Month: September 2016

Lebesgue’s Dominated Convergence Theorem for Convergence in Measure

Lebesgue’s Dominated Convergence Theorem for Convergence in Measure

If \{f_k\} satisfies f_k\xrightarrow{m}f on E and |f_k|\leq\phi\in L(E), then f\in L(E) and \int_E f_k\to\int_E f.

Proof

Let \{f_{k_j}\} be any subsequence of \{f_k\}. Then f_{k_j}\xrightarrow{m}f on E. Thus there is a subsequence f_{k_{j_l}}\to f a.e.\ in E. Clearly |f_{k_{j_l}}|\leq\phi\in L(E).

By the usual Lebesgue’s DCT, f\in L(E) and \int_E f_{k_{j_l}}\to\int_E f.

Since every subsequence of \{\int_E f_k\} has a further subsequence that converges to \int_E f, we have \int_E f_k\to\int_E f.

Basel Problem using Fourier Series

A very famous mathematical problem known as the “Basel Problem” is solved by Euler in 1734. Basically, it asks for the exact value of \sum_{n=1}^\infty\frac{1}{n^2}.

Three hundred years ago, this was considered a very hard problem and even famous mathematicians of the time like Leibniz, De Moivre, and the Bernoullis could not solve it.

Euler showed (using another method different from ours) that \displaystyle \sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}, bringing him great fame among the mathematical community. It is a beautiful equation; it is surprising that the constant \pi, usually related to circles, appears here.

Squaring the Fourier sine series

Assume that \displaystyle f(x)=\sum_{n=1}^\infty b_n\sin nx.

Then squaring this series formally,
\begin{aligned} (f(x))^2&=(\sum_{n=1}^\infty b_n\sin nx)^2\\ &=\sum_{n=1}^\infty b_n^2\sin^2 nx+\sum_{n\neq m}b_nb_m\sin nx\sin mx. \end{aligned}

To see why the above hold, see the following concrete example:
\begin{aligned} (a_1+a_2+a_3)^2&=(a_1^2+a_2^2+a_3^2)+(a_1a_2+a_1a_3+a_2a_1+a_2a_3+a_3a_1+a_3a_2)\\ &=\sum_{n=1}^3 a_n^2+\sum_{n\neq m}a_na_m. \end{aligned}

Integrate term by term

We assume that term by term integration is valid.
\displaystyle \frac 1\pi\int_{-\pi}^\pi (f(x))^2\,dx=\frac 1\pi\int_{-\pi}^{\pi}\sum_{n=1}^\infty b_n^2\sin^2{nx}\,dx+\frac{1}{\pi}\int_{-\pi}^\pi\sum_{n\neq m}b_nb_m\sin nx\sin mx\,dx.

Recall that \displaystyle \int_{-\pi}^\pi \sin nx\sin mx\,dx=\begin{cases}0 &\text{if }n\neq m\\ \pi &\text{if }n=m \end{cases}.

So
\begin{aligned} \frac 1\pi\int_{-\pi}^{\pi}\sum_{n=1}^\infty b_n^2\sin^2{nx}\,dx&=\frac 1\pi\sum_{n=1}^\infty b_n^2(\int_{-\pi}^\pi\sin^2 nx\,dx)\\ &=\frac 1\pi\sum_{n=1}^\infty b_n^2 (\pi)\\ &=\sum_{n=1}^\infty (b_n)^2. \end{aligned}

Similarly
\begin{aligned} \frac{1}{\pi}\int_{-\pi}^\pi\sum_{n\neq m}b_nb_m\sin nx\sin mx\,dx&=\frac 1\pi\sum_{n\neq m}b_nb_m(\int_{-\pi}^{\pi}\sin nx\sin mx\,dx)\\ &=\frac 1\pi\sum_{n\neq m}b_nb_m(0)\\ &=0. \end{aligned}

So \displaystyle \frac 1\pi\int_{-\pi}^\pi (f(x))^2\,dx=\sum_{n=1}^\infty (b_n)^2. (Parseval’s Identity)

Apply Parseval’s Identity to f(x)=x

By Parseval’s identity,
\displaystyle \frac{1}{\pi}\int_{-\pi}^\pi x^2\,dx=\sum_{n=1}^\infty(\frac{2(-1)^{n+1}}{n})^2.

