Ngô Quốc Anh

Tháng Chín 13, 2009

QE in Department of Mathematics, National University of Singapore, August 2009

Chuyên mục: Giải Tích Cổ Điển, Giải Tích Hiện Đại, Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 14:53

I have just passed QE held in August 2009 for my first attendance, I hereby show you the analysis paper

Question 1 [10 marks]. Suppose $f$ and $g$ are both measurable functions on the interval $(0,1)$ such that for all $t \in \mathbb R^1$

$| \{ x \in(0,1):f(x)\geq t\}| = |\{ x\in (0,1):g(x)\geq t\}|$

Assume that $f$ and $g$ both are monotone decreasing and continuous from left. Can you conclude that $f(x)=g(x)$ for all $x \in (0,1)$ ? Give the reason to support your answer.

Question 2 [10 marks]. Compute the volume of the region bounded by

${\left( {{a_{11}}x + {a_{12}}y + {a_{13}}z} \right)^2} + {\left( {{a_{21}}x + {a_{22}}y + {a_{23}}z} \right)^2} + {\left( {{a_{31}}x + {a_{32}}y + {a_{33}}z} \right)^2} = 1$

where the determinant of the $3 \times 3$ matrix $(a_{ij})$ is NOT equal to zero.

Question 3 [10 marks]. Let $D$ be a measureable set in $\mathbb R^n$ with finite measure. Suppose $\phi(x,t)$ is a real valued continuous function on $D \times \mathbb R^1$ such that for almost every $x \in D$ , $\phi(x,t)$ is a continuous function of $t$ and for every real number $t$ , $\phi(x,t)$ is measurable function of $x$ . If $\{f_n\}$ is a sequence of measurable functions on $D$ that converges to $f$ in measure, show that $\{\phi(x,f_n(x))\}$ converges to $\phi(x,f(x))$ in measure.

Question 4 [10 marks]. Find the function

$I\left( y \right) =\displaystyle\int\limits_0^\infty {{e^{ - a{x^2}}}\cos \left( {yx} \right)dx}$

if $a>0$ is a constant. Justify your answer.

Question 5 [10 marks]. Compute the intergal

$\displaystyle\int\limits_0^\pi {\frac{{x\sin x}}{{1 + {a^2} - 2a\cos x}}dx}$

where $a>0$ is a constant.

Question 6 [10 marks]. Supposet $f(z)$ is a holomorphic function on the complex plane $\mathbb C$ . If $f$ locally keeps the area invariant, what will the function $f$ be?

Question 7 [10 marks]. Is there an analytic function $f$ on $\Delta$ (unit disk in the complex plane with center $0$ ) such that $|f(z)|<1$ for $|z|<1$ with $f(0)=\frac{1}{2}$ and $f'(0)=\frac{3}{4}$ ? If so, find such an $f$ . Is it unique?

Question 8 [10 marks]. Let $m<n$ be two positive integers and $\Omega$ and $G$ be open subsets in $\mathbb R^n$ and $\mathbb R^m$ respectively. Does there exist a map $f :\Omega \to G$ such that $f$ and the inverse of $f$ are both $C^1$ ? Justify your answer.

Question 9 [10 marks]. Is there a square integrable function $f$ on $[0,\pi]$ such that both inequalities

$\displaystyle\int\limits_0^\pi{{{\left( {f\left( x \right)-\sin x} \right)}^2}dx}\leq\frac{4}{9}$

and

$\displaystyle\int\limits_0^\pi{{{\left( {f\left( x \right)-\cos x} \right)}^2}dx}\leq\frac{1}{9}$

hold? Justify your answer.

Question 10 [10 marks]. Let $\alpha_k$ for $k=1,2,...,n$ be $n$ real numbers such that $0<\alpha_k<\pi$ for any $k$ . Define

$\alpha=\displaystyle\frac{1}{n}\sum\limits_{k=1}^{n}{\alpha_k}.$

Show that

$\displaystyle{\left( {\prod\limits_{k = 1}^n {\frac{{\sin {\alpha _k}}}{{{\alpha _k}}}} } \right)^{\frac{1}{n}}} \leq \frac{{\sin \alpha }}{\alpha }.$

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Tháng tám 29, 2009

On a polynomials of degree n, having all its zeros in the unit dis

Chuyên mục: Các Bài Tập Nhỏ, Giải tích 7 (MA4247), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 21:07

I found a very useful inequality involving a polynomials of degree $n$ , having all its zeros in the unit disk from the following paper, doi:10.1016/j.jmaa.2009.07.049, published in J. Math. Anal. Appl.

