Ngô Quốc Anh

October 29, 2007

Dirichlet-to-Neumann operator

Filed under: Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 23:29

You have a domain \Omega\subset \mathbb R^n and a partial differential equation (such as \Delta u=0) in that domain. Take a function \phi defined on \partial\Omega, solve the boundary value problem with u=\phi on the boundary, and compute \psi=\frac{\partial u}{\partial n}, the normal derivative of solution on the boundary. The map \phi\mapsto\psi is the D-n-N operator.

A typical problem (motivated by tomography) is to recover the PDE from the D-n-N operator.

Mountain Pass Theorem

Filed under: Nghiên Cứu Khoa Học, PDEs — Ngô Quốc Anh @ 23:00

Suppose F \in C^1(V) satisfies (PS) condition with F(0)=0. There exist \rho >0, \alpha >0 and e \in V such that

\displaystyle\mathop {\inf }\limits_{\left\| u \right\| = \rho } F\left( u \right) \geqslant \alpha, \left\| e \right\| \geqslant \rho

and

F\left( e \right) < \alpha.

Then

\displaystyle \beta = \mathop {\inf }\limits_{\Sigma \in \Gamma } \mathop {\sup }\limits_{u \in \Sigma } F\left( u \right)

is a critical value.

October 27, 2007

Đề thi cuối kỳ GT5 của K49

Filed under: Đề Thi — Ngô Quốc Anh @ 18:42

Những đề thi này chỉ có tính tham khảo, kể từ năm học 2007-2008 này, hình thức ra đề sẽ được thay đổi nhằm mục đích phù hợp với chương trình đạo tạo theo tín chỉ đang được áp dụng.

kthk_k49-thtsptt_gt5_deso1.pdf

kthk_k49-thtsptt_gt5_deso2.pdf

October 18, 2007

Đề thi giữa kỳ GT5 của K51

Filed under: Đề Thi — Ngô Quốc Anh @ 16:07

ktgk_k51-a2_gt5_1.pdf

ktgk_k51-a3_gt5_1.pdf

Một số dạng bài tập xét sự hội tụ của chuỗi số dương

Filed under: Giải Tích 4 — Ngô Quốc Anh @ 14:54

1. Let \sum a_n be a convergent series, with a_n \geq 0. What can we say about the nature of each of the following series?

(a) \displaystyle\sum\sqrt {a_n} (b) \displaystyle\sum a_n^2 (c) \displaystyle\sum \frac {a_n}{1 + a_n}
(d) \displaystyle\sum \frac {a_n^2}{1 + a_n^2} (e) \displaystyle\sum \frac {\sqrt {a_n}}{n} (f) \displaystyle\sum \frac {a_n}{\sqrt {n}}
(g) \displaystyle\sum \frac {a_n}{n} (h) \displaystyle\sum na_n (i) \displaystyle\sum \cos(a_n)
(j) \displaystyle\sum \sin(a_n) (k) \displaystyle\sum \tan(a_n) (l) \displaystyle\sum \cos(\sqrt {a_n})
(m) \displaystyle\sum [1 - \cos(\sqrt {a_n})] (n) \displaystyle\sum [1 - \cos(a_n)] (o) \displaystyle\sum \arcsin(a_n)
(p) \displaystyle\sum \arccos(a_n) (q) \displaystyle\sum a_n^{3/2} (r) \displaystyle\sum a_n^3
(s) \displaystyle\sum \frac {a_1 + a_2 + ... + a_n}{n} (t) \displaystyle\sum \arctan(a_n) (u) \displaystyle\sum \frac {1}{1 + a_n}
(v) \displaystyle\sum ( - 1)^na_n (w) \displaystyle\sum \sqrt [n]{a_n} (x) \displaystyle\sum (a_n - a_{n + 1})
(y) \displaystyle\sum [\ln(1 + a_n)]^{5/4}

2) Regarding the same series of the previous question, how about if \sum a_n diverges?

Hint.

\displaystyle\left( a \right) \to a_n = \frac{1}{{n^2 }}.

\displaystyle\left( b \right) \to a_n^2 \leqslant a_n \sum\limits_{n = 0}^{ + \infty } {a_n } .

\displaystyle\begin{gathered}\left( d \right) \to \frac{{a_n^2 }}{{1 + a_n^2 }} \leqslant \frac{{a_n^2 }}{{2a_n }} = a_n \hfill \\\left( c \right) \to \frac{{a_n }}{{1 + a_n }} \leqslant a_n . \hfill \\\end{gathered}

\displaystyle\left( e \right) \to \frac{{\sqrt {a_n } }}{n} \leqslant \frac{1}{2}\left( {a_n + \frac{1}{{n^2 }}} \right)

\displaystyle\left( f \right) \to \mathop {\lim }\limits_{n \to \infty } \frac {{\frac {{a_n }} {{\sqrt n }}}} {{a_n }} = \mathop {\lim }\limits_{n \to \infty } \frac {1} {{\sqrt n }} = 0= \mathop {\lim }\limits_{n \to \infty } \frac {1} {n} = \mathop {\lim }\limits_{n \to \infty } \frac {{\frac {{a_n }} {n}}} {{a_n }} \leftarrow \left( g \right)

\displaystyle\left( h \right) \to a_n = \frac{1}{{n^2 }}.

