Linh Tinh « Ngô Quốc Anh

Tháng Mười Hai 2, 2009

MuPAD

Chuyên mục: Linh Tinh — 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.

1. We start with the following function

$f(x,y)=-(x^2+y^2)(\cos (3y) + \sin (7y))$ .

To plot its graph, I use the following

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)

and this is what we get

exp(-x^2-y^2)*(cos(3*y)+sin(7*x))

To obtain the following picture

you need

f := plot::Function3d(exp(-x^2-y^2)*(cos(3*y)+sin(7*x)), x=-2..2, y=-2..2, Submesh=[1,1]): graphLines:=plot::modify(f, Filled = FALSE, XLinesVisible = FALSE, YLinesVisible = FALSE, ZContours = [Automatic, 15], LineColorFunction=((x, y, z) -> [(z+2)/4, 0.5, (2-z)/4])): linesProjection := plot::Transform3d( [0, 0, -1.5], // shift vector [1, 0, 0, // transformation matrix 0, 1, 0, 0, 0, 0], graphLines): plot(f,linesProjection)

2. With the following function

${\rm Re}(\cos (x+iy)$ .

To plot its graph, I use the following

f := plot::Function3d(Re(cos(x+I*y)), x=-2..2, y=-2..2, Submesh=[1,1], FillColorFunction=((x, y, z) -> [(z+2)/4, 0.5, (2-z)/4])): plot(f)

and this is what we get Re(cos(x+I*y) 3. If our function has singularities, for example, Gamma function $f(x,y)=|\Gamma (x+iy)|$ , we have

Re(cos(x+I*y) f := plot::Function3d(abs(gamma(x+I*y)), x=-4.5..4.5, y=-3..3, Submesh=[3,3], XSubmesh = 1, YSubmesh = 2): plot(f)

4. If you intend to plot sereval graphs, the following is a good example.

sin(x^2 + y^2), cos(x^2 - y^2)

Here we used functions $\sin(x^2 + y^2)$ , $\cos(x^2 - y^2)$ . The code I used is

plotfunc3d(sin(x^2 + y^2), cos(x^2 - y^2), x = 0..1, y = 0..2)

5. The Mobius band can be drawn by using the following

x := cos(alpha) * (1 + r * cos(alpha/2)): y := sin(alpha) * (1 + r * cos(alpha/2)): z := r * sin(alpha/2): plot(plot::Surface([x,y,z], r = -0.5 .. 0.5, alpha = -PI .. PI, Mesh = [35, 31], LineColor = RGB::Black.[0.2]), Axes = None, Scaling = Constrained)

6. If you need a transparent picture, probably you need to add a factor like abs(sin(z)) in the FillColorFunction. Following is an example

K := (x,y,z)->[abs(sin(x)), abs(sin(y)), abs(sin(z)), abs(sin(z))]: F1 := plot::Function3d( 1.7*x*exp(-1/2*(x^2+y^2)), x=-4..4, y=-4..4, FillColorFunction=K): F2 := plot::Function3d(-1, x=-4..4, y=-4..4): plot(F1, F2)

The original one is

7. I will end this topic by showing some more pictures obtained by zooming a little bit closed.

sin(x

plotfunc3d(sin(x - -PI)*sin(y - -PI), Submesh = [4, 4], x = -PI .. PI, y = -PI .. PI)

sin(x

In conclusion, if you want to plot a general surface in 3D, you need its parametrization in order to draw. For details, I prefer the reader to this paper.

Phản hồi (4)

Tháng Mười 30, 2009

A characteristic of essentially bounded functions

Chuyên mục: Các Bài Tập Nhỏ, Giải Tích 6 (MA5205), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 11:53

In this topic, we prove the following statement

Statement: Let $(X,\mathcal B, m)$ be a probability space. Let $h \in L^2(m)$ . Then $h$ is essentially bounded iff $h \cdot f \in L^2(m)$ for all $f \in L^2(m)$ .

