# Ngô Quốc Anh

## August 31, 2008

### Tập hoàn hảo không chứa số hữu tỉ nào

Filed under: Các Bài Tập Nhỏ, Linh Tinh — Ngô Quốc Anh @ 15:03

Does there exist a nonempty perfect subset of that doesn’t contain any rational number?

Solution. Let be an enumeration of the rationals and let $S = \bigcup_{k = 1}^\infty \left(q_k - 2^{ - k},q_k + 2^{ - k}\right)$. Its complement is closed, contains no rationals, and is uncountable because $\mu (S) < 2\Rightarrow \mu (S') = \infty$. It follows from the Cantor-Bendixson Theorem that has a nonempty perfect subset.

### Tập trong R^3 liên thông, compắc, không đâu trù mật với độ đo Lebesgue dương

Filed under: Các Bài Tập Nhỏ, Linh Tinh — Ngô Quốc Anh @ 14:59

Is there a subset of $E\subset [0,1]\times [0,1] \times [0,1]$ such that and is connected, compact, and nowhere dense?

Solution. Let be a “fat” Cantor set in – construct it by deleting middle intervals, but arrange the sizes of the deleted intervals so that Now let $x\in A \text{ or }y\in A\text{ or }z\in A\}.$ This should be path-connected, hence connected, and compact and nowhere dense should be true as well.

_________________

Về tập Cantor, xem thêm ở đây http://en.wikipedia.org/wiki/Cantor_set. Đây là tập không đếm được nhưng lại có độ đo Lebesgue bằng 0.

### L^1 hội tụ hầu khắp nơi

Suppose . Show that the integral $\phi(x)=\int_{\mathbb{R}^3}\frac{f(y)dy}{|x-y|}$

converges for a.e. .

Solution. Let $\psi(x)=\int_{\mathbb{R}^3}\frac{|f(y)|}{|x-y|}\,dy.$

Let be the ball centered at the origin of radius  $\int_{B_r}\psi(x)\,dx\le\int_{B_r}\int_{\mathbb{R}^3} \frac{|f(y)|}{|x-y|} \,dy\,dx$ $=\int_{\mathbb{R}^3}|f(y)|\int_{B_r}\frac{1}{|x-y|}\,dx\,dy$

We can show that $\int_{B_r}\frac{1}{|x-y|}\,dx\le\int_{B_r}\frac{1}{|x|}\,dx= Cr^2.$

(The constant can be computed – it’s actually – but the exact value is immaterial.) Hence, $\int_{B_r}\psi(x)\,dx\le Cr^2\int_{\mathbb{R}^3}|f(y)|\,dy$.

Since this is finite, we must have almost everywhere on Since we can repeat this for any (choose a countable sequence of such tending to ) we can say that is finite almost everywhere on Hence the integral that defines converges absolutely for almost every Note the importance here of having the function be locally integrable – in fact, uniformly locally integrable.

## August 18, 2008

### Partial Differential Equations through Examples and Exercises (Texts in the Mathematical Sciences)

Filed under: Sách Hay — Ngô Quốc Anh @ 12:47 This book examines the complicated subject of Partial Differential Equations (PDEs). It involves the reader throughout by presenting theory, examples and exercises together. Both the classical and abstract aspects of the theory are dealt with, so that, for example, classical and generalised solutions in Sobolev and distribution spaces are treated. Most of the work is devoted to second or higher order PDEs; part of the distribution theory is included, covering Dirac’s delta distribution delta function. Many practical tools are offered for solving important problems with the basic three PDEs, namely the wave equation, the Laplace equation, the heat equation and their generalisations. The majority of the problems are mathematical in character, though often physical interpretations are given.
Audience: This volume is intended for undergraduate and graduate students in mathematics, physics technology and economics interested in PDEs for modelling complex systems
• Hardcover: 416 pages
• Publisher: Springer; 1 edition (October 31, 1997)
• Language: English
• ISBN-10: 0792347242
• ISBN-13: 978-0792347248
• Product Dimensions: 9.3 x 6.3 x 1.3 inches

Tôi có bản copy của sách này…

## August 9, 2008

### Tính tích phân suy rộng loại I

Filed under: Các Bài Tập Nhỏ, Giải Tích 2, Giải Tích 4 — Ngô Quốc Anh @ 13:16

Tính $\displaystyle \int_{0}^{+\infty}{\frac{e^{-x^2}}{\left(x^2+\frac{1}{2}\right)^2}\; dx}$.

Lời giải. Đặt $\displaystyle I(\alpha) = - \int_{0}^{\infty} \frac {e^{ - x^2}}{x^2 + \alpha} \, dx$.

Khi đó ta có $\displaystyle\begin{gathered} \int_0^\infty {\frac{{{e^{ - {x^2}}}}}{{{x^2} + \alpha }}} \,dx = \int_0^\infty {{e^{ - {x^2}}}} \int_0^\infty {{e^{ - ({x^2} + \alpha )t}}} \,dtdx \hfill \\\qquad \qquad \qquad \; \,= \int_0^\infty {{e^{ - \alpha t}}} \int_0^\infty {{e^{ - (1 + t){x^2}}}} \,dxdt \hfill \\ \qquad \qquad \qquad \; \,= \int_0^\infty {\frac{{\sqrt \pi }}{2}} \frac{{{e^{ - \alpha t}}}}{{\sqrt {1 + t} }}\,dt, \hfill \\ \end{gathered}$

ta thấy $\displaystyle\int_{0}^{\infty} \frac {e^{ - x^2}}{(x^2 + \frac {1}{2})^2} \, dx = I'(\tfrac{1}{2}) = \frac {\sqrt {\pi}}{2} \int_{0}^{\infty} \frac {t}{\sqrt {1 + t}} \, e^{ - \frac {t}{2}} \, dt$.

Nhưng $\displaystyle \begin{gathered} \int_0^\infty {\frac{t}{{\sqrt {1 + t} }}} \,{e^{ - \frac{t}{2}}}\,dt = \int_0^\infty {\sqrt {1 + t} } \,{e^{ - \frac{t}{2}}}\,dt - \int_0^\infty {\frac{1}{{\sqrt {1 + t} }}} \,{e^{ - \frac{t}{2}}}\,dt \hfill \\ \qquad \qquad\qquad \qquad= \left[ { - 2\sqrt {1 + t} \,{e^{ - \frac{t}{2}}}} \right]_0^\infty + \int_0^\infty {\frac{1}{{\sqrt {1 + t} }}} \,{e^{ - \frac{t}{2}}}\,dt - \int_0^\infty {\frac{1}{{\sqrt {1 + t} }}} \,{e^{ - \frac{t}{2}}}\,dt \hfill \\ \qquad \qquad\qquad \qquad= 2. \hfill \\\end{gathered}$

Vậy $\displaystyle\int_{0}^{\infty} \frac {e^{ - x^2}}{(x^2 + \frac {1}{2})^2} \, dx = \sqrt {\pi}$.