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:
1. Redemption is valid from 15 Jan – 25 Jan or while stocks last. Limited Stocks only!
2. Only the hard copy of this email will be accepted.
3. Limited to 1 redemption per email address per name.
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.

View original post

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

Screen Shot 2016-01-12 at 8.26.51 PM

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.

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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

CW Approximation

A weak homotopy equivalence is a map $f:X\to Y$ that induces isomorphisms $\pi_n(X,x_0)\to\pi_n(Y,f(x_o))$ for all $n\geq 0$ and all choices of basepoint $x_0$ .

In other words, Whitehead’s theorem says that a weak homotopy equivalence between CW complexes is a homotopy equivalence. Just to recap, a map $f:X\to Y$ is said to be a homotopy equivalence if there exists a map $g:Y\to X$ such that $fg\cong id_Y$ and $gf\cong id_X$ . The spaces $X$ and $Y$ are called homotopy equivalent.

It turns out that for any space $X$ there exists a CW complex $Z$ and a weak homotopy equivalence $f:Z\to X$ . This map $f:Z\to X$ is called a CW approximation to $X$ .

Excision for Homotopy Groups

According to Hatcher (Chapter 4.2), the main difficulty of computing homotopy groups (versus homology groups) is the failure of the excision property. However, under certain conditions, excision does hold for homotopy groups:

Theorem (4.23): Let $X$ be a CW complex decomposed as the union of subcomplexes $A$ and $B$ with nonempty connected intersection $C=A\cap B$ . If $(A,C)$ is m-connected and $(B,C)$ is n-connected, $m,n\geq 0$ , then the map $\pi_i(A,C)\to\pi_i(X,B)$ induced by inclusion is an isomorphism for $i<m+n$ and a surjection for $i=m+n$ .

Miscellaneous Definitions

Suspension: Let $X$ be a space. The suspension $SX$ is the quotient of $X\times I$ obtained by collapsing $X\times\{0\}$ to one point and $X\times\{1\}$ to another point.

The definition of suspension is similar to that of the cone in the following way. The cone $CX$ is the union of all line segments joining points of $X$ to one external vertex. The suspension $SX$ is the union of all line segments joining points of $X$ to two external vertices.

The classical example is $X=S^n$ , when $SX=S^{n+1}$ with the two “suspension points” at the north and south poles of $S^{n+1}$ , the points $(0,\dots,0,\pm 1)$ .

Here are some graphical sketches of the case where $X$ is the 0-sphere and the 1 sphere respectively.

Linear Fractional Transformation (Mobius Transformation)

The transformation $w=\frac{az+b}{cz+d}$ , with $ad-bc\neq 0$ , and $a,b,c,d$ are complex constants, is called a linear fractional transformation, or Mobius transformation.

One key property of linear fractional transformations is that it transforms circles and lines into circles and lines.

Let us find the linear fractional transformation that maps the points $z_1=2$ , $z_2=i$ , $z_3=-2$ onto the points $w_1=1$ , $w_2=i$ , $w_3=-1$ . (Question taken from Complex Variables and Applications (Brown and Churchill))

Solution: $w=\frac{3z+2i}{iz+6}$

What we have to do is basically solve the three simultaneous equations arising from $w=\frac{az+b}{cz+d}$ , namely $1=\frac{2a+b}{2c+d}$ , $i=\frac{ia+b}{ic+d}$ and $-1=\frac{-2a+b}{-2c+d}$ .

Eventually we can have all the variables in terms of $c$ : $a=-3ic$ , $b=2c$ , $d=-6ic$ . Substituting back into the Mobius Transformation gives us the answer.

Donate to Singapore Charity @ Giving.sg

URL: https://www.giving.sg/

Just to introduce this website to readers who haven’t heard of it. Donations above $50 are tax deductible, and it also features ways to volunteer for the charity organisations. Do check it out!

Q: What is Giving.sg? (taken from their FAQ page)

A: Giving.sg is Singapore’s very own one-stop portal for empowering all of us on our Giving Journey, whether we are looking to help local non-profit organisations (NPOs) by giving our TIME, by general volunteering; our TALENT, by skills volunteering; or [our] TREASURE, by donations.

Giving.sg brings together its predecessors sggives.org and sgcares.org, both of which have helped raise over S$51 million for more than 350 [local] non-profits, as well as seen over 40,000 volunteers generously gift their time to these [local] non-profits.

