# Ngô Quốc Anh

## August 29, 2009

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

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.

## August 18, 2009

### An other limit supremum of sin function

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

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

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

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

Indeed for any $x \in A$, since

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

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

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

Since

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

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

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

## August 17, 2009

### The use of equivalence in measure

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 be a measurable space, and let be two measures. Then is said to be equivalent to if

for measurable sets in , i.e. the two measures have precisely the same null sets. Equivalence is often denoted or .

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

Equivalence of measures is an equivalence relation on the set of all measures .

Examples.

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

Application. Let be a finite measure on , and define

Show that is finite a.e. with respect to the Lebesgue measure on .

Proof. Let

then where

and

Clearly

and by Fubini’s Theorem

then

Since

Thus the following function

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

## August 16, 2009

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

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

$\displaystyle \begin{gathered}{K_{1}}= f\left({{t_{i}},{y_{i}}}\right),\hfill\\ {K_{2}}= f\left({{t_{i}}+\frac{1}{2}\Delta t,{y_{i}}+\frac{1}{2}\Delta t{K_{1}}}\right),\hfill\\ {K_{3}}= f\left({{t_{i}}+\frac{1}{2}\Delta t,{y_{i}}+\frac{1}{2}\Delta t{K_{2}}}\right),\hfill\\ {K_{4}}= f\left({{t_{i}}+\Delta t,{y_{i}}+\Delta t{K_{3}}}\right),\hfill\\ \end{gathered}$

then

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

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

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

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

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

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

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

Applying the above discussion to RK4 method, we see that

$\displaystyle\begin{gathered}{K_{4}}=\lambda\left({{y_{i}}+\Delta t{K_{3}}}\right),\hfill\\ {K_{3}}=\lambda\left({{y_{i}}+\frac{1}{2}\Delta t{K_{2}}}\right),\hfill\\ {K_{2}}=\lambda\left({{y_{i}}+\frac{1}{2}\Delta t{K_{1}}}\right),\hfill\\ {K_{1}}=\lambda{y_{i}},\hfill\\ \end{gathered}$

which implies

$\displaystyle\begin{gathered}{K_{4}}=\lambda\left({{y_{i}}+\Delta t\lambda\left({{y_{i}}+\frac{1}{2}\Delta t\lambda\left({{y_{i}}+\frac{1}{2}\Delta t{y_{i}}}\right)}\right)}\right) =\lambda{y_{i}}\left({1+\Delta t\lambda\left({1+\frac{1}{2}\Delta t\lambda\left({1+\frac{1}{2}\Delta t}\right)}\right)}\right),\hfill\\ {K_{3}}=\lambda\left({{y_{i}}+\frac{1}{2}\Delta t\lambda\left({{y_{i}}+\frac{1}{2}\Delta t{y_{i}}}\right)}\right) =\lambda{y_{i}}\left({1+\frac{1}{2}\Delta t\lambda\left({1+\frac{1}{2}\Delta t}\right)}\right),\hfill\\ {K_{2}}=\lambda\left({{y_{i}}+\frac{1}{2}\Delta t\lambda{y_{i}}}\right) =\lambda{y_{i}}\left({1+\frac{1}{2}\Delta t\lambda }\right),\hfill\\ {K_{1}}=\lambda{y_{i}},\hfill\\ \end{gathered}$

which yields

$\displaystyle\begin{gathered}{K_{4}}=\lambda\left({{y_{i}}+\Delta t\lambda\left({{y_{i}}+\frac{1}{2}\Delta t\lambda\left({{y_{i}}+\frac{1}{2}\Delta t{y_{i}}}\right)}\right)}\right) =\lambda{y_{i}}\left({1+\Delta t\lambda\left({1+\frac{1}{2}\Delta t\lambda\left({1+\frac{1}{2}\Delta t}\right)}\right)}\right),\hfill\\ {K_{3}}=\lambda\left({{y_{i}}+\frac{1}{2}\Delta t\lambda\left({{y_{i}}+\frac{1}{2}\Delta t{y_{i}}}\right)}\right) =\lambda{y_{i}}\left({1+\frac{1}{2}\Delta t\lambda\left({1+\frac{1}{2}\Delta t}\right)}\right),\hfill\\ {K_{2}}=\lambda\left({{y_{i}}+\frac{1}{2}\Delta t\lambda{y_{i}}}\right) =\lambda{y_{i}}\left({1+\frac{1}{2}\Delta t\lambda }\right),\hfill\\ {K_{1}}=\lambda{y_{i}},\hfill\\ \end{gathered}$

Thus

$\displaystyle\frac{{{y_{i+1}}}}{{{y_{i}}}}= 1+\frac{{\Delta t}}{6}\left[{\lambda+2\lambda\left({1+\frac{1}{2}\Delta t\lambda }\right)+2\lambda\left({1+\frac{1}{2}\Delta t\lambda\left({1+\frac{1}{2}\Delta t}\right)}\right)+\lambda\left[{1+\Delta t\lambda\left({1+\frac{1}{2}\Delta t\lambda\left({1+\frac{1}{2}\Delta t}\right)}\right)}\right]}\right]$

which is nothing but

$\displaystyle\frac{{{y_{i+1}}}}{{{y_{i}}}}= 1+\Delta t\lambda+\frac{{{{\left({\Delta t\lambda }\right)}^{2}}}}{2}+\frac{{{{\left({\Delta t\lambda }\right)}^{3}}}}{6}+\frac{{{{\left({\Delta t\lambda }\right)}^{4}}}}{{24}}$.

