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)<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

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

Filed under: 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 }.

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