Simplifying, we get \displaystyle \frac 1\pi\cdot\left[\frac{x^3}{3}\right]_{-\pi}^\pi=\sum_{n=1}^\infty\frac{4}{n^2}.
\begin{aligned} \frac 1\pi(\frac{2\pi^3}{3})&=\sum_{n=1}^\infty \frac{4}{n^2}\\ \frac{\pi^2}{6}&=\sum_{n=1}^\infty\frac{1}{n^2}. \end{aligned}

China Eastern website not working

Currently, all versions of China Eastern Airlines 东方航空 websites (e.g. http://sg.ceair.com/, hk.ceair, etc) are not working.

I tried searching for a ticket in December and an error message popped out: We apologize that there are insufficient seats on ## segment of your searched flight. Please change the search options. Thank you for your cooperation!

I called the customer service and they confirmed that it is an error (their system shows that there are indeed still plenty of seats). Hope they fix it soon.

Circle in Different Representations

tomcircle's avatarMath Online Tom Circle

Affine Line: $latex {mathbb {A}^1}&s=3$

Six Representations of a Circle: $latex {mathbb {S}^1}&s=3$
1) Euclidean Geometry
Unit Circle : $latex x^2 + y^2 = 1$

2) Curve:
Transcendental Parameterization :
$latex boxed { e(theta) = (cos theta, sin theta) qquad
0 leq theta leq 2pi }&fg=aa0000&s=3
$

Rational Parameterisation :
$latex boxed {
e(h) = left(frac {1-h^2} {1+h^2} : , : frac {2h} {1+h^2}right) quad text { h any number or } infty
} &fg=aa0000&s=2
$

image

3) Affine Plane $latex {mathbb {A}^2}&s=3$
1-dim sub-Space = Lines thru Origin

image
4) Polygonal Representation

5) Identifying Intervals: (closed loop)
image

6) $latex text {Translation } (tau, {tau}^{-1}) text { on a Line } $ $latex {mathbb {A}^1}&s=3$

$latex boxed {
{mathbb {S}^1} = {mathbb {A}^1 } Big/ { langle tau , {tau}^{-1} rangle}
}&fg=aa0000&s=3
$
$latex {mathbb {S}^1} = text { Space of all orbits} $

image

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Inspirational Chinese Phrase 宠辱不惊

Source: http://www.ypzihua.com/product-2793.html

chongrubujing

Beautiful calligraphy and meaningful words.

Chinese Characters: 宠辱不惊 闲看庭前花开花落 去留无意 漫随天外云卷云舒

Translation: “Don’t be disturbed by fortune or misfortune. Be relaxed no matter how flowers bloom and wilt. To be or not to be needs no hard decision. Take it natural no matter how clouds flow high and low.” (Source: http://www.en84.com/dianji/minyan/201410/00015499.html)

Generalized Lebesgue Dominated Convergence Theorem Proof

This key theorem showcases the full power of Lebesgue Integration Theory.

Generalized Lebesgue Dominated Convergence Theorem

Let \{f_k\} and \{\phi_k\} be sequences of measurable functions on E satisfying f_k\to f a.e. in E, \phi_k\to \phi a.e. in E, and |f_k|\leq\phi_k a.e. in E. If \phi\in L(E) and \int_E \phi_k\to\int_E \phi, then \int_E |f_k-f|\to 0.

