆板，才导致没能给”相对论”应有的荣耀。

所以同一个故事可以从不同的方面去解读，那么学生再遇到与此有关的作文题目，无论是关于坚持，不放弃，又或者换个角度看问题等主题，都可以用这同一个故事当作论据。学生在平时做阅读题目时，要做个有心人。遇到一些名句俗语，或者有深刻含义的故事，都可以记在心里。说不定写作文的时候会派上用场！

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The disk quota for wordpress.com is full

Strangely, I received an error notification that “The disk quota for mathtuition88.wordpress.com is full”, even though I have barely used 5% of my 3 GB storage space under the free plan.

Anyone experienced this?

Screen Shot 2016-01-13 at 4.13.35 PM

Seems everything is alright, I can still post. Very strange.

Mapping of Infinite Vertical Strip by the Exponential Function

This post is continued from a previous post on how to map an open region between two circles to a vertical strip. We wish to map the infinite vertical strip $0<\text{Re}(z)<1$ onto the upper half plane. Reference book is Complex Variables and Applications, Example 3 page 44.

First, we need to map the vertical strip onto the horizontal strip $0<y<\pi$ . This is easily accomplished by $w=\pi iz$ . The factor $i$ is responsible for the rotation (90 degree anticlockwise), while the factor $\pi$ is responsible for the scaling.

Let us consider the exponential mapping $w=e^z=e^{x+iy}=e^x\cdot e^{iy}$ . Consider line $y=k$ . This line will be mapped onto the ray (from the origin) with argument $k$ . Since $0<y<\pi$ , the image is a collection of rays of arguments $0<\theta<\pi$ that “sweeps” across and covers the entire upper half plane.

In conclusion, the map $f(z)=e^{\pi i z}$ will do the job for mapping the vertical strip $0<x<1$ onto the upper half plane $y>0$ .

Let A be a simple R-algebra, then M_n(A) is simple

We let $A$ be a simple $R$ -algebra. Then $M_n(A)$ , the n by n matrix algebra over $A$ is simple. In particular if $D$ is a division algebra, then $M_n(D)$ is simple.

Reference: Associative Algebras (Graduate Texts in Mathematics)

We will split our proof into 2 Lemmas.

Lemma 1: If $I\lhd A$ , then $M_n(I)\lhd M_n(A)$ .

Proof: Let $X=(x_{ij}), Y=(y_{ij})\in M_n(I), Z=(z_{ij})\in M_n(A)$ .

$(X-Y)_{ij}=x_{ij}-y_{ij}\in I$ . Therefore $X-Y\in M_n(I)$ .

$(XY)_{ij}=\sum_{k=1}^n x_{ik}y_{kj}\in I$ , thus $XY\in M_n(I)$ . Similarly $YX\in M_n(I)$ . Thus $M_n(I)$ is indeed an ideal of $M_n(A)$ .

Lemma 2 (Tricky part, takes some time to digest): If $J\lhd M_n(A)$ , then $J=M_n(I)$ for some $I\lhd A$ .

A technical thing we need to know is the $E_{ij}$ matrix, which is a n by n matrix with 1 in the (i,j) entry, 0 elsewhere. A key property is that $E_{ij}E_{kl}=E_{il}$ if $j=k$ , and 0 otherwise (zero matrix).

The key idea here is that $I$ can be taken to be $I=\{r\in A\mid rE_{11}\in J\}$ . We can check that $I$ is indeed an ideal. We need to know that it is possible to premultiply and postmultiply a given matrix by “elementary matrices” such that any entry of the given matrix is “shifted” to the (1,1) entry. An example will be $\begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}0&0\\1&0\end{pmatrix}=\begin{pmatrix}d&0\\0&0\end{pmatrix}$ .

