# Ngô Quốc Anh

## April 19, 2008

### Bài tập hay về metric

Filed under: Giải Tích 3 — Ngô Quốc Anh @ 0:00

Let $(X,d)$ be a metric space and $f: \mathbb R \to \mathbb R$ a function with $f(0)=0$, $f'(t) >0$, $f''(t) \leq 0$ for $t>0$. Prove that $f \circ d$ is a metric on $X$.

Solution. The function $f$ is increasing and concave down. Let $d'=fd$

1. Since $d'(x,y)=0$ then $d(x,y)=0$ (because $f$ is increasing), so $x=y$. Of course, $d'(x,y)>0$ for $x\ne y$ (because $f$ is increasing) and $d'(x,x)=0$ because $f(0)=0$.
2. Symmetry is easy because $d$ is so.
3. $d'(x,z)=f(d(x,z)) \leq f(d(x,y)+d(y,z))$ (because $f$ is increasing) $\leq f(d(x,y))+f(d(y,z))$ (because $f(0)=0$ and $f$ is concave down), so triangle inequality is verified.

Remark. You want to prove $f(a+b)\leq f(a)+f(b)$ (then you can substitute $a = d(x,y)$, $b = d(y,z)$).

By concavity,

$\displaystyle f(a) \ge \frac{a}{a+b} f(a+b) + \frac{b}{a+b} f(0)$

and

$\displaystyle f(b) \ge \frac{b}{a+b} f(a+b) + \frac{a}{a+b}f(0)$.

Sum them up, you are done.

## April 14, 2008

### Toàn bộ đề thi cuối kỳ môn Giải tích của K52-A1T A1S A2 A3

Filed under: Đề Thi — Ngô Quốc Anh @ 21:50

kthk_k52-gt1_deso1

kthk_k52-gt1_deso2

kthk_k52-gt2_deso1

kthk_k52-gt2_deso2

kthk_k52-gt3_deso1

kthk_k52-gt3_deso2

kthk_k52-gt4_deso1

kthk_k52-gt4_deso2

kthk_k52-gt5_deso1

kthk_k52-gt5_deso2

### Toàn bộ đề thi cuối kỳ môn Giải tích của K51-A1T A1S A2 A3

Filed under: Đề Thi — Ngô Quốc Anh @ 14:08

### Toàn bộ đề thi cuối kỳ môn Giải tích của K50-A1T A1S A2 A3

Filed under: Đề Thi — Ngô Quốc Anh @ 14:00

kthk_k50_gt1_deso1

kthk_k50_gt1_deso2

kthk_k50_gt2_deso1

kthk_k50_gt2_deso2

kthk_k50_gt3_deso1

kthk_k50_gt3_deso2

kthk_k50_gt4_deso1

kthk_k50_gt4_deso2

kthk_k50_gt5_deso1

kthk_k50_gt5_deso2

## April 7, 2008

### Integral with fraction of Min and Max of a function

Filed under: Giải Tích 2 — Ngô Quốc Anh @ 2:37

Let  be a positive non-zero integer. Prove that:

$\int_0^1 \frac {\min \left(x^n,(1 - x)^n\right)}{\max \left(x^n,(1 - x)^n\right)} \, dx = 2 n ( - 1)^{n - 1} \left(\log (2) + \sum _{i = 1}^{n - 1} \frac {( - 1)^i}{i}\right) - 1$.

Hint. Simplify the integral to:

$\int_0^1 \frac {\min \left(x^n,(1 - x)^n\right)}{\max \left(x^n,(1 - x)^n\right)} \, dx = \int_0^{\frac {1}{2}} \left(\frac {1}{x} - 1\right)^{ - n} \, dx + \int_{\frac {1}{2}}^1 \left(\frac {1}{x} - 1\right)^n \, dx$.

Use the Symmetry

$\int_0^{\frac {1}{2}} \left(\frac {1}{x} - 1\right)^{ - n} \, dx = \int_{\frac {1}{2}}^1 \left(\frac {1}{x} - 1\right)^n \, dx$.

Use newton

$\left(\frac {1}{x} - 1\right)^n = ( - 1)^n + \frac {( - 1)^{n - 1} n}{x} + \sum _{i = 2}^n x^{ - i} ( - 1)^{n - i} \left( \begin{array}{c} n \\ i \end{array} \right)$.