Therefore, the stability condition is given as follows

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

### The eigenvalues of a common tridiagonal matrix

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

The eigenvalues of then tridiagonal matrix

are

and its conresponding eigenvector is

Proof. Let present an eigenvalue of the given matrix (denoted by ) and the corresponding eigenvector with components . Then

If we defined , then we have

The solution of the above equation is of the form

where are constants and are the roots of the equation

Since , then and . Hence,

or

We also have

Finally,

The -component of the eigenvector is

## August 15, 2009

### Summation by parts (Abel sum formula)

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

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

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

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

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

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

We also have the following identity

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

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

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

### Several questions involving the Vitali set

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

We say that are equivalent, and write , 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 be a sequence of all rationals in with and define (mod 1).

Now we show that the are pairwise disjoint and

.

Indeed, if , then (mod 1) and (mod 1), with and belonging to . Consequently, , which means that and therefore . This shows that if . Since each is in some equivalence class, differs modulo 1 from an element in by a rational number, say , in . Thus , which proves that

.

The opposite inclusion is obvious.

Question 1. Show that there exist sets such that , and and

with strict inequality.

Solution. We put . Clearly, is a decreasing sequence. Since the are pairwise disjoint, we see that and . Moreover,

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

and the proof is complete.

Question 2. Show that there exist disjoint such that

with strict inequality.

Solution. We put then are pairwise disjoint and obviously

.

Moreover, all the are of the same outer measure. Thus which completes the proof.

Question 3. Show that each of the sets

is non-measurable.

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

Question 5. Show that there exist sets and such that

but

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

## August 14, 2009

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

In the previous topic I show you by the following map

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

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

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

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

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

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

$\displaystyle\frac{{z+1}}{{1-z}}=\frac{{\left({x+1}\right)+iy}}{{\left({1-x}\right)-iy}}=\frac{{\left[{\left({x+1}\right)+iy}\right]\left[{\left({1-x}\right)+iy}\right]}}{{{{\left({1-x}\right)}^{2}}+{y^{2}}}}$

which yields

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

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

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

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

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

The map

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

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

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

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

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

The map

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

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

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

## August 13, 2009

### Two examples of analytic function on a punctured unit disk which has a removable singularity at the origin

Filed under: Các Bài Tập Nhỏ, Giải tích 7 (MA4247), Linh Tinh — Ngô Quốc Anh @ 15:54

The following question was proposed in NUS under the QE in AY 2007-2008:

Consider the punctured disk $D=\{ z \in \mathbb C | 0 <|z| <1\}$. Suppose $f : D \to \mathbb C$ is an analytic function such that

$\displaystyle |f''(z)| \leq \frac{2}{|z|^2}$

for all $z \in D$. Is it true that f has a removable singular point at $z=0$?

Proof. Denote the Laurent expansion of $f$ by

$\displaystyle f\left( z \right) = \sum\limits_{n = - \infty }^{ + \infty } {{a_n}{z^n}}$

where

$\displaystyle {a_n} = \frac{1} {{2\pi i}}\int\limits_{\left| z \right| = r < 1} {\frac{{f\left( z \right)dz}} {{{z^{n + 1}}}}}$.

Then from

$\displaystyle f''\left( z\right) =\sum\limits_{n =-\infty }^{+\infty }{\underbrace{\left({n+2}\right)\left({n+1}\right){a_{n+2}}}_{{b_{n}}}{z^{n}}}$

we get

$\displaystyle {b_n} = \frac{1} {{2\pi i}}\int\limits_{\left| z \right| = r < 1} {\frac{{f''\left( z \right)dz}} {{{z^{n + 1}}}}}$.

Thus,

$\displaystyle\left|{{b_{n}}}\right|\leqslant\frac{1}{{2\pi }}\int\limits_{\left| z\right| = r < 1}{\left|{\frac{{f''\left( z\right)}}{{{z^{n+1}}}}}\right|}dz\leqslant\frac{1}{\pi }\int\limits_{\left| z\right| = r < 1}{\left|{\frac{1}{{{z^{n+3}}}}}\right|}dz\leqslant\frac{1}{{\pi{r^{n+3}}}}2\pi r =\frac{2}{{{r^{n+2}}}}$.

When $n \leq -3$, let $r \to 0$ we see that $z=0$ is a removable singularity of $f$ since $b_n=0 (n \leq -3)$ implies $a_n=0 (n \leq -1)$.

Remark. The second derivative can be replaced by an $\mathbb Z\ni m \geq 1$and therefore $\frac{2}{|z|^2}$ should be $\frac{2}{|z|^m}$. This is a question of UCLA QE in Winter 2007.

We also have a similar question proposed in a QE of Indiana University. It says that if $f : D \to \mathbb C$ is an analytic function such that

$\displaystyle |f(z)|\leq\log\frac{1}{|z|}$

for all $z \in D$. Then $f \equiv 0$.

Proof
. As above, one has

$\displaystyle \left|{{b_{n}}}\right| =\left|{\frac{1}{{2\pi i}}\int\limits_{\left| z\right| = r < 1}{\frac{{f\left( z\right)dz}}{{{z^{n+1}}}}}}\right|\leqslant\frac{1}{{{r^{n}}}}\log\frac{1}{r}$.

When $n<0$, letting $r \to 0$ we have $a_n= 0$ for all $n \leq -1$ which implies $z=0$ is a removable singularity of $f$. In other words, $f$ can be extended to an analytic function of the unit disk. Since $\log \frac{1}{|z|} =0$ when $|z|=1$, by the Maximum Modules Principle we obtain $f \equiv 0$.

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