Proof

We have |f_k-f|\leq|f_k|+|f|\leq\phi_k+\phi. Applying Fatou’s lemma to the non-negative sequence \displaystyle h_k=\phi_k+\phi-|f_k-f|, we get \displaystyle 2\int_E\phi\leq\liminf_{k\to\infty}\int_E (\phi_k+\phi-|f_k-f|).
That is, \displaystyle 2\int_E \phi\leq2\int_E\phi-\limsup_{k\to\infty}\int_E |f_k-f|.

Since \int_E\phi<\infty, we get \limsup_{k\to\infty}\int_E |f_k-f|\leq 0. Since \liminf_{k\to\infty}\int_E |f_k-f|\geq 0, this implies \lim_{k\to\infty}\int_E |f_k-f|=0.

Simplicial Complex

tomcircle's avatarMath Online Tom Circle

Simplices:
0-dim (Point) $latex triangle_0 $
1-dim (Line) $latex triangle_1 $
2-dim (Triangle) $latex triangle_2 $
3-dim (Tetrahedron) $latex triangle_3 $

Simplicial Complex: built by various Simplices under some rules.

Definitions of Simplex (S)
Face
Orientation
Boundary ($latex delta $)
$latex displaystyle boxed {
delta(S) = sum_{i=0}^{n} (-1)^i (v_0 …hat v_i …v_n)}&fg=aa0000&s=3
$

image

Theorem: $latex boxed { delta ^2 (S) = 0}&fg=00bb00&s=4 $

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Why We Should Stop Grading Students on a Curve

A very nice article against the philosophy of the bell curve, which is a prominent feature of examinations all over the world, including Singapore. I am sure that when Gauss invented the bell curve, he didn’t intend it to be used for examinations!

Source: http://www.nytimes.com/2016/09/11/opinion/sunday/why-we-should-stop-grading-students-on-a-curve.html?_r=0

Excerpts:

The goal is to fight grade inflation, but the forced curve suffers from two serious flaws. One: It arbitrarily limits the number of students who can excel. If your forced curve allows for only seven A’s, but 10 students have mastered the material, three of them will be unfairly punished. (I’ve found a huge variation in overall performance among the classes I teach.)

The more important argument against grade curves is that they create an atmosphere that’s toxic by pitting students against one another. At best, it creates a hypercompetitive culture, and at worst, it sends students the message that the world is a zero-sum game: Your success means my failure.

Exhibit B: I spent a decade studying the careers of “takers,” who aim to come out ahead, and “givers,” who enjoy helping others. In the short run, across jobs in engineering, medicine and sales, the takers were more successful. But as months turned into years, the givers consistently achieved better results.

The results: Their average scores were 2 percent higher than the previous year’s, and not because of the bonus points. We’ve long knownthat one of the best ways to learn something is to teach it. In fact, evidence suggests that this is one of the reasons that firstborns tend to slightly outperform younger siblings on grades and intelligence tests: Firstborns benefit from educating their younger siblings. The psychologists Robert Zajonc and Patricia Mullally noted in a review of the evidence that “the teacher gains more than the learner in the process of teaching.”

Finite group generated by two elements of order 2 is isomorphic to Dihedral Group

Suppose G=\langle s,t\rangle where both s and t has order 2. Prove that G is isomorphic to D_{2m} for some integer m.

Note that G=\langle st, t\rangle since (st)t=s. Since G is finite, st has a finite order, say m, so that (st)^m=1_G. We also have [(st)t]^2=s^2=1.

We claim that there are no other relations, other than (st)^m=t^2=[(st)t]^2=1.

Suppose to the contrary sts=1. Then sstss=ss, i.e. t=1, a contradiction. Similarly if ststs=1, tsststsst=tsst implies s=1, a contradiction. Inductively, (st)^ks\neq 1 and (ts)^kt\neq 1 for any k\geq 1.

Thus \displaystyle G\cong D_{2m}=\langle a,b|a^m=b^2=(ab)^2=1\rangle.