\displaystyle\left( i \right) \to .

\displaystyle\left( j \right) \to |\sin(a_n)| \leqslant a_n.

\displaystyle\left( k \right) \to .

\displaystyle\left( l \right) \to .

\displaystyle\left( m \right) \to .

\displaystyle\left( n \right) \to .

\displaystyle\left( o \right) \to .

\displaystyle\left( p \right) \to .

\displaystyle\begin{gathered} \left( q \right) \to a_n^{\frac{3}{2}} \leqslant a_n \hfill \\\left( r \right) \to a_n^3 \leqslant a_n \hfill \\\end{gathered}

\displaystyle\left( s \right) \to \frac{{a_1 }}{n} \leqslant \frac{{a_1 + ... + a_n }}{n}.

\displaystyle\left( t \right) \to .

\displaystyle\left( u \right) \to a_n = \frac{1}{{n^2 }}.

\displaystyle\left( v \right) \to .

\displaystyle\left( w \right) \to a_n = \frac{1}{{n^n }}.

\displaystyle\left( x \right) \to |a_n-a_{n+1}| \leqslant a_n + a_{n+1}.

\displaystyle\left( y \right) \to .

October 16, 2007

1 = …?

Filed under: Các Bài Tập Nhỏ — Ngô Quốc Anh @ 23:12

S=1
    =1\left(\boxed{\frac{1}{2}}+\frac{1}{2}\right) 
    =1\left(\frac{1}{2}+\frac{1}{2}\left(\boxed{\frac{1}{3}}+\frac{2}{3}\right)\right) 
    =1\left(\frac{1}{2}+\frac{1}{2}\left(\frac{1}{3}+\frac{2}{3}\left(\boxed{\frac{1}{5}}+\frac{4}{5}\right)\right)\right) 
    =1\left(\frac{1}{2}+\frac{1}{2}\left(\frac{1}{3}+\frac{2}{3}\left(\frac{1}{5}+\frac{4}{5}\left(\boxed{\frac{1}{7}}+\frac{6}{7}\right)\right)\right)\right) 
    =1\left(\frac{1}{2}+\frac{1}{2}\left(\frac{1}{3}+\frac{2}{3}\left(\frac{1}{5}+\frac{4}{5}\left(\frac{1}{7}+\frac{6}{7}\left(\boxed{\frac{1}{11}}+\frac{10}{11}\right)\right)\right)\right)\right) 
    =1\left(\frac{1}{2}+\frac{1}{2}\left(\frac{1}{3}+\frac{2}{3}\left(\frac{1}{5}+\frac{4}{5}\left(\frac{1}{7}+\frac{6}{7}\left(\frac{1}{11}+\frac{10}{11}\left(\boxed{\frac{1}{13}}+\frac{12}{13}\right)\right)\right)\right)\right)\right)\right) 
    =1\left(\frac{1}{2}+\frac{1}{2}\left(\frac{1}{3}+\frac{2}{3}\left(\frac{1}{5}+\frac{4}{5}\left(\frac{1}{7}+\frac{6}{7}\left(\frac{1}{11}+\frac{10}{11}\left(\frac{1}{13}+\frac{12}{13}\left(\boxed{\frac{1}{17}}+\frac{16}{17}\right)\right)\right)\right)\right)\right)\right)\right) 
    \ldots\;\;\;\ldots 

S=\sum_{n=1}^\infty\;\left\{\frac{1}{p_n-1}\prod_{k=1}^n\;\left(1-\frac{1}{p_k}\right)\right\} 

 
the thing is anyone can basically start from 1 and go on to concoct any type of series and then ask for it’s sum.

Đề thi giữa kỳ GT4 của K51

Filed under: Đề Thi — Ngô Quốc Anh @ 22:42

ktgk_k51-a1t_gt3.pdf

Đề thi giữa kỳ GT3 của K51

Filed under: Đề Thi — Ngô Quốc Anh @ 22:40

ktgk_k51-a1t_gt2b.pdf

Đề thi giữa kỳ GT2 của K51

Filed under: Đề Thi — Ngô Quốc Anh @ 22:38

ktgk_k51-a1s_gt2a.pdf

ktgk_k51-a1t_gt2a.pdf

ktgk_k51-a23_gt2a_1.pdf

Đề thi giữa kỳ GT1 của K51

Filed under: Đề Thi — Ngô Quốc Anh @ 22:37

ktgk_k51-a1s_gt1.pdf

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