# Ngô Quốc Anh

## February 10, 2013

### 2013 Tết Holiday

Filed under: Linh Tinh — Ngô Quốc Anh @ 14:24

Tết Holiday

Vietnamese New Year, more commonly known by its shortened name Tết or Tết Nguyên Đán, is the most important and popular holiday and festival in Vietnam. It is the Vietnamese New Year marking the arrival of spring based on the Chinese calendar, a lunisolar calendar. For those who do not know about Tết, please read an article in wikipedia for details.

At the first moment of the new year, I wish you a good health and prosperity all year round and thank you for your interest in my blog.

## January 29, 2013

Filed under: Linh Tinh — Tags: — Ngô Quốc Anh @ 5:37

I found this interesting note regarding to Matlab. In that note, they plotted the word HI using Matlab. Here I try to use MuPAD in order to get a slightly better picture.

As mentioned in the note, the full function we need to use is

$\displaystyle e^{-x^2-\frac{1}{2}y^2} \cos(4x) + e^{-3\big( (x+\frac{1}{2})^2+\frac{1}{2}y^2 \big)}.$

If you plot that full function, what you are going to have is the following picture

## January 22, 2013

### PhD Thesis Defense

Filed under: Linh Tinh, Luận Văn — Ngô Quốc Anh @ 15:46

I just passed my PhD defense on 18 Jan, 2013. Following is the front page of the slides I used during the defense.

The committee of my defense consists of

Since the thesis contains some unpublished results, I cannot provide the slides here but you can email me if interested.

Title page of the slides used in my PhD thesis defense

## March 10, 2012

### An integral of 1/(1+|x|^2) over the whole Euclidean space

Filed under: Giải Tích 2, Giải Tích 3, Linh Tinh — Ngô Quốc Anh @ 2:46

Sometimes, we need a precise value for following

$\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^\alpha }}}}, \quad \alpha>\frac{n}{2}.$

As such, I am going to calculate it and place the result here for future works.

In order to evaluate the above integral, we need to use the so-called co-area formula. We first write

$\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^\alpha }}}} = \int_0^{ + \infty } {\left( {\int_{\partial {B_0}(r)} {\frac{{dS}}{{{{(1 + {r^2})}^\alpha }}}} } \right)dr}.$

Note that

$\displaystyle \begin{gathered} \int_{\partial {B_0}(r)} {\frac{{dS}}{{{{(1 + {r^2})}^\alpha }}}} = \frac{1}{{{{(1 + {r^2})}^\alpha }}}\int_{\partial {B_0}(r)} {dS} \hfill \\ \qquad\qquad\qquad= \frac{1}{{{{(1 + {r^2})}^\alpha }}}\text{Area}({B_0}(r)) = \frac{{2{\pi ^{\frac{n}{2}}}}}{{\Gamma \left( {\frac{n}{2}} \right)}}\frac{{{r^{n - 1}}}}{{{{(1 + {r^2})}^\alpha }}}\end{gathered}.$

Therefore,

$\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^\alpha }}}} = \frac{{2{\pi ^{\frac{n}{2}}}}}{{\Gamma \left( {\frac{n}{2}} \right)}}\int_0^{ + \infty } {\frac{{{r^{n - 1}}}}{{{{(1 + {r^2})}^\alpha }}}} dr = {\pi ^{\frac{n}{2}}}\frac{{\Gamma \left( {\alpha - \frac{n}{2}} \right)}}{{\Gamma (\alpha )}}$

## April 8, 2011

### 100k hits

Filed under: Linh Tinh — Ngô Quốc Anh @ 20:52

Hello everyone,

This is a great news and I should share this immediately !

My blog has just gone through 100,000 hits at this moment . I take this opportunity to thank all of you for your interest in my blog, for your suggestion, for your beautiful comments, and for everything…

As can be seen, I have started using this wordpress blog since May, 2007. However, there was not much information here until the time I came to Singapore to pursue my PhD degree. That was August, 2008. During this period of time, I have learnt much and of course I have some spare time to write down things that I think it may be useful.

