# Ngô Quốc Anh

## September 26, 2009

### How to calculate limit by using definition of definite integral?

Filed under: Các Bài Tập Nhỏ, Giải Tích 2 — Ngô Quốc Anh @ 14:23

Let $f : [a, b] \to \mathbb R$ be a continuous function, not necessarily nonnegative. Partition $[a, b]$ into $n$ consecutive sub-intervals $[x_{i-1}, x_i]$ ($i = 1, 2, ..., n$) each of length $\Delta x = \frac{b-a}{n}$, where we set $a=x_0$, $b=x_n$ and $x_1, x_2,...,x_{n-1}$ to be successive points between $a$ and $b$ with $x_k-x_{k-1}=\Delta x$. Let $c_k$ be any intermediate point in the sub-interval $[x_{k-1},x_k]$. Then the sum

$\displaystyle\sum\limits_{k = 1}^n {f\left( {{c_k}} \right)\Delta x}$

is called a Riemann sum for $f$ on $[a, b]$.

Suppose we let the number of partition in $P$ tends to infinity.

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {f\left( {{c_k}} \right)\Delta x} = I.$

We call $I$ the Riemann integral (or definite integral) of $f$ over $[a, b]$ and we write

$\displaystyle I = \int_a^b {f(x)dx} .$

In other words,

$\displaystyle\int_a^b {f(x)dx} = \frac{{b - a}}{n}\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {f\left( {{c_k}} \right)}$

if the limit on the right side exists.

If we put $c_k=x_{k-1}$ we the obtain

$\displaystyle\int_a^b {f(x)dx} = \frac{{b - a}}{n}\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {f\left( {a + (k - 1)\frac{{b - a}}{n}} \right)} .$

Example 1. Find

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{n + 1}} + \cdots + \frac{1}{{2n - 2}} + \frac{1}{{2n - 1}}} \right).$

Solution. Clearly

$\displaystyle\frac{1}{n} + \frac{1}{{n + 1}} + \cdots + \frac{1}{{2n - 2}} + \frac{1}{{2n - 1}} = \sum\limits_{k = 1}^n {\frac{1}{{n + (k - 1)}}} = \frac{1}{n}\sum\limits_{k = 1}^n {\frac{1}{{1 + \frac{{k - 1}}{n}}}} .$

Then if we choose $a=0$, $b=1$ we then get

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{n + 1}} + \cdots + \frac{1}{{2n - 2}} + \frac{1}{{2n - 1}}} \right) = \int_0^1 {\frac{{dx}}{{1 + x}}} .$

With this it is easy to see that

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{n + 1}} + \cdots + \frac{1}{{2n - 2}} + \frac{1}{{2n - 1}}} \right) = \ln 2$

since

$\displaystyle\int_0^1 {\frac{{dx}}{{1 + x}}} = \ln 2.$

If we put $c_k=x_k$ we the obtain

$\displaystyle\int_a^b {f(x)dx} = \frac{{b - a}}{n}\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {f\left( {a + k\frac{{b - a}}{n}} \right)} .$

Example 2. Find

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\frac{n}{{{n^2} + {1^2}}} + \frac{n}{{{n^2} + {2^2}}} + \cdots + \frac{n}{{{n^2} + {{(n - 1)}^2}}} + \frac{n}{{{n^2} + {n^2}}}} \right).$

Solution. Clearly,

$\displaystyle\frac{n}{{{n^2} + {1^2}}} + \frac{n}{{{n^2} + {2^2}}} + \cdots + \frac{n}{{{n^2} + {n^2}}} = \sum\limits_{k = 1}^n {\frac{n}{{{n^2} + {k^2}}}} = \frac{1}{n}\sum\limits_{k = 1}^n {\frac{1}{{1 + {{\left( {\frac{k}{n}} \right)}^2}}}}$

which yields

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\frac{n}{{{n^2} + {1^2}}} + \frac{n}{{{n^2} + {2^2}}} + \cdots + \frac{n}{{{n^2} + {n^2}}}} \right) = \int_0^1 {\frac{{dx}}{{1 + {x^2}}}} = \frac{\pi }{4}.$

Remark. It is worth mentioning that in general it is not true that

$\displaystyle\lim \left( {summation} \right) = summation\left( {\lim } \right).$

For example, we all know that for each fixed $k$

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \frac{1}{{n + k}} = 0$

but

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\sum\limits_{k = 1}^n {\frac{1}{{n + k}}} } \right) = \ln 2 \ne 0 = \sum\limits_{k = 1}^n {\left( {\mathop {\lim }\limits_{n \to \infty } \frac{1}{{n + k}}} \right)} .$

The point is

$\displaystyle\lim \left( {summation} \right) = summation\left( {\lim } \right)$

holds true only for finite summation.