Let us begin the proof to first show $J\subseteq M_n(I)$ . Let $X=(x_{ij})\in J$ . We write $X=\sum_{i,j}^n x_{ij}E_{ij}$ . For any $1\leq p,q\leq n$ ,

$\begin{aligned}E_{1p}XE_{q1}&=E_{1p}(\sum x_{ij}E_{ij})E_{q1}\\ &=(\sum x_{pj}E_{1j})E_{q1}\\ &=x_{pq}E_{11}\in J \end{aligned}$

Therefore $x_{pq}\in I$ . Therefore $X\in M_n(I)$ . What we are actually doing here is first pick an arbitrary matrix $X$ in $J$ . Then, we do the “shifting” process to show that any (p,q) entry in $X$ can be “shifted” to the (1,1) entry, thus it is inside our ideal $I$ . Since any arbitrary (p,q) entry in $X$ is inside the ideal $I$ , we conclude that $X$ is an element of $M_n(I)$ .

Next part is to show $M_n(I)\subseteq J$ . Let $X=(x_{ij})\in M_n(I)$ . We have $x_{ij}E_{11}\in J$ for all $i,j$ . We compute that

$\begin{aligned}E_{i1}(x_{ij}E_{11})E_{1j}&=x_{ij}E_{i1}E_{1j}\\ &=x_{ij}E_{ij}\in J \end{aligned}$

Therefore $X=\sum x_{ij}E_{ij}\in J$ . We will illustrate what we are doing here by an example in the 2 by 2 case. Let us have $X=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ . Since $X$ is in $M_n(I)$ , what we have, by definition of $I$ , is that $\begin{pmatrix}a&0\\0&0\end{pmatrix}$ , $\begin{pmatrix}b&0\\0&0\end{pmatrix}$ , $\begin{pmatrix}c&0\\0&0\end{pmatrix}$ , $\begin{pmatrix}d&0\\0&0\end{pmatrix}$ are all in $J$ . We then proceed to “shift” them all into their correct positions: $\begin{pmatrix}a&0\\0&0\end{pmatrix}$ , $\begin{pmatrix}0&b\\0&0\end{pmatrix}$ , $\begin{pmatrix}0&0\\c&0\end{pmatrix}$ , $\begin{pmatrix}0&0\\0&d\end{pmatrix}$ , all of which are still in the ideal $J$ . $X$ is the sum of all of them, thus also in $J$ .

Final Conclusion

Since $A$ is simple, its ideals are 0 and itself. Thus, the ideals of $M_n(A)$ will be also the 0 matrix and itself, and thus is simple. A division algebra $D$ is simple (Any ideal containing a nonzero element $x$ , when multiplied by its inverse will give 1. Thus the ideal will contain 1 and thus the entire ring $D$ ), thus $M_n(D)$ is also simple.

1000 Posts

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Finally, reached 1000 posts!

Better To Try And Fail Than Never To Try At All – Poem by William F. O’Brien

Source: http://www.poemhunter.com/poem/better-to-try-and-fail-than-never-to-try-at-all/

Some say risk nothing, try only for the sure thing,
Others say nothing gambled nothing gained,
Go all out for your dream.
Life can be lived either way, but for me,
I’d rather try and fail, than never try at all, you see.

Some say “Don’t ever fall in love,
Play the game of life wide open,
Burn your candle at both ends.”
But I say “No! It’s better to have loved and lost,
Than never to have loved at all, my friend.”

When many moons have gone by,
And you are alone with your dreams of yesteryear,
All your memories will bring you cheer.
You’ll be satisfied, succeed or fail, win or lose,
Knowing the right path you did choose.

William F. O’Brien

Recommended Functional Analysis Book (Graduate Level)

Functional Analysis is a subject that combines Linear Algebra with Analysis. I researched online, and it seems one of the best Functional Analysis Book for Graduate level is Functional Analysis by Peter Lax. This book is ideal for a second course in Functional Analysis. For a first course in Functional Analysis, I would recommend Kreyszig, which is listed on my Recommended Undergraduate Math Books page.

This is Theorem 5 in the book: Let $X$ be a linear space over the reals.

The following 9 properties hold:

The empty set is convex.
A subset consisting of a single point is convex.
Every linear subspace of $X$ is convex.
The sum of two convex subsets is convex.
If $K$ is convex, so is $-K$ .
The intersection of an arbitrary collection of convex sets is convex.
Let $\{K_j\}$ be a collection of convex subsets that is totally ordered by inclusion. Then their union $\cup K_j$ is convex.
The image of a convex set under a linear map is convex.
The inverse image of a convex set under a linear map is convex.