Integrate the last sum and prove that:

$\sum _{i = 2}^n \frac {( - 1)^{n - i} \left(2^i - 2\right) }{2 (i - 1)}\left( \begin{array}{c} n \\ i \end{array} \right) = n ( - 1)^{n - 1} \sum _{i = 1}^{n - 1} \frac {( - 1)^i}{i} - \frac{1}{2} \left((-1)^n+1\right)$.

## April 5, 2008

### More integral involving Gamma function

Filed under: Giải Tích 2 — Ngô Quốc Anh @ 14:58

Show that

$\int_0^1\;e^{ - \alpha\,x}\;\ln\,x\;\textbf dx\; = \;\boxed{ - \dfrac{1}{\alpha}\left(\ln\,\alpha + \gamma + \Gamma(0,\alpha)\right)}$

where  represents the incomplete gamma function.

Solution. With  , we have
$\mathcal{I} = \frac {1}{\alpha} \int_0^{\alpha} e^{ - t} \ln t \, dt - \frac {\ln \alpha}{\alpha} \int_0^{\alpha} e^{ - t} \, dt =$
$= \frac {1}{\alpha} \left\{\left( \int_0^{ + \infty} - \int_{\alpha}^{ + \infty} \right) e^{ - t} \ln t dt - (1 - e^{ - \alpha})\ln \alpha \right\}$
$= \frac {1}{\alpha} \left\{ - \gamma - \ln \alpha + e^{ - \alpha} \ln \alpha - \underbrace{\int_{\alpha}^{ + \infty}e^{ - t}\ln t \, dt}_{\mathcal{J}} \right\}$
To compute J we integrate by parts:
$\mathcal{J} = \left[ - e^{ - t} \ln t \right]_{\alpha}^{ + \infty} + \int_{\alpha}^{ + \infty} \frac {e^{ - t}}{t} dt = e^{ - \alpha} \ln \alpha + \underbrace{\int_{ - \infty}^{ - \alpha} \frac {e^{x}}{x} \, dx}_{\mbox{Ei}( - \alpha) = - \Gamma (0, \alpha)}$
And Finally:
$\mathcal{I} = - \frac {1}{\alpha} \left\{\gamma + \ln \alpha + \Gamma (0, \alpha) \right\}$.

Done.

## April 1, 2008

### Find all derivable functions

Filed under: Các Bài Tập Nhỏ, Giải Tích 1 — Ngô Quốc Anh @ 22:25

Find all derivable functions $\mathbb R\to \mathbb R$, with the property

$f(x+y+z)+3xyz=f(x)+f(y)+f(z)+3(x+y+z)(xy+yz+zx)-2$

for every $x,y,z\in \mathbb R$.

Solution. $'_x$ with respect to $x$ to both sides

$f'\left( {x + y + z} \right) - 3\left( {x + y + z} \right)^2 = f'\left( x \right) - 3x^2$.

Thus

$f'\left( x \right) - 3x^2$

is a constant.

Note. Actually with only ”$f$ is derivable in one point” you obtain the same result

$x=y=z=0 \Longrightarrow f(0)=1$.

Suppose that $f$ is derivable in $t_0$

$(x=t_0)\& (z=0) \Longrightarrow \frac {f(y)-1}y = \frac {f(t_0+y)-f(t_0)}y-3t_0(t_0+y)$,

so $f$ is derivable in $0$. But

$\frac {f(x+y)-f(x)}y = 3x(x+y)+\frac {f(y)-1}y$

shows that $f(x)$ is derivable for all $x\in \mathbb R$: $f'(x) = 3x^2 + f'(0)$ and so on.

### Klein bottle

Filed under: Linh Tinh — Ngô Quốc Anh @ 17:52

In mathematics, the Klein bottle is a certain non-orientable surface, i.e., a surface (a two-dimensional topological space) with no distinct “inner” and “outer” sides. Other related non-orientable objects include the Möbius strip and the real projective plane. Whereas a Möbius strip is a two dimensional object with one side and one edge, a Klein bottle is a two dimensional object with one side and no edges. (For comparison, a sphere is a two dimensional object with no edges and two sides.)

The Klein bottle was first described in 1882 by the German mathematician Felix Klein. It was originally named the Kleinsche Fläche “Klein surface”; however, this was incorrectly interpreted as Kleinsche Flasche “Klein bottle”, which ultimately led to the adoption of this term in the German language as well.

The Klein bottle immersed in three-dimensional space.