## September 25, 2009

### An introduction to Yamabe problem

Filed under: Nghiên Cứu Khoa Học, PDEs, Riemannian geometry — Tags: — Ngô Quốc Anh @ 15:51

Hidehiko Yamabe, in his famous paper entitled On a deformation of Riemannian structures on compact manifolds, Osaka Math. J. 12 (1960), pp. 21-37,  wanted to solve the Poincaré conjecture. For this he thought, as a first step, to exhibit a metric with constant scalar curvature. He considered conformal metrics (the simplest change of metric is a conformal one), and gave a proof of the following statement:

On a compact Riemannian manifold $(M, g)$, there exists a metric $g'$ conformal to $g$, such that the corresponding scalar curvature $R'$ is constant”.

The Yamabe problem was born, since there is a gap in Yamabe’s proof. Now there are many proofs of this statement.

Let us recall the question. Let $(M_n, g)$ be a compact $C^\infty$-Riemannian manifold of dimension $n \geq 3$, is its scalar curvature. The problem is:

Does there exists a metric $g'$, conformal to $g$, such that the scalar curvature $R'$ of the metric $g$ is constant?”.

Let us consider the conformal metric $g'=e^fg$ with $f \in C^\infty$. If $\Gamma'^l_{ij}$ and $\Gamma_{ij}^l$ denote the Chrisoffel symbols relating to $g'$ and $g$, respectively, then

$\displaystyle\Gamma '^l_{ij} - \Gamma _{ij}^l = \frac{1} {2}\left( {{g_{kj}}\frac{{\partial f}} {{\partial {x_i}}} + {g_{ki}}\frac{{\partial f}} {{\partial {x_j}}} - {g_{ij}}\frac{{\partial f}} {{\partial {x_k}}}} \right){g^{kl}} = \frac{1} {2}\left( {\delta _j^l{\partial _i}f + \delta _i^l{\partial _j}f - {g_{ij}}{\nabla ^l}f} \right).$

Clearly,

$\displaystyle R'_{ij}=R'^k_{ikj}=R_{ij}-\frac{n-2}{2}\nabla _{ij}f+\frac{n-2}{4}\nabla _if\nabla _jf-\frac{1} {2}\left(\nabla _\nu^\nu f+\frac{n-2}{2}\nabla^\nu f\nabla _\nu f\right)g_{ij}$

so

$\displaystyle R' = {e^{ - f}}\left( {R - \left( {n - 1} \right)\nabla _\nu ^\nu f - \frac{{\left( {n - 1} \right)\left( {n - 2} \right)}} {4}{\nabla ^\nu }f{\nabla _\nu }f} \right).$

If we consider the conformal deformation in the form $g'=\varphi^\frac{4}{n-2}g$ (with $\varphi \in C^\infty$, $\varphi>0$), the scalar curvature satisfies the equation

$\displaystyle \frac{{4\left( {n - 1} \right)}} {{n - 2}}\Delta \varphi + R\varphi = R'{\varphi ^{\frac{{n + 2}} {{n - 2}}}}$

where $\Delta \varphi = - {\nabla ^\nu }{\nabla _\nu }\varphi$. So, Yamabe problem is equivalent to solving the above equation with $R'=const$ and the solution $\varphi$ must be smooth and strictly positive.

Link to PDF file of the paper Osaka Math. J. 12 (1960), pp. 21-37 can be found here http://projecteuclid.org/euclid.ojm/1200689814

## September 18, 2009

### The method of moving planes: An example inspired by Gidas, Ni, and Nirenberg

Filed under: Linh Tinh, Nghiên Cứu Khoa Học, PDEs — Tags: — 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) for any $x_1>0$ and $x_1^\star 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, 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

## September 13, 2009

### An algebraic identity for 9th level

Filed under: 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 the 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,

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

### QE in Department of Mathematics, National University of Singapore, August 2009

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 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

$\displaystyle\int\limits_0^\pi{{{\left( {f\left( x \right)-\cos x} \right)}^2}dx}\leq\frac{1}{9}$

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