Statement. If $P(z)$ is a polynomial of degree $n$ , having all its zeros in the disk $|z| \leq 1$ , then

$\left| {zP'(z)} \right| \geq \displaystyle\frac{n} {2}\left| {P(z)} \right|$

for $|z|=1$ .

Proof. Since all the zeros of $P(z)$ lie in $latex|z| \leq 1$. Hence if $z_1, z_2,...,z_n$ are the zeros of $P(z)$ , then $|z_j| \leq 1$ for all $j =1,2,...,n$ . Clearly,

$\displaystyle\Re \frac{{{e^{i\theta }}P'\left( {{e^{i\theta }}} \right)}}{{P\left( {{e^{i\theta }}} \right)}} = \sum\limits_{j = 1}^n {\Re \frac{{{e^{i\theta }}}}{{{e^{i\theta }} - {z_j}}}} ,$

for every point $e^{i\theta}$ , $0 \leq \theta < 2 \pi$ which is not a zero of $P(z)$ . Note that

$\displaystyle\sum\limits_{j = 1}^n {\Re \frac{e^{i\theta }}{e^{i\theta } - z_j}}\geq\sum\limits_{j = 1}^n{\frac{1}{2}}=\frac{n}{2}.$

This implies

$\displaystyle\left| {\frac{{{e^{i\theta }}P'\left( {{e^{i\theta }}} \right)}}{{P\left( {{e^{i\theta }}} \right)}}} \right| \geq \Re \frac{{{e^{i\theta }}P'\left( {{e^{i\theta }}} \right)}}{{P\left( {{e^{i\theta }}} \right)}} \geq \frac{n}{2},$

for every point $e^{i \pi}$ , $0 \leq \theta < 2\pi$ . Hence

$\left| {zP'(z)} \right| \geq \displaystyle \frac{n}{2}\left| {P(z)} \right|$

for $|z|=1$ and this completes the proof.

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Tháng tám 18, 2009

An other limit supremum of sin function

Chuyên mục: Các Bài Tập Nhỏ, Giải Tích 1, Giải Tích 6 (MA5205), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 12:35

In the topic we showed that for any irrational $\alpha$ the limit $\mathop {\lim }\limits_{n \to \infty } \sin \left( {n\alpha \pi } \right)$ does not exist. In this topic, we consider the following limit

$\mathop {\overline {\lim } }\limits_{n \to \infty } \sin \left( {nx} \right) .$

To be precise, we prove that

$\mathop {\overline {\lim } }\limits_{n \to \infty } \sin \left( {nx} \right) = 1$

for almost every $x \in [0,2\pi]$ .

Solution. Let

$A = \left\{ {x \in \left( {0,2\pi } \right): \frac{x} {\pi } \notin \mathbb{Q}} \right\}.$

Then is a measurable set of measure $2\pi$ . Moreover, for any $x \in A$ ,

$\mathop {\overline {\lim } }\limits_{n \to \infty } \sin \left( {nx} \right) = 1.$

Indeed for any $x \in A$ , since

$\left\{ {k\frac{x} {\pi } - 2l: l \in \mathbb{Z}} \right\}$

is dense subgroup of $\mathbb R$ there are sequences $\{k_n\}$ and $\{l_n\}$ of $\mathbb Z$ such that

$\mathop {\lim }\limits_{n \to \infty } \left( {{k_n}\frac{x} {\pi } - {l_n}} \right) = \frac{1} {2}.$

Since

$\frac{1} {2} \notin \left\{ {k\frac{x} {\pi } - 2l: k,l \in \mathbb{Z}} \right\}$

$\{k_n\}$ admits a subsequence $\{k'_n\}$ either increasing to $+\infty$ or decreasing to $-\infty$ . If $\mathop {\lim }\limits_{n \to \infty } {{k'}_n} = + \infty$ then

$\mathop {\lim }\limits_{n \to \infty } \sin \left( {{{k'}_n}x} \right) = \mathop {\lim }\limits_{n \to \infty } \sin \left( {...$

Otherwise $\mathop {\lim }\limits_{n \to \infty } \left( { - 3{{k'}_n}} \right) = + \infty$ and

$\mathop {\lim }\limits_{n \to \infty } \sin \left( { - 3{{k'}_n}x} \right) = \mathop {\lim }\limits_{n \to \infty } \sin \lef...$

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Tháng tám 17, 2009

The use of equivalence in measure

Chuyên mục: Các Bài Tập Nhỏ, Giải Tích 6 (MA5205), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 14:19

In mathematics, and specifically in measure theory, equivalence is a notion of two measures being “the same”. Two measures are equivalent if they have the same null sets.