News article featuring Giving.sg: http://www.straitstimes.com/singapore/online-platform-among-initiatives-to-promote-giving-in-singapore

Analytic Isomorphism (Complex Analysis)

An analytic isomorphism $f:U\to V$ (where $U$ and $V$ are open sets) is an analytic function such that there exists another analytic function $g:V\to U$ , satisfying $f\circ g=id_V$ and $g\circ f=id_U$ .

Interesting Blog Post on Mathematical Conversations

Source: http://www.theliberatedmathematician.com/2015/12/why-i-do-not-talk-about-math/

A honest opinion on the nature of mathematical conversations, by this blog post author Piper Harron. (Also see our previous blog post on her interesting PhD Thesis) Very interesting read, for those who are in the mathematical community.

Secondary Chinese Tuition

For those interested in Secondary Level (O Level) Chinese Tuition, do check out http://chinesetuition88.com/.

The tutor (Ms Gao) will be taking in a few more students in 2016 for Chinese Tuition. Do check out http://chinesetuition88.com/ for more info!

(1/2)! = (√π)/2

Math Online Tom Circle

Richard Feynman (Nobel Physicist) proved it in high school using a funny Calculus: “Differentiating under Integral” — is it legitimate to do so ? Of course it is by “The Fundamental Theorem of Calculus”

Note: We were thought in high school the “HOW” of calculating (such as integration and differentiation), but not the “WHY” (the Theorem behind). Richard Feynman was unique in exploring the WHY since high school, it helped later he was assigned by President Reagan to investigate the 1986 ‘Challenger’ disaster ?

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Cellular Approximation for Pairs

(This is Example 4.11 in Hatcher’s book).

Cellular Approximation for Pairs: Every map $f:(X,A)\to (Y,B)$ of CW pairs can be deformed through maps $(X,A)\to (Y,B)$ to a cellular map $g:(X,A)\to (Y,B)$ .

What “map of CW pairs” mean, is that $f$ is a map from $X$ to $Y$ , and the image of $A\subseteq X$ under $f$ is contained in $B$ . CW pair $(X,A)$ means that $X$ is a cell complex, and $A$ is a subcomplex.

First, we use the ordinary Cellular Approximation Theorem to deform the restriction $f:A\to B$ to be cellular. We then use the Homotopy Extension Property to extend this to a homotopy of $f$ on all of $X$ . Then, use Cellular Approximation Theorem again to deform the resulting map to be cellular staying stationary on $A$ .

We use this to prove a corollary: A CW pair $(X,A)$ is n-connected if all the cells in $X-A$ have dimension greater than $n$ . In particular the pair $(X,X^n)$ is n-connected, hence the inclusion $X^n\hookrightarrow X$ induces isomorphisms on $\pi_i$ for $i<n$ and a surjection on $\pi_n$ .

First we note that being n-connected means that the space is non-empty, path-connected, and the first n homotopy groups are trivial, i.e. $\pi_i(X)\cong 0$ for $1\leq i\leq n$ .

Proof: First, we apply cellular approximation to maps $(D^i,\partial D^i)\to (X,A)$ with $i\leq n$ , thus the map is homotopic to a cellular map of pairs $g$ . Since all the cells in $X-A$ have dimension greater than $n$ , the n-skeleton of $X$ must be inside $A$ . Therefore $g$ is homotopic to a map whose image is in $A$ , and thus it is 0 in the relative homotopy group $\pi_i(X,A)$ . This proves that the CW pair $(X,A)$ is n-connected. Note that 0-connected means path-connected.

Consider the long exact sequence of the pair $(X,X^n)$ :

$\dots\to\pi_n(X^n,x_0)\xrightarrow{i_*}\pi_n(X,x_0)\xrightarrow{j_*}\pi_n(X,X^n,x_0)\xrightarrow{\partial}\pi_{n-1}(X^n,x_0)\to\dots\to\pi_0(X,x_0)$

Since it is an exact sequence, the image of any map equals the kernel of the next. Thus, $\text{Im}(i_*)=\ker j_*=\pi_n(X,x_0)$ (since $\pi_n(X,X^n,x_0)=0$ ). Thus $i_*$ is surjective. Since $\pi_n(X,X^n,x_0)=0$ , the later terms in the long exact sequence are also 0, thus, the inclusion $X^n\hookrightarrow X$ induces isomorphisms on $\pi_i$ for $i<n$ , since the first n homotopy groups all vanish.