A bit for the future, I will keep writing things that I usually face during my mathematics career. Further than that, it totally depends on how much time I will have. Anyway, please do enjoy what we have here.

Again, thank you and have a nice day ,

Ngo Quoc Anh.

## April 1, 2011

### Several interesting limits from a paper by Chang-Qing-Yang

Recently, I have learnt from my friend, ZJ, the following result

Assume that $F:\mathbb R \to \mathbb R$ is absolutely integrable. Then

$\displaystyle\begin{gathered} \mathop {\lim }\limits_{t \to \pm \infty } {e^{2t}}\int_t^{ + \infty } {F(x){e^{ - 2x}}dx} = 0, \hfill \\ \mathop {\lim }\limits_{t \to \pm \infty } {e^{ - 2t}}\int_{ - \infty }^t {F(x){e^{ - 2x}}dx} = 0. \hfill \\ \end{gathered}$

The result seems reasonable by the following observation, for example, we consider the first identity when $t \to +\infty$. Then the factor

$\displaystyle\int_t^{ + \infty } {F(x){e^{ - 2x}}dx}$

decays faster then the exponent function $\exp (2t)$. This may be true, of course we need to prove mathematically, because the integrand contains the term $\exp (-2x)$ which turns out to be a good term since $x \geqslant t$. So here is the trick in order to solve such a problem.

## October 18, 2010

### 1/infinity = 0 is equivalent to 1/0=infinity?

Filed under: Các Bài Tập Nhỏ, Linh Tinh — Ngô Quốc Anh @ 12:05

It is now the time to discuss some funny thing. I just learn from GR class this morning a proof of the following statement

$\displaystyle \frac{1}{\infty}=0 \quad \Longleftrightarrow \quad \frac{1}{0}=\infty$.

Okay, let us start with the left hand side. By rotating 90 degrees counter-clockwise both sides of

$\displaystyle \frac{1}{\infty}=0$

we get

$\displaystyle -18=0$.

Now adding both sides by 8 we arrive at

$\displaystyle -10=8$.

Again, rotating 90 degrees clockwise both sides we reach to

$\displaystyle \frac{1}{0}=\infty$.

The reverse case can be treated similarly.

## April 9, 2010

### Kelvin transform: Laplacian

Filed under: Các Bài Tập Nhỏ, Linh Tinh, Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 0:01

For each point $x \ne 0$, denote $x=(x_1,...,x_n)$ and

$\displaystyle \xi = {x^\sharp } = \left( {\frac {{{x_1}}} {{{{\left| x \right|}^2}}},...,\frac {{{x_n}}} {{{{\left| x \right|}^2}}}} \right)$

is the inversion of  $x$ with respect to the unit sphere. We have the following identities

$\displaystyle\frac {{\partial {\xi _j}}} {{\partial {x_k}}} = \frac {1} {{{{\left| x \right|}^2}}}\left( {{\delta _{jk}} - 2\frac {{{x_j}{x_k}}} {{{{\left| x \right|}^2}}}} \right)$

and

$\displaystyle\sum\limits_{l = 1}^n {\frac {{\partial {\xi _l}}} {{\partial {x_j}}}\frac {{\partial {\xi _l}}} {{\partial {x_k}}}} = \frac {1} {{{{\left| x \right|}^4}}}{\delta _{jk}}$.

Thus,

$\displaystyle\sum\limits_{l = 1}^n {\frac {{\partial {x_l}}} {{\partial {\xi _j}}}\frac {{\partial {x_l}}} {{\partial {\xi _k}}}} = \frac {1} {{{{\left| \xi \right|}^4}}}{\delta _{jk}}$.