Brief sketch of proofs:

We give a brief sketch of the idea behind the proofs.

We are using the definition of convex as follows: $X$ is a linear space over the reals; a subset $K$ of $X$ is called convex if, whenever $x$ and $y$ belong to $K$ , all points of the form $ax+(1-a)y$ , $0\leq a\leq 1$ also belong to $K$ .

Property 1 is vacuously true.

Property 2 is true because of $ax+(1-a)x\equiv x$ .

Property 3 is true because $ax+(1-a)y$ is a linear combination and is thus in the linear subspace.

Property 4) Let $C_1$ and $C_2$ be the two convex subsets. Let $x_1+y_1$ and $x_2+y_2$ be points in $C_1+C_2=\{x+y:x\in C_1, y\in C_2\}$

$\begin{aligned} a(x_1+y_1)+(1-a)(x_2+y_2)&=ax_1+ay_1+x_2+y_2-ax_2-ay_2\\ &=[ax_1+(1-a)x_2]+[ay_1+(1-a)y_2]\\ &\in C_1+C_2 \end{aligned}$

Property 5) We just need to know that $-K=\{-x:x\in K\}$ and this algebraic observation: $a(-x)+(1-a)(-y)=-(ax+(1-a)y)$ .

Property 6) Let $x,y\in\bigcap_{i\in I}C_i$ . $ax+(1-a)y\in C_i$ for all $i\in I$ , thus $ax+(1-a)y\in\bigcap_{i\in I}C_i$ .

Property 7) Let $x,y\in\bigcup K_j$ , where $K_i\subseteq K_{i+1}$ . Let $x\in K_n$ , $y\in K_m$ , then either $K_n\subseteq K_m$ or $K_m\subseteq K_n$ . If $K_n\subseteq K_m$ , $ax+(1-a)y\subseteq K_m\subseteq \bigcup K_j$ . Similarly for the other case $K_m\subseteq K_n$ .

Property 8) Observe that $af(x)+(1-a)f(y)=f(ax+(1-a)y)\in f(K)$ .

Property 9) The only tricky thing about this part is that we cannot assume that the inverse $f^{-1}$ exists. We can only talk about the pre-image.

Let $w,z\in f^{-1}(K)$ . $f(w)\in K$ and $f(z)\in K$ .

We have $f(aw+(1-a)z)=af(w)+(1-a)f(z)\in K$ .

Thus $aw+(1-a)z\in f^{-1}(K)$ .

The End!

Image of Vertical Strip under Inversion

This is a slight generalisation of Example 3 in Churchill’s Complex Variables and Applications.

Consider the infinite vertical strip $c_1<x<c_2$ , under the transformation $w=1/z$ , where $c_1, c_2$ are of the same sign.

When $x=c_1$ , note that by arguments similar to earlier analysis, we have $x=\frac{u}{u^2+v^2}$ .

$u^2+v^2=\frac{1}{c_1}u$ , upon completing the square, becomes

$(u-\frac{1}{2c_1})^2+v^2=(\frac{1}{2c_1})^2$ — Circle 1

Similarly the line $x=c_2$ is transformed into:

$(u-\frac{1}{2c_2})^2+v^2=(\frac{1}{2c_2})^2$ — Circle 2

Note that as $x$ gets larger, the radius of the resultant circle gets smaller.

Thus, the resultant image is the (open) region between the two circles.

The converse holds too, i.e. the region between two circles will be mapped back to the vertical strip by inversion.

Example

Find an analytic isomorphism from the open region between the two circles $|z|=1$ and $|z-\frac{1}{2}|=\frac{1}{2}$ to the vertical strip $0<\text{Re}(z)<1$ .

First, we “shift” the circles to the left by 1 unit via the map $w=z-1$ . Now we are in the situation of our above analysis, thus inversion $w=1/z$ maps the region to the vertical strip $-1<x<-1/2$ . The map $w=-z$ (reflection), followed by $w=z-\frac{1}{2}$ , finally finishing up with a scaling of factor 2 brings us to the vertical strip desired.