Proof. If $h$ is bounded, then by using the Holder inequality one has

$\displaystyle\int_X {{{\left| {h \cdot f} \right|}^2}dm}\leq \underbrace {\sqrt {\int_X {{{\left| h \right|}^2}dm} } }_{ \leqslant c}\sqrt {\int_X {{{\left| f \right|}^2}dm} }<+\infty$

for all $f \in L^2(m)$ . Conversely, we suppose $h$ is such that $h \cdot f \in L^2(m)$ whenever $f \in L^2(m)$ . Let

$\displaystyle X_n = \{ x \in X : n-1 \leq |h(x)| < n\}, \quad \forall n>0$ .

Then $\{X_n\}_1^\infty$ partitions $X$ . Let

$\displaystyle f\left( x \right) =\sum\limits_{n = 1}^\infty{\frac{1}{{n\sqrt {m\left( {{X_n}} \right)} }}{\chi_{{X_n}}}\left( x \right)} ,$

where it is understood that the $n$ -term is omiited if $m(X_n)=0$ . Then

$\displaystyle\int_X {{{\left| f \right|}^2}dm}=\int_X {{{\left({\sum\limits_{n = 1}^\infty {\frac{1}{{n\sqrt {m\left( {{X_n}}\right)} }}{\chi _{{X_n}}}\left( x \right)} } \right)}^2}dm}\leq \sum\limits_{n = 1}^\infty{\frac{1}{{{n^2}}}}<\infty$

which implies $f \in L^2(m)$ . Since

$\displaystyle\int_X {{{\left| {hf} \right|}^2}dm}=\sum\limits_{n \in F} {\int_{{X_n}} {{{\left| {hf} \right|}^2}dm}}\geq\sum\limits_{n \in F}{{{\left( {\frac{{n - 1}}{n}}\right)}^2}}$

where $F = \left\{ {n:m\left( {{X_n}} \right) \ne 0} \right\}$ . Sincc $h \cdot f \in L^2(m)$ we have that $F$ is finite and therefore $h$ is essentially bounded.

Để lại phản hồi

Tháng Mười 28, 2009

The weak and weak* topologies: A few words

Chuyên mục: Giải tích 7 (MA4247), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 2:48

The weak and weak* topologies are the weakest in which certain linear functionals are continuous.

We start with a normed linear space $X$ . The dual space of $X$ , denoted by $X'$ , is the collection of all continuous linear functionals, i.e., the set of all mapping $\ell : X \to \mathbb R$ satisfying

$\ell(ax)=a \ell (x)$ , $\ell(x+y)=\ell(x)+\ell(y)$

and

$\displaystyle\lim_{n \to \infty} \ell(x_n) = \ell(x)$ when $\displaystyle\lim_{n \to \infty} \|x_n - x\|=0$ .

Definition 1. In $X$ , the strong topology is the norm topology, i.e., we can talk about an open set of $X$ , for example $U$ in the following sense: $U \subset X$ is said to be open if and only if for each $x_0 \in U$ , there exists $\varepsilon>0$ such that $\{ x \in X: \|x-x_0\|<\varepsilon\} \subset U$ .

Claim 1. Bounded linear functionals are continuous in the strong topology.

Proof. We first recall that a linear functional $\ell$ is said to be bounded if there is a positive number $c$ such that $|\ell (x)| \leq c\|x\|$ for all $x \in X$ .

Now we assume $\ell$ is continuous but not bounded; then for any choice of $c=n$ , one has $\ell(x_n) > n \|x_n\|$ . Clearly, $x_n$ can be replaced by any multiple of $x_n$ ; if we normalize $x_n$ so that

$\displaystyle \|x_n\|=\frac{1}{\sqrt{n}}$

then $x_n \to 0$ but $\ell (x_n) \to \infty$ . This shows the lack of boundedness implies the lack of continuity.

Now we assume $\ell$ is bounded. For arbitray $x_n$ and $x$ , one gets

$|\ell(x_n)-\ell (x)| = |\ell (x_n-x)| \leq c\|x_n-x\|$ ;

this shows that boundedness implies continuity.