Definition. Let $(X, \Sigma)$ be a measurable space, and let $\mu, \nu : \Sigma \to [0, +\infty]$ be two measures. Then $\mu$ is said to be equivalent to $\nu$ if

$\mu (A) = 0 \iff \nu (A) = 0$

for measurable sets in $\Sigma$ , i.e. the two measures have precisely the same null sets. Equivalence is often denoted $\displaystyle{\mu \sim \nu}$ or $\mu \approx \nu$ .

In terms of absolute continuity of measures, two measures are equivalent if and only if each is absolutely continuous with respect to the other:

$\mu \sim \nu \iff \mu \ll \nu \ll \mu.$

Equivalence of measures is an equivalence relation on the set of all measures $\Sigma \to [0, +\infty]$ .

Examples.

Gaussian measure and Lebesgue measure on the real line are equivalent to one another.
Lebesgue measure and Dirac measure on the real line are inequivalent.

Application. Let $\mu$ be a finite measure on $\mathbb R$ , and define

$f\left( x \right) = \int\limits_{ - \infty }^{ + \infty } {\frac{{\ln \left| {x - t} \right|}} {{\sqrt {\left| {x - t} \right...$

Show that f(x) is finite a.e. with respect to the Lebesgue measure on $\mathbb R$ .

Proof. Let

$g\left( x \right) = \left\{ \begin{gathered} \frac{{\ln \left| x \right|}} {{\sqrt {\left| x \right|} }},x \ne 0, \hfill \\...$

then $g \in L^1(\mathbb R, d\nu)$ where

$d\nu \left( x \right) = \frac{{dx}} {{1 + {x^2}}}$

and

$f\left( x \right) = \int_{ - \infty }^{ + \infty } {g\left( {x - t} \right)d\mu \left( t \right)} ,x \in \mathbb{R}.$

Clearly

$\int_{ - \infty }^{ + \infty } {\left| {f\left( x \right)} \right|d\mu \left( x \right)} \leqslant \int_{ - \infty }^{ + \in...$

and by Fubini’s Theorem

$\int_{ - \infty }^{ + \infty } {\left( {\int_{ - \infty }^{ + \infty } {\left| {g\left( {x - t} \right)} \right|d\mu \left( t...$

then

$\int_{ - \infty }^{ + \infty } {\left| {f\left( x \right)} \right|d\mu \left( x \right)} \leqslant \int_{ - \infty }^{ + \in...$

Since

$\begin{gathered} \int\limits_{ - \infty }^{ + \infty } {\left( {\int\limits_{ - \infty }^{ + \infty } {\left| {g\left( {x -...$

Thus the following function

$x \mapsto \int\limits_{ - \infty }^{ + \infty } {\left| {g\left( {x - t} \right)} \right|d\mu \left( t \right)}$

finite a.e. with respect to the measure $\nu$ . The conclusion follows from that the measure $\nu$ and the Lebesgue measure are equivalent.

Source: http://en.wikipedia.org/wiki/Equivalence_(measure_theory)

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Tháng tám 16, 2009

On the stability of the Runge-Kutta 4 (RK4)

Chuyên mục: Các Bài Tập Nhỏ, Giải tích 9 (MA5265), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 14:11

In the literature, the so-called RK4 is given as following: we first define the following coefficients

$\begin{gathered} {K_1} = f\left( {{t_i},{y_i}} \right), \hfill \\ {K_2} = f\left( {{t_i} + \frac{1} {2}\Delta t,{y_i} + \...$

then

${y_{i + 1}} = {y_i} + \frac{{\Delta t}} {6}\left( {{K_1} + 2{K_2} + 2{K_3} + {K_4}} \right).$

This is the most important iterative method for the approximation of solutions of ordinary differential equations

$y' = f(t, y), \quad y(t_0) = y_0.$

This technique was developed around 1900 by the German mathematicians C. Runge and M.W. Kutta.