Composition of all the above functions leads us to the desired transformation $f(z)=2(\frac{-1}{z-1}-\frac 12)=\frac{-1-z}{z-1}$ to the very nice strip $0<\text{Re}(x)<1$ .

We will mention an analytic isomorphism of this vertical strip to the upper half plane $\text{Im} z>0$ in a subsequent blog post.

Ultimate Tic-Tac-Toe

Math with Bad Drawings

Updated 7/16/2013 – See Original Here

Once at a picnic, I saw mathematicians crowding around the last game I would have expected: Tic-tac-toe.

As you may have discovered yourself, tic-tac-toe is terminally dull. There’s no room for creativity or insight. Good players always tie. Games inevitably go something like this:

But the mathematicians at the picnic played a more sophisticated version. In each square of their tic-tac-toe board, they’d drawn a smaller board:

As I watched, the basic rules emerged quickly.

View original post 434 more words

Clash of Clans Math – Using Linear Inequalities to Save Elixir

This is quite a nice video on Clash of Clans and Math. Only got 70 views so far, but it is definitely a well prepared video.

Another video:

Farming in Clash of Clans has just gotten harder, due to nerf on Town Hall not granting shield, and very powerful defenses for TH10 and TH11. Watch these videos and you may gain some idea on a better way to farm.

Definition of Tensor Product: M Tensor N

Let $R$ be a commutative ring with 1 and let $M$ and $N$ be $R$ -modules. This blog post will be about what is the $R$ -module $M\otimes_R N$ . The source of the material will primarily come from Abstract Algebra, 3rd Edition (by Dummit and Foote) which is a highly recommended Algebra book for undergraduates.

Motivation

The tensor product is a construction that, roughly speaking, allows us to take “products” $mn$ of elements $m\in M$ and $n\in N$ .

Definition

(Following Dummit; there is another equivalent definition using “universal property”, see Wikipedia)

$M\otimes_R N$ is the quotient of the free $\mathbb{Z}$ -module over $M\times N$ (also called module of the formal linear combinations of elements of $M\times N$ ) by the subgroup generated by elements of the form:

$(m_1+m_2,n)-(m_1,n)-(m_2,n)$

$(m,n_1+n_2)-(m,n_1)-(m,n_2)$

$(mr,n)-(m,rn)$

The outcome of the above definition is that the following nice properties hold:

$(m_1+m_2)\otimes n=m_1\otimes n+m_2\otimes n$

$m\otimes (n_1+n_2)=m\otimes n_1+m\otimes n_2$

$mr\otimes n=m\otimes rn$

Recent News about MOE (Ministry of Education)

The following are some interesting news on Singapore’s MOE Ministry of Education. Readers interested may want to check them out:

Motivational: Kung Fu Panda 3 Song

The new Kung Fu Panda 3 song lyrics is out!

Very motivational lyrics. The title itself <Try> is very motivational.

“Only those who dare to fail greatly can ever achieve greatly.” ― Robert F. Kennedy

The full lyrics are as follows (includes some Chinese):