Definition 2. In $X$ , the weak topology is the weakest topology in which all bounded linear functionals are continuous.

The open sets in the weak topology are unions of finite intersections of sets of the form

$\{ x : a< \ell(x) < b\}$ .

Clearly, in an infinite-dimensional space the intersection of a finite number of sets of the above form is unbounded. This shows that every set that is open in the weak topology is unbounded. In particular, the balls

$\{ x : \|x\|<R\}$

opens in the strong topology, are not open is the weak topology.

Definition 3. In $X'$ the dual space of $X$ , the weak* topology is the crudest topology in which all linear functionals

$x: X' \to \mathbb R, x(\ell) := \ell(x)$

are continuous.

If $X'$ is nonreflexive, the weak* topology is genuinely coarser than the weak topology, as will be clear from the following theorem due to Alaoglu

Theorem (Alaoglu). The closed unit ball in $X'$ is compact in the weak* topology.

We end this topic by the following theorem

Theorem. The closed unit ball in $X$ is compact in the weak topology if and only if $X$ is reflexive.

Để lại phản hồi

Tháng Chín 18, 2009

Moving plane method via a small example inspired by Gidas, Ni, and Nirenberg

Chuyên mục: Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 1:41

In this section we will use the moving plane method to discuss the symmetry of solutions. The following result was first proved by Gidas, Ni and Nirenberg.

Theorem. Suppose $u \in C(\bar B_1) \cap C^2(B_1)$ is a positive solution of

$-\Delta u = f(u)$ in $B_1$ and $u=0$ on $\partial B_1$

where $f$ is locally Lipschitz in $\mathbb R$ . Then $u$ is radialy symmetric in $B_1$ and $\frac{\partial u}{\partial r}(x)<0$ for $x \ne 0$ .

The original proof requires that solutions be $C^2$ up to the boundary. Here we give a method which does not depend on the smoothness of domains nor the smoothness of solutions up to the boundary.

Statement. Suppose that is a bounded domain which is convex in $x_1$ direction and symmetric with respect to the plane $\{x_1 = 0\}$ . Suppose $u \in C^2(\Omega)\cap C(\bar\Omega)$ is a positive solution of

$-\Delta u = f(u)$ in $\Omega$ and $u=0$ on $\partial \Omega$

where $f$ is locally Lipschitz in $\mathbb R$ . Then $u$ is radialy symmetric in $x_1$ and $D_{x_1}u(x)<0$ for $x \ne 0$ .

Idea of proof. Write $x=(x_1, y)\in\Omega$ for $y \in \mathbb R^{n-1}$ . We will prove that

$u(x_1,y)<u(x_1^\star,y)$ for any $x_1>0$ and $x_1^\star<x_1$ with $x_1^\star+x_1>0$ .

Then by letting $x_1^\star \to -x_1$ , we get $u(x_1,y)\leq u(-x_1,y)$ for any $x_1$ . Then by changing the direction $x_1 \to -x_1$ we get the symmetry.

Proof. Let $a=\sup x_1$ for $(x_1, y)\in\Omega$ . For $0<\lambda<a$ , define

$\Sigma_\lambda=\{x \in \Omega: x_1>\lambda\}$

$T_\lambda = \{x_1=\lambda\}$

$\Sigma'_\lambda=$ the reflection of $\Sigma_\lambda$ with the respect to $T_\lambda$

$x_\lambda=(2\lambda-x_1,x_2,...,x_n)$ for $x=(x_1,x_2,...,x_n)$ .

In $\Sigma_\lambda$ we define

$w_\lambda(x)=u(x)-u(x_\lambda)$ for $x\in \Sigma_\lambda$ .

Then we have by Mean Value Theorem

$\Delta w_\lambda+c(x,\lambda)w_\lambda=0$ in $\Sigma_\lambda$

$w_\lambda \leq 0$ and $w_\lambda \not \equiv 0$ on $\partial \Sigma_\lambda$ .

where $c(x,\lambda)$ is a bounded function in $\Sigma_\lambda$ .