In order to study its stability, we use the model problem

$y' = \lambda y, \quad \Re \lambda < 0.$

In other words, we replace f(t,y) by $\lambda y$ . Then the stability condition for time step $\Delta$ comes from the following condition

$\left| \frac{y_{i+1}}{y_i} \right| \leq 1.$

Applying the above discussion to RK4 method, we see that

$\begin{gathered} {K_4} = \lambda \left( {{y_i} + \Delta t{K_3}} \right), \hfill \\ {K_3} = \lambda \left( {{y_i} + \frac{...$

which implies

$\begin{gathered} {K_4} = \lambda \left( {{y_i} + \Delta t\lambda \left( {{y_i} + \frac{1} {2}\Delta t\lambda \left( {{y_i} ...$

which yields

${y_{i + 1}} = {y_i} + \frac{{\Delta t}} {6}\left( {\lambda {y_i} + 2\lambda {y_i}\left( {1 + \frac{1} {2}\Delta t\lambda } \r...$

Thus

$\frac{{{y_{i + 1}}}} {{{y_i}}} = 1 + \frac{{\Delta t}} {6}\left[ {\lambda + 2\lambda \left( {1 + \frac{1} {2}\Delta t\lambda...$

which is nothing but

$\frac{{{y_{i + 1}}}} {{{y_i}}} = 1 + \Delta t\lambda + \frac{{{{\left( {\Delta t\lambda } \right)}^2}}} {2} + \frac{{{{\left...$

Therefore, the stability condition is given as follows

$\left| {1 + z + \frac{{{z^2}}} {2} + \frac{{{z^3}}} {6} + \frac{{{z^4}}} {{24}}} \right| \leq 1, \quad \Re z < 0.$

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The eigenvalues of a common tridiagonal matrix

Chuyên mục: Các Bài Tập Nhỏ, Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 3:36

The eigenvalues of then $N \times N$ tridiagonal matrix

$\left( {\begin{array}{*{20}{c}} a & b & {} & {} & {} & {} & {} \\ c & a & b & {} &...$

are

${\lambda _s} = a + 2\sqrt {bc} \cos \frac{{s\pi }} {{N + 1}}$

and its conresponding eigenvector is

${v_s}^T = \left( {{{\left( {\frac{c} {b}} \right)}^{\frac{1} {2}}}\sin \frac{{s\pi }} {{N + 1}},{{\left( {\frac{c} {b}} \righ...$

Proof. Let $\lambda$ present an eigenvalue of the given matrix (denoted by ) and the corresponding eigenvector with components v_1,...,v_N . Then

$\left( {\begin{array}{*{20}{c}} a & b & {} & {} & {} & {} & {} \\ c & a & b & {} &...$

If we defined $v_0 = v_{N+1}=0$ , then we have

$c{v_{i - 1}} + \left( {a - \lambda } \right){v_i} + b{v_{i + 1}} = 0,i = \overline {1,N} .$

The solution of the above equation is of the form

${v_i} = Bm_1^i + Cm_2^i$

where B, C are constants and m_1, m_2 are the roots of the equation

$c+(a-\lambda)m + bm^2=0.$

Since $v_0 = v_{N+1}=0$ , then 0=B+C and $0 = Bm_1^{N + 1} + Cm_2^{N + 1}$ . Hence,

${\left( {\frac{{{m_1}}} {{{m_2}}}} \right)^{N + 1}} = 1 = {e^{\sqrt { - 1} i2\pi }}$

$\frac{{{m_1}}} {{{m_2}}} = {e^{\frac{{\sqrt { - 1} i2\pi }} {{N + 1}}}}.$

We also have

${m_1}{m_2} = \frac{c} {b}, \quad {m_1} + {m_2} = \frac{{\lambda - a}} {b}.$

Finally,

$\lambda = a + b\sqrt {\frac{c} {b}} \left( {{e^{\frac{{\sqrt { - 1} i2\pi }} {{N + 1}}}} + {e^{ - \frac{{\sqrt { - 1} i2\pi ...$

The -component of the eigenvector is

${v_j} = Bm_1^j + Cm_2^j = B\sqrt {{{\left( {\frac{c} {b}} \right)}^j}} \left( {{e^{\frac{{\sqrt { - 1} i2\pi }} {{N + 1}}}} -...$