Next, think of $\xi$ as a system of orthogonal curvilinear coordinates for $x$, we deduce that the metric tensor of the Euclidean space in curvilinear coordinates

$\displaystyle {g_{j,k}}\left( {{\xi _j},{\xi _k}} \right) = \sum\limits_{l = 1}^n {\frac {{\partial {x_l}}} {{\partial {\xi _j}}}\frac {{\partial {x_l}}} {{\partial {\xi _k}}}} = \frac {1} {{{{\left| \xi \right|}^4}}}{\delta _{jk}}$.

This implies that the so-called Lame coefficients is

$\displaystyle {h_j} = \sqrt {{g_{j,j}}\left( {{\xi _j},{\xi _j}} \right)} = \frac {1} {{{{\left| \xi \right|}^2}}}$.

## January 13, 2010

### R-G: A few words about writing covariant derivative of tensors in a short form

Filed under: Linh Tinh, Nghiên Cứu Khoa Học, Riemannian geometry — Ngô Quốc Anh @ 14:21

Before going further, I would like to mention a convention in writing covariant derivative of tensors in a short form. What I mean is how to understand the following notation $\nabla_i g_{\alpha\beta}$ where $g$ is a Riemannian metric.

We all know that we can apply covariant derivative to a scalar function, for example, $\nabla_i f$ is nothing but the partial derivative with respect to $x^i$. However, the notation $\nabla_i g_{\alpha\beta}$ is a little bit different. For $g$, a Riemannian metric, which is also a $(0,2)$ tensor, in the full form, we can write

$\displaystyle g=g_{\alpha\beta}dx^\alpha \otimes dx^\beta$.

The sub indexes $\alpha\beta$ in the notation $\nabla_i g_{\alpha\beta}$ tell us that we are talking about the $\alpha\beta$-component of the covariant derivative of $(0,2)$ tensor $g$ with respect to the vector field $\frac{\partial}{\partial x^i}$.

If you have $\nabla_i h^{\alpha\beta}$, then you are working on some $(2,0)$-tensor $h$ of the form

$\displaystyle h=h^{\alpha\beta} \frac{\partial}{\partial x^\alpha} \otimes \frac{\partial}{\partial x^\beta}$.

We now go back to the case $\nabla_i g_{\alpha\beta}$. Precisely, one should write

$\displaystyle {\nabla _i}{g^{\alpha \beta }} = ({\nabla _{\frac{\partial }{{\partial {x^i}}}}}g)\left( {\frac{\partial }{{\partial {x^\alpha }}},\frac{\partial }{{\partial {x^\beta }}}} \right)$.

Since $\{dx^\alpha\}$ is a basis for the dual space of a space spanned by the basis $\{\frac{\partial}{\partial x^\beta}\}$, it is clear to see that the right hand side of the above convention is just the $\alpha\beta$-component, that is, the coefficient of the term

$\displaystyle {\nabla _i}{g^{\alpha \beta }}d{x^\alpha } \otimes d{x^\beta }$.

Now we show that $\nabla_i g^{\alpha\beta}=0$ whenever $g$ is a Riemannian metric. This is equivalent to show that every coefficients of $\nabla_i g$ equal to zero, in other word, $\nabla_i g=0$. Since $g$ is a $(0,2)$-tensor, then we can compute covariant derivative of $g$ as follows