We need to show $w_\lambda<0$ in $\Sigma_\lambda$ for any $\lambda \in (0,a)$ . This implies in particular that $w_\lambda$ assumes along $\partial \Sigma_\lambda \cap \Omega$ its maximum in $\Sigma_\lambda$ . By Hopf Lemma we have for any such $\lambda \in (0,a)$

$D_{x_1}w_\lambda\bigg|_{x_1=\lambda} = 2D_{x_1}u\bigg|_{x_1=\lambda}<0$ .

For any $\lambda$ close to $a$ , we have $w_\lambda<0$ by the maximum principle for narrow domain. Let $(\lambda_0, a)$ be the largest interval of values of $\lambda$ such that $w_\lambda<0$ in $\Sigma_\lambda$ . We want to show that $\lambda_0=0$ . If $\lambda_0>0$ , by continuity, $w_\lambda \leq 0$ in $\Sigma_{\lambda_0}$ and $w_{\lambda_0} \not \equiv 0$ on $\partial \Sigma_{\lambda_0}$ . Then the Strong Maximum Principle implies $w_{\lambda_0}<0$ in $\Sigma_{\lambda_0}$ . We will show that for any small $\varepsilon>0$

$w_{\lambda_0-\varepsilon} <0$ in $\Sigma_{\lambda_0-\varepsilon}$ .

Fix $\delta>0$ (to be determined). Let $K$ be a closed subset in $\Sigma_{\lambda_0}$ such that $|\Sigma_{\lambda_0} - K| <\frac{\delta}{2}$ . The fact that $w_{\lambda_0}<0$ in $\Sigma_{\lambda_0}$ implies

$w_{\lambda_0}(x)\leq -\eta <0$ for any $x \in K$ .

By continuity we have

$w_{\lambda_0-\varepsilon}<0$ in $K$ .

For $\varepsilon>0$ small, $|\Sigma_{\lambda_0-\varepsilon}-K|<\delta$ . We choose $\delta$ in such a way that we may apply the Maximum Principle for domain with small volume to $w_{\lambda_0-\varepsilon}$ in $\Sigma_{\lambda_0-\varepsilon}-K$ . Hence we get

$w_{\lambda_0-\varepsilon} \leq 0$ in $\Sigma_{\lambda_0-\varepsilon}-K$

and then

$w_{\lambda_0-\varepsilon}(x) <0$ in $\Sigma_{\lambda_0-\varepsilon} -K$ .

Therefore we obtain for any small $\varepsilon>0$

$w_{\lambda_0-\varepsilon}(x) <0$ in $\Sigma_{\lambda_0-\varepsilon}$ .

This contradicts the choice of $\lambda_0$ .

latex

Để lại phản hồi

Tháng Chín 13, 2009

An algebraic identity for 9th level

Chuyên mục: Các Bài Tập Nhỏ, Linh Tinh — Ngô Quốc Anh @ 21:16

In this topic I will show you how to prove

$\sqrt[3]{2} + \sqrt[3]{{20}} - \sqrt[3]{{25}} = 3\sqrt {\sqrt[3]{5} - \sqrt[3]{4}}$

In order to prove that fact, we just do as following: by using

$(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)$

we obtain

${\Big( {\sqrt[3]{2} + \underbrace {\sqrt[3]{{20}}}_{\sqrt[3]{{{2^2}}}\sqrt[3]{5}} - \underbrace {\sqrt[3]{{25}}}_{\sqrt[3]{{{5^2}}}}} \Big)^2}=\sqrt[3]{{{2^2}}} + \sqrt[3]{{{2^4}}}\sqrt[3]{{{5^2}}} + \sqrt[3]{{{5^4}}} + 2\left( {\underbrace {\sqrt[3]{2}\sqrt[3]{{{2^2}}}}_2\sqrt[3]{5} - \sqrt[3]{2}\sqrt[3]{{{5^2}}} - \sqrt[3]{{{2^2}}}\underbrace {\sqrt[3]{5}\sqrt[3]{{{5^2}}}}_5} \right).$