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Tháng tám 15, 2009

Summation by parts (Abel sum formula)

Chuyên mục: Các Bài Tập Nhỏ, Giải tích 9 (MA5265), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 22:29

Suppose $\{f_k\}$ and $\{g_k\}$ are two sequences. Then,

$\sum_{k=m}^n f_k(g_{k+1}-g_k) = \left[f_{n+1}g_{n+1} - f_m g_m\right] - \sum_{k=m}^n g_{k+1}(f_{k+1}- f_k).$

Using the forward difference operator $\Delta$ , it can be stated more succinctly as

$\sum_{k=m}^n f_k\Delta g_k = \left[f_{n+1} g_{n+1} - f_m g_m\right] - \sum_{k=m}^n g_{k+1}\Delta f_k,$

Note that summation by parts is an analogue to the integration by parts formula,

$\int f\,dg = f g - \int g\,df.$

We also have the following identity

$\sum_{k=m}^n f_k(g_k}-g_{k-1}) = \left[f_{n+1}g_n - f_m g_{m-1}\right] - \sum_{k=m}^n g_k(f_{k+1}- f_k).$

Using the backward difference operator $\Delta$ , it can be stated more succinctly as

$\sum_{k=m}^n f_k\Delta g_{k-1} = \left[f_{n+1}g_n - f_m g_{m-1}\right] - \sum_{k=m}^n g_k \Delta f_k,$

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Several questions involving the Vitali set

Chuyên mục: Các Bài Tập Nhỏ, Giải Tích 6 (MA5205), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 14:36

We denote by the Vitali set which is defined as follows:

We say that $x, y \in [0, 1)$ are equivalent, and write $x \sim y$ , if and only if is a rational number. This equivalence relation partitions into an uncountable family of disjoint equivalence classes. By the axiom of choice there is a set which contains exactly one element from each equivalence class.

Now let $\{ \tau_n\}$ be a sequence of all rationals in [0, 1) with $\tau_0 =0$ and define $V_n = V + \tau_n$ (mod 1).

Now we show that the V_n are pairwise disjoint and

$\bigcup\limits_n {V_n } = \left[ {0,1} \right)$ .

Indeed, if $x \in V_i \cap V_j$ , then $x = v_i+\tau_i$ (mod 1) and $x = v_j+\tau_j$ (mod 1), with v_i and v_j belonging to V = V_0 . Consequently, $v_i - v_j \in \mathbb Q$ , which means that $v_i \sim v_j$ and therefore i = j . This shows that $V_i \cap V_j = 0$ if $i \ne j$ . Since each $x \in [0, 1)$ is in some equivalence class, differs modulo 1 from an element in by a rational number, say , in [0, 1) . Thus $x \in V_k$ , which proves that

$[0, 1) \subset \bigcup\limits_n {V_n }$ .

The opposite inclusion is obvious.

Question 1. Show that there exist sets E_1, E_2, ...,E_k,... such that $E_k \searrow E$ , and $|E_k|_e < \infty$ and

$\lim_{k \to \infty} |E_k|_e > |E|_e$

with strict inequality.

Solution. We put $E_n = \bigcup\limits_{k \geq n} {V_k }$ . Clearly, $\{E_n\}$ is a decreasing sequence. Since the V_k are pairwise disjoint, we see that $E: = \bigcap\limits_n {E_n } = \emptyset$ and $E_n \searrow E$ . Moreover,

$\left| {E_n } \right|_e \geq \left| {V_n } \right|_e = \left| V \right|_e > 0$

(the last inequality comes from the fact that is not measurable). It is now enough to show that

$\mathop {\lim }\limits_{n \to \infty } \left| {E_n } \right|_e \geq \left| V \right|_e > 0 = \left| E \right|_e$

and the proof is complete.

Question 2. Show that there exist disjoint E_1, E_2, ...,E_k,... such that

$\left| { \bigcup_k E_k } \right|_e < \sum_k {\left| {E_k } \right|_e }$

with strict inequality.

Solution. We put E_n = V_n then E_n are pairwise disjoint and obviously

$\left| {\bigcup_n {E_n } } \right|_e = \left| {\left[ {0,1} \right)} \right|_e = 1$ .