$\displaystyle\begin{gathered}{\nabla _i}g = {\nabla _i}\left( {{g_{\alpha \beta }}d{x^\alpha } \otimes d{x^\beta }} \right) \hfill \\\qquad = \left( {\frac{\partial }{{\partial {x^i}}}{g_{\alpha \beta }}} \right)d{x^\alpha } \otimes d{x^\beta } + {g_{\alpha \beta }}{\nabla _i}\left( {d{x^\alpha } \otimes d{x^\beta }} \right) \hfill \\ \qquad= \left( {\frac{\partial }{{\partial {x^i}}}{g_{\alpha \beta }}} \right)d{x^\alpha } \otimes d{x^\beta } + {g_{\alpha \beta }}{\nabla _i}\left( {d{x^\alpha }} \right) \otimes d{x^\beta } + {g_{\alpha \beta }}d{x^\alpha } \otimes {\nabla _i}\left( {d{x^\beta }} \right) \hfill \\ \qquad= \left( {\frac{\partial }{{\partial {x^i}}}{g_{\alpha \beta }}} \right)d{x^\alpha } \otimes d{x^\beta } - \left( {{g_{\alpha \beta }}\Gamma _{ik}^\alpha+ {g_{\alpha \beta }}\Gamma _{ik}^\beta } \right)d{x^k} \otimes d{x^\beta } \hfill \\ \qquad= \left( {\frac{\partial }{{\partial {x^i}}}{g_{\alpha \beta }} - \Gamma _{i\alpha }^k{g_{k\beta }} - \Gamma _{i\beta }^k{g_{k\alpha }}} \right)d{x^\alpha } \otimes d{x^\beta }. \hfill \\ \end{gathered}$

Since $g$ is a metric connection, i.e.

$\displaystyle Xg\left( {Y,Z} \right) = g\left( {{\nabla _X}Y,Z} \right) + g\left( {Y,{\nabla _X}Z} \right)$

one has

$\displaystyle \frac{\partial }{{\partial {x^i}}}g\left( {\frac{\partial }{{\partial {x^\alpha }}},\frac{\partial }{{\partial {x^\beta }}}} \right) = g\left( {{\nabla _i}\frac{\partial }{{\partial {x^\alpha }}},\frac{\partial }{{\partial {x^\beta }}}} \right) + g\left( {\frac{\partial }{{\partial {x^\alpha }}},{\nabla _i}\frac{\partial }{{\partial {x^\beta }}}} \right)$.

Thus

$\displaystyle\frac{\partial }{{\partial {x^i}}}{g_{\alpha \beta }} = g\left( {\Gamma _{i\alpha }^k\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^\beta }}}} \right) + g\left( {\frac{\partial }{{\partial {x^\alpha }}},\Gamma _{i\beta }^k\frac{\partial }{{\partial {x^k}}}} \right) = \Gamma _{i\alpha }^k{g_{k\beta }} + \Gamma _{i\beta }^k{g_{k\alpha }}$.

In other word the $\alpha\beta$-component of $\nabla_i g$ equals to zero. At the last word, in order to calculate covariant derivative of some $(p,q)$-tensors, you need to use the following three properties

1. ${\nabla _X}\left( {fT} \right) = X(f)T + f{\nabla _X}T$.
2. ${\nabla _X}\left( {T \otimes Q} \right) = {\nabla _X}T \otimes Q + T \otimes {\nabla _X}Q$.
3. For $(1,0)$- or $(0,1)$-tensors:

$\displaystyle {\nabla _{\frac{\partial }{{\partial {x^i}}}}}\frac{\partial }{{\partial {x^j}}} = \Gamma _{ij}^k\frac{\partial }{{\partial {x^k}}}$

and

$\displaystyle {\nabla _{\frac{\partial }{{\partial {x^i}}}}}d{x^j}=-\Gamma _{ik}^jd{x^k}$.

## December 2, 2009

Filed under: Linh Tinh — Tags: — Ngô Quốc Anh @ 22:21

MuPAD was a Computer algebra system (CAS). Originally developed by the MuPAD research group at the University of Paderborn, Germany, it was developed by the company SciFace Software GmbH & Co. KG in cooperation with the MuPAD research group and partners from some other universities since 1997.

I found MuPAD a very cool tool for plotting graph of functions. I will show you in details some examples.

$f(x,y)=-(x^2+y^2)(\cos (3y) + \sin (7y))$.
f := plot::Function3d(exp(-x^2-y^2)*(cos(3*y)+sin(7*x)), x=-2..2, y=-2..2, Submesh=[1,1], FillColorFunction=((x, y, z) -> [(z+2)/4, 0.5, (2-z)/4])): plot(f)