In case you haven’t heard what’s going on in Leicester …

Math teachers / students / Math lovers do sign this petition to stop Leicester university from cutting 20% of their math researchers/lecturers. #mathisimportant

gowers's avatarGowers's Weblog

Strangely, this is my second post about Leicester in just a few months, but it’s about something a lot more depressing than the football team’s fairytale winning of the Premier League (but let me quickly offer my congratulations to them for winning their first Champions League match — I won’t offer advice about whether they are worth betting on to win that competition too). News has just filtered through to me that the mathematics department is facing compulsory redundancies.

The structure of the story is wearily familiar after what happened with USS pensions. The authorities declare that there is a financial crisis, and that painful changes are necessary. They offer a consultation. In the consultation their arguments appear to be thoroughly refuted. But this is ignored and the changes go ahead.

Here is a brief summary of the painful changes that are proposed for the Leicester mathematics department. There are…

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Java Family

tomcircle's avatarMath Online Tom Circle

YearVerCode NameDescription
19951.0 Java Applets
1977 1.1JavaEvent, Beans, Internationalization
Dec 19781.2Java 2J2SE, J2EE, J2ME, Java Card
2000 1.3Java 2J2SE 1.3
2002 1.4Java 2J2SE 1.4
2004 1.5Java 5J2SE 1.5
Nov 2006 6Java 6Open-Source Java SE 6. “Multithreading” by Doug Lea
May 2007OpenJDK free software
2010Oracle acquired Sun
Jul 20117Java 7“Dolphin”
Mar 20148Java 8Lambda Function

Javac: Java Compiler

Java Distributions:

1. JDK (Java Developer Kit )
◇JRE & Javac & tools

2. JRE (Java Runtime Environment)
◇ JVM & core class libraries
◇ Windows / Mac / Linux

Java is Object-Oriented Programming (OOP):
1. Class
public class Employee {

public int age;
public double salary;

public Employee () { [<– constructor with no arg]
}

public Employee (int ageValue, double salaryValue) { [<- constructor with args]

age…

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Leibniz Integral Rule (Differentiating under Integral) + Proof

“Differentiating under the Integral” is a useful trick, and here we describe and prove a sufficient condition where we can use the trick. This is the Measure-Theoretic version, which is more general than the usual version stated in calculus books.

Let X be an open subset of \mathbb{R}, and \Omega be a measure space. Suppose f:X\times\Omega\to\mathbb{R} satisfies the following conditions:
1) f(x,\omega) is a Lebesgue-integrable function of \omega for each x\in X.
2) For almost all w\in\Omega, the derivative \frac{\partial f}{\partial x}(x,\omega) exists for all x\in X.
3) There is an integrable function \Theta: \Omega\to\mathbb{R} such that \displaystyle \left|\frac{\partial f}{\partial x}(x,\omega)\right|\leq\Theta(\omega) for all x\in X.

Then for all x\in X, \displaystyle \frac{d}{dx}\int_\Omega f(x,\omega)\,d\omega=\int_\Omega\frac{\partial}{\partial x} f(x,\omega)\,d\omega.

Proof:
By definition, \displaystyle \frac{\partial f}{\partial x}(x,\omega)=\lim_{h\to 0}\frac{f(x+h,\omega)-f(x,\omega)}{h}.

Let h_n be a sequence tending to 0, and define \displaystyle \phi_n(x,\omega)=\frac{f(x+h_n,\omega)-f(x,\omega)}{h_n}.

It follows that \displaystyle \frac{\partial f}{\partial x}(x,\omega)=\lim_{n\to\infty}\phi_n(x,\omega) is measurable.

Using the Mean Value Theorem, we have \displaystyle |\phi_n(x,\omega)|\leq\sup_{x\in X}|\frac{\partial f}{\partial x}(x,\omega)|\leq\Theta(w) for each x\in X.