Moreover, all the E_n are of the same outer measure. Thus $\sum_n {\left| {E_n } \right|_e } = + \infty$ which completes the proof.

Question 3. Show that each of the sets

${E_n} = \bigcup\limits_{k = 0}^n {{V_k}}$

is non-measurable.

Question 4. Show that if is a measurable subset of the Vitali set , then |E|=0 .

Question 5. Show that there exist sets and such that

${\left| {A \cup B} \right|_i} = {\left| A \right|_i} + {\left| B \right|_i}$

but

${\left| {A \cup B} \right|_e} < {\left| A \right|_e} + {\left| B \right|_e}.$

Question 6. Show that any set of positive outer measure contains a non-measurable subset.

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Tháng tám 14, 2009

How to find a conformal mapping between the quadrants and the semidisc

Chuyên mục: Các Bài Tập Nhỏ, Giải tích 7 (MA4247), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 21:34

In the previous topic I show you by the following map

$f : z \mapsto \frac{z+1}{z-1}$

maps $\{z : \Re z < 0\}$ onto $\{w : |w|<1\}$ , and is conformal. Therefore the map

$g : z \mapsto \frac{z+1}{1-z}$

maps $\{z : \Re z > 0\}$ onto $\{w : |w|<1\}$ , and is conformal. What I am going to do is to prove that

$\Re z > 0\quad \Leftrightarrow \quad \Re \left( {\frac{{z + 1}} {{1-z}}} \right) > 0.$

To this purpose, we assume z=x+iy , i.e., $\Re z = y$ . Now by a simple calculation

$\frac{{z + 1}} {{1 - z}} = \frac{{\left( {x + 1} \right) + iy}} {{\left( {1 - x} \right) - iy}} = \frac{{\left[ {\left( {x + ...$

which yields

$\Re \left( {\frac{{z + 1}} {{1 - z}}} \right) = \frac{{2y}} {{{{\left( {1 - x} \right)}^2} + {y^2}}}.$

Having this fact we can easily see that under the map the first and fourth quadrants maps to upper and lower semidisks, respectively.

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An easy way to construct a conformal mapping between upper half plane and the open unit disk

Chuyên mục: Các Bài Tập Nhỏ, Giải tích 7 (MA4247), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 20:54

The locus |z + 1|= | z -1| is the perpendicular bisector of the line segment joining to , that is, the imaginary axis. The set |z + 1|< | z -1| is then the set of points closer to than to , that is, the left half-plane $\Re z <0$ . Hence, $\Re z <0$ if and only if

$\frac{|z+1|}{|z-1|}<1.$

The map

$f : z \mapsto \frac{z+1}{z-1}$

maps $\{z : \Re z < 0\}$ onto $\{w : |w|<1\}$ , and is conformal as $f' \ne 0$ . The inverse map is easily seen to be

$w \mapsto z=\frac{w+1}{w-1}.$

The locus |z + i|= | z - i| is the perpendicular bisector of the line segment joining to , that is, the real axis. The set |z + i|< | z -i| is then the set of points closer to than to , that is, the lower half-plane $\Im z <0$ . Hence, $\Im z <0$ if and only if

$\frac{|z+i|}{|z-i|}<1.$

The map

$g : z \mapsto \frac{z+i}{z-i}$

maps $\{z : \Im z < 0\}$ onto $\{w : |w|<1\}$ , and is conformal as $g' \ne 0$ . The inverse map is easily seen to be

$w \mapsto z=\frac{w+1}{w-1}i.$

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	Ngô Quốc Anh trong bài Các công trình gần đây
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Ngô Quốc Anh

Tháng Chín 13, 2009

QE in Department of Mathematics, National University of Singapore, August 2009

Tháng tám 29, 2009

On a polynomials of degree n, having all its zeros in the unit dis

Tháng tám 18, 2009

An other limit supremum of sin function

Tháng tám 17, 2009

The use of equivalence in measure

Tháng tám 16, 2009

On the stability of the Runge-Kutta 4 (RK4)

The eigenvalues of a common tridiagonal matrix

Tháng tám 15, 2009

Summation by parts (Abel sum formula)

Several questions involving the Vitali set

Tháng tám 14, 2009

How to find a conformal mapping between the quadrants and the semidisc

An easy way to construct a conformal mapping between upper half plane and the open unit disk

Đề mục

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