Therefore

${\left( {\sqrt[3]{2} + \sqrt[3]{{20}} - \sqrt[3]{{25}}} \right)^2} = \sqrt[3]{{{2^2}}} + 2\sqrt[3]{2}\sqrt[3]{{{5^2}}} + 5\sqrt[3]{5} + 2\left( {2\sqrt[3]{5} - \sqrt[3]{2}\sqrt[3]{{{5^2}}} - 5\sqrt[3]{{{2^2}}}} \right).$

Clearly

$\sqrt[3]{{{2^2}}} + 2\sqrt[3]{2}\sqrt[3]{{{5^2}}} + 5\sqrt[3]{5} + 2\left( {2\sqrt[3]{5} - \sqrt[3]{2}\sqrt[3]{{{5^2}}} - 5\sqrt[3]{{{2^2}}}} \right) = 9\left( {\sqrt[3]{5} - \sqrt[3]{{{2^2}}}} \right).$

Thus

${\left( {\sqrt[3]{2} + \sqrt[3]{{20}} - \sqrt[3]{{25}}} \right)^2} = 9\left( {\sqrt[3]{5} - \sqrt[3]{4}} \right)$

which yields

$\left| {\sqrt[3]{2} + \sqrt[3]{{20}} - \sqrt[3]{{25}}} \right| = 3\sqrt {\sqrt[3]{5} - \sqrt[3]{4}} .$

Finally, by using the fact that $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ we get

${\left( {\sqrt[3]{2} + \sqrt[3]{{20}}} \right)^3} - 25 = \underbrace {\left( {22 + 3\sqrt[3]{{{2^2}}}\sqrt[3]{{20}} + 3\sqrt[3]{2}\sqrt[3]{{{{20}^2}}}} \right) - 25}_{3\left( {\sqrt[3]{{{2^2}}}\sqrt[3]{{20}} + \sqrt[3]{2}\sqrt[3]{{{{20}^2}}} - 1} \right)} > 0$

which implies

${\sqrt[3]{2} + \sqrt[3]{{20}} - \sqrt[3]{{25}}}>0.$

In other words,

$\sqrt[3]{2} + \sqrt[3]{{20}} - \sqrt[3]{{25}} = 3\sqrt {\sqrt[3]{5} - \sqrt[3]{4}}.$

Phản hồi (3)

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, Đề Thi — 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

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

Để lại phản hồi

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.

Để lại phản hồi

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

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

Để lại phản hồi

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)

Để lại phản hồi

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

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

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

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

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

Để lại phản hồi

Bài viết cũ hơn »

	Ngô Quốc Anh trong bài Tích phân bội liên quan đến hà…
	R-G: Gauss and Codaz… trong bài R-G: Codazzi equations in clas…
	Ngô Quốc Anh trong bài Solving initial value problem …
	vo khanh nguyen trong bài Solving initial value problem …
	Ngô Quốc Anh trong bài Tính tích phân suy rộng loại…
	Ngô Quốc Anh trong bài Tính tích phân suy rộng loại…
	hoang cuong trong bài Tính tích phân suy rộng loại…
	Ngô Quốc Anh trong bài Một số dạng bài tập xét sự hội…
	Ngô Quốc Anh trong bài An algebraic identity for 9th…
	tamddao trong bài An algebraic identity for 9th…

Ngô Quốc Anh

Tháng Mười Hai 2, 2009

MuPAD

Tháng Mười 30, 2009

A characteristic of essentially bounded functions

Tháng Mười 28, 2009

The weak and weak* topologies: A few words

Tháng Chín 18, 2009

Moving plane method via a small example inspired by Gidas, Ni, and Nirenberg

Tháng Chín 13, 2009

An algebraic identity for 9th level

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)

Đề mục

Phân loại

Lưu trữ

Liên Kết

Bài gửi gần đây

Lịch

Bình luận gần đây

Số người đang xem

Thống kê