Thus for each x\in X, by the Dominated Convergence Theorem, we have \displaystyle \lim_{n\to\infty}\int_\Omega \phi_n(x,\omega)\,d\omega=\int_\Omega\lim_{n\to\infty}\phi_n(x,\omega)\,dw which implies \displaystyle \lim_{h_n\to 0}\frac{\int_\Omega f(x+h_n,\omega)\,d\omega-\int_\Omega f(x,\omega)\,d\omega}{h_n}=\int_\Omega \frac{\partial f}{\partial x}(x,\omega)\,d\omega.

That is, \displaystyle \frac{d}{dx}\int_\Omega f(x,\omega)\,d\omega=\int_\Omega \frac{\partial}{\partial x}f(x,\omega)\,d\omega.

Habitica: Productivity that Grants XP

Very interesting productivity app that resembles a game. Do check it out!

Matthew Bowman's avatarThe Catholic Geeks

A few months ago, I noticed someone in one of my Facebook groups posting about an interesting app called Habitica. It’s one of a host of time-management and productivity-increasing applications, both web- and mobile-based. What sets it apart, however, is that it turns your efforts at organizing your life into a game. Specifically, it turns your life into something reminiscent of a classic, pixelated, 8-bit RPG.

So no, the title of this post is not a metaphor. You literally get XP (and gold) for doing your tasks in real life. 

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Three Properties of Galois Correspondence

The Fundamental Theorem of Galois Theory states that:

Given a field extension E/F that is finite and Galois, there is a one-to-one correspondence between its intermediate fields and subgroups of its Galois group.
1) H\leftrightarrow E^H where H\leq\text{Gal}(E/F) and E^H is the corresponding fixed field (the set of those elements in E which are fixed by every automorphism in H).
2) K\leftrightarrow\text{Aut}(E/K) where K is an intermediate field of E/F and \text{Aut}(E/K) is the set of those automorphisms in \text{Gal}(E/F) which fix every element of K.

This correspondence is a one-to-one correspondence if and only if E/F is a Galois extension.

Three Properties of the Galois Correspondence

  1. It is inclusing-reversing. The inclusion of subgroups H_1\subseteq H_2 holds iff the inclusion of fields E^{H_2}\subseteq E^{H_1} holds.
  2. If H is a subgroup of \text{Gal}(E/F), then |H|=[E:E^H] and |\text{Gal}(E/F)/H|=[E^H:F].
  3. The field E^H is a normal extension of F (or equivalently, Galois extension, since any subextension of a separable extension is separable) iff H is a normal subgroup of \text{Gal}(E/F).

3 Zika Methods that do NOT work #Zika

As the Zika/Dengue virus has spread to Singapore, I have been researching (in my spare time) on ways to prevent Zika/Dengue. The following are some interesting ways that however eventually do not work (or worse, attract more mosquitoes), so do not waste your money trying these!

1) Bug Zappers that use Ultraviolet Light as a Lure

In theory, bug zappers that use electricity to kill mosquitoes sound like a great idea. However, the problem comes from the fact that bug zappers use Ultraviolet Light (UV) to attract insects. Mosquitoes are attracted by Carbon Dioxide, not UV light, so you will end up killing 99% other insects, some of which are beneficial insects. See http://insects.about.com/od/StingingBitingInsects/a/Do-Bug-Zappers-Kill-Mosquitoes.htm.

2) Ultrasonic Buzzing devices that sound like Male Mosquitoes, since Female Mosquitoes with fertilized eggs actively avoid Male Mosquitoes

This seems to be complete BS. See http://www.mosquitoreviews.com/ultrasonic-mosquito-app.html

3) Pitcher Plants to eat Mosquitoes

This sounds like a genius idea at first, since Pitcher Plants eat insects, and grow in tropical climates, exactly where Aedes Mosquitoes live. The problem is mosquito larvae can survive very well in Pitcher Plants.

4) P.S. If you have a pond with large koi fish in it, you may want to add smaller fish, since koi fish (and other large fish in general) are known not to eat mosquito larvae. “Koi may be beautiful, but they are generally too large to prey on mosquito larvae and are known for their mellow nature.”