Try – 派伟俊/周杰伦
(《功夫熊猫3》电影全球主题曲)
中文词：方文山
英文词：冼佩瑾
曲：派伟俊
小派：You always have to do something
Just to show the world that you exist
So you try
You hope they’ll see
If on this brand new day you’ll look
On the bright side of the same old street
You will see
What you deserve
Jay：Let’s go
我说几华里我送别了过去
他们说人生的结局非常的戏剧
塞外羌笛孤城马蹄
在武侠的世界里谁与谁来为敌
合：La la la la la la la la la
黄沙里用竹笔写下的字叫勇气
Jay：You just have to try
To be who you are
And you ought to fly
Step into the light
小派：And soon you will find
Be yourself
Somewhere deep inside
There’s a universe right there waiting to be unlocked
The key lies in looking into yourself
Jay：Oh Try try try try
Just do what is right
You’ll fly so high
Let go of the brakes
Be who you are
Be yourself ’cause your power is on
合： When you believe in what you’ve got
You know you’re perfect just be who you are
小派：So they don’t see what you’re made of
But I like you and I know they’re wrong
Now it’s time
To show them what you got
Let the blue skies cheer you on
Embrace the wind we’ll ride along
You’re perfect when you’re who you are
Jay：这世界有些事有些人凭感觉
别管他旌旗密布遍野狼烟霜雪
那故事在穿越而我也在翻页
一行行做好准备敏锐而直接
合：La la la la la la la la la
爱不灭真实的一切废话全收回
小派：You just have to try
To be who you are
Jay：And you ought to fly
Step into the light
小派：And soon you will find
Be yourself
Somewhere deep inside
There’s a universe right there waiting to be unlocked
The key lies in looking into yourself
Jay：Oh Try try try try
Just do what is right
You’ll fly so high
Let go of the brakes
合： Be who you are
Be yourself ’cause your power is on
When you believe in what you’ve got
You know you’re perfect just be who you are
小派：You just have to try
To be who you are
Jay：And you ought to fly
Step into the light
小派：And soon you will find
Be yourself
Somewhere deep inside
There’s a universe right there waiting to be unlocked
The key lies in looking into yourself
Jay：Oh Try try try try
Just do what is right
You’ll fly so high
Let go of the brakes

合： Be who you are
Be yourself ’cause your power is on
When you believe in what you’ve got

Chinese Tuition

Chinese Tuition: http://chinesetuition88.com/

Chinese Tuition Singapore

新加坡华文补习老师

Tutor: Ms Gao (高老师）

Ms Gao is a patient tutor, and also effectively bilingual in both Chinese and English.

A native speaker of Mandarin, she speaks clearly with perfect accent and pronunciation. She is also well-versed in Chinese history, idioms and proverbs.

Ms Gao is able to teach Chinese at the Primary and Secondary school level. She will teach in an exam-oriented style, but will also try her best to make the lesson interesting for the student.

Ms Gao graduated from Huaqiao University, which is founded by late Chinese premier Zhou Enlai.

Contact:

Email: chinesetuition88@gmail.com

Website: http://chinesetuition88.com/

(Preferably looking for students staying in the West side of Singapore, e.g. Clementi / Dover / Jurong East / Boon Lay / Queenstown)

Image of Infinite Strip under Transformation w=1/z

Q: Find the image of the infinite strip $0<y<1/(2c)$ under the transformation $w=1/z$ .

(This question is taken from Complex Variables and Applications (Brown and Churchill))

Answer: $u^2+(v+c)^2>c^2$ , $v<0$ .

Solution:

We are using the standard notation $w=u+iv$ , $z=x+iy$ . From the equation $\displaystyle z=\frac{1}{w}=\frac{1}{u+iv}=\frac{u-iv}{u^2+v^2}$ , we can conclude that $\displaystyle y=\frac{-v}{u^2+v^2}$ .

With some algebra and completing the square, one can obtain $u^2+(v+\frac{1}{2y})^2=(\frac{1}{2y})^2$ , which is the equation of a circle centered at $(0,-\frac{1}{2y})$ with radius $1/2y$ . When $y=1/(2c)$ , the equation of the circle is $u^2+(v+c)^2=c^2$ . As $y$ gets smaller (closer to zero), the radius of the circle becomes larger, while still remaining tangent to the horizontal axis.

Thus, the image of the strip is $u^2+(v+c)^2>c^2$ , $v<0$ .

The Shape of Space

Just came across this book: The Shape of Space (Chapman & Hall/CRC Pure and Applied Mathematics). It is a very unique book, in the sense that it is aimed at high school students, but even a undergraduate or graduate student can benefit from it. It has a lot of diagrams, that are missing in most textbooks, presumably because it takes a lot of effort to draw a mathematical (3D) diagram.

It will be useful to students who want to learn more about topology. This book can be read casually, it is not like a textbook, yet it has substantial mathematical content.

Example of an illustration in the book:

illustration topology