Piper Harron discusses her artistic and wonderful math Ph.D. thesis

This is the most “unique” PhD thesis I have ever seen. Very special, and humorous to read, and coming from the most elite institution Princeton, under the guidance of Fields Medalist Manjul Bhargava.

mathbabe

Piper Harron is a mathematician who is very happy to be here, and yes, is having a great time, despite the fact that she is standing alone awkwardly by the food table hoping nobody will talk to her.

Piper, would you care to write a mathbabe post describing your thesis, and yourself, and anything else you’d care to mention?

When Cathy (Cathy? mathbabe?) asked if I would like to write a mathbabe post describing my thesis, and myself, and anything else I’d care to mention, I said “sure!” because that is objectively the right answer. I then immediately plunged into despair.

Describe my thesis? My thesis is this thing that was initially going to be a grenade launched at my ex-prison, for better or for worse, and instead turned into some kind of positive seed bomb where flowers have sprouted beside the foundations I thought I wanted to crumble…

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2015 in review

The WordPress.com stats helper monkeys prepared a 2015 annual report for this blog.

Here’s an excerpt:

The Louvre Museum has 8.5 million visitors per year. This blog was viewed about 170,000 times in 2015. If it were an exhibit at the Louvre Museum, it would take about 7 days for that many people to see it.

Click here to see the complete report.

Rouche’s Theorem and Applications

This blog post is on Rouche’s Theorem and some applications, namely counting the number of zeroes in an annulus, and the fundamental theorem of algebra.

Rouche’s Theorem: Let $f(z)$ , $g(z)$ be holomorphic inside and on a simple closed contour $K$ , such that $|g(z)|<|f(z)|$ on $K$ . Then $f$ and $f+g$ have the same number of zeroes (counting multiplicities) inside $K$ .

Rouche’s Theorem is useful for scenarios like this: Determine the number of zeroes, counting multiplicities, of the polynomial $f(z)=2z^5-6z^2-z+1=0$ in the annulus $1\leq |z|\leq 2$ .

Solution:

Let $K_1$ be the unit circle $|z|=1$ . We have

$\begin{aligned}|2z^5-z+1|&\leq |2z^5|+|z|+|1|\\ &=2+1+1\\ &=4\\ &<6\\ &=|-6z^2| \end{aligned}$

on $K_1$ .

Since $-6z^2$ has 2 zeroes in $K_1$ , therefore $f$ has 2 zeroes inside $K_1$ , by Rouche’s Theorem.

Let $K_2$ be the circle $|z|=2$

$\begin{aligned} |-6z^2-z+1|&\leq |-6z^2|+|-z|+|1|\\ &=6(2^2)+2+1\\ &=27\\ &<64\\ &=|2z^5| \end{aligned}$

on $K_2$ . Therefore $f$ has 5 zeroes inside $K_2$ .

Therefore $f$ has 5-2=3 zeroes inside the annulus.

We do a computer check using Wolfram Alpha (http://www.wolframalpha.com/input/?i=2z%5E5-6z%5E2-z%2B1%3D0). The moduli of the five roots are (to 3 significant figures): 0.489, 0.335, 1.46, 1.45, 1.45. This confirms that 3 of the zeroes are in the given annulus.

Fundamental Theorem of Algebra Using Rouche’s Theorem

Rouche’s Theorem provides a rather short proof of the Fundamental Theorem of Algebra: Every degree n polynomial with complex coefficients has exactly n roots, counting multiplicities.

Proof: Let $f(z)=a_0+a_1z+a_2z^2+\dots+a_nz^n$ . Chose $R\gg 1$ sufficiently large so that on the circle $|z|=R$ ,

$\begin{aligned} |a_0+a_1z+a_2z^2+\dots+a_{n-1}z^{n-1}|&\leq|a_0|+|a_1|R+|a_2|R^2+\dots+|a_{n-1}|R^{n-1}\\ &<(\sum_{i=0}^{n-1}|a_i|)R^{n-1}\\ &<|a_n|R^n\\ &=|a_nz^n| \end{aligned}$

Since $a_nz^n$ has $n$ roots inside the circle, $f$ also has $n$ roots in the circle, by Rouche’s Theorem. Since $R$ can be arbitrarily large, this proves the Fundamental Theorem of Algebra.