# Ngô Quốc Anh

## September 26, 2008

Filed under: Giải Tích 6 (MA5205) — Ngô Quốc Anh @ 14:58

Giới thiệu với các bạn problem sets của MA5205 mà tôi phải học ở học kỳ này.

1. Basic topology: ma5205t1
2. Functions of bounded varitation: ma5205t2
3. Measure (Outer, Inner, Lebesgue): ma5205t3
4. Measurable functions: ma5205t4
5. Lebesgue integral: ma5205t5
6. Repeated integration: ma5205t6
7. Lebesgue’s differentiation theorem: ma5205t7
8. $L^p$ spaces: ma5205t8
9. Approximations of the identity and maximal functions: ma5205t9

Đề thi:

Giáo trình:

• R. L. Wheeden and A. Zygmund, Measure and integral – An Introduction to Real Analysis, Maecel Dekker, Inc.

## September 17, 2008

### L^2 differentiable?

Filed under: Các Bài Tập Nhỏ, Giải Tích 2, Giải Tích 3 — Ngô Quốc Anh @ 13:04

Let  and define .

a) Must  be differentiable at 0?
b) Must  have any differentiable points?
c) Let , show that  exists and determine what it is.

Solutions.

a) No. For example, let  for  so that  near zero. This  but  does not exist.

b) Yes. In fact,  must be differentiable almost everywhere, by the Lebesgue theorem on the differentiation of the integral. This theorem requires only that  which is true.

c) By the Schwarz inequality,

$f^2(x) = \left(\int_0^x}g(t)\,dt\right)^2\le \left(\int_0^x 1\,dt\right) \left(\int_0^xg^2(t)\,dt\right).$

At least that’s it for  Being careful about the other side, we determine that

$0\le\phi(x) = f^2(x)\le|x|\left|\int_0^xg^2(t)\,dt\right|.$

But since  is an integrable function we have (by an argument that uses the Dominated Convergence Theorem) that

$\lim_{x\to 0}\int_0^xg^2(t)\,dt = 0.$

Hence $\lim_{x\to0}\frac {\phi(x)}{x} = 0,$ so 

### Lời giải của Euler cho bài toán Basel về tính tổng của một chuỗi số

Filed under: Các Bài Tập Nhỏ, Giải Tích 4 — Ngô Quốc Anh @ 12:50

The Basel problem is a famous problem in number theory, first posed by Pietro Mengoli in 1644, and solved by Leonhard Euler in 1735. Since the problem had withstood the attacks of the leading mathematicians of the day, Euler’s solution brought him immediate fame when he was twenty-eight. Euler generalised the problem considerably, and his ideas were taken up years later by Bernhard Riemann in his seminal 1859 paper On the Number of Primes Less Than a Given Magnitude, in which he defined his zeta function and proved its basic properties. The problem is named after Basel, hometown of Euler as well as of the Bernoulli family, who unsuccessfully attacked the problem.

The Basel problem asks for the precise summation of the reciprocals of the squares of the natural numbers, i.e. the precise sum of the infinite series:

$\displaystyle\sum_{n=1}^\infty \frac{1}{n^2} = \lim_{n \to +\infty}\left(\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2}\right)$.

The series is approximately equal to 1.644934 (sequence A013661 in OEIS). The Basel problem asks for the exact sum of this series (in closed form), as well as a proof that this sum is correct. Euler found the exact sum to be $\frac{\pi^2}{6}$ and announced this discovery in 1735. His arguments were based on manipulations that were not justified at the time, and it was not until 1741 that he was able to produce a truly rigorous proof.

Proof and comments. Euler’s original “derivation” of the value $\frac{\pi^2}{6}$ is clever and original. He essentially extended observations about finite polynomials and assumed that these same properties hold true for infinite series. Of course, Euler’s original reasoning requires justification, but even without justification, by simply obtaining the correct value, he was able to verify it numerically against partial sums of the series. The agreement he observed gave him sufficient confidence to announce his result to the mathematical community.

To follow Euler’s argument, recall the Taylor series expansion of the sine function

$\displaystyle\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$.

Dividing through by $x$, we have

$\displaystyle\frac{\sin(x)}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \cdots$.

Now, the roots (zeros) of $\frac{\sin x}{x}$ occur precisely at $x = n\cdot\pi$ where $n = \pm1, \pm2, \pm3, \dots$. Let us assume we can express this infinite series as a product of linear factors given by its roots, just as we do for finite polynomials

$\displaystyle\begin{gathered}\frac{{\sin (x)}}{x} = \left( {1 - \frac{x}{\pi }} \right)\left( {1 + \frac{x}{\pi }} \right)\left( {1 - \frac{x}{{2\pi }}} \right)\left( {1 + \frac{x}{{2\pi }}} \right)\left( {1 - \frac{x}{{3\pi }}} \right)\left( {1 + \frac{x}{{3\pi }}} \right) \cdots\hfill \\ \qquad \quad= \left( {1 - \frac{{{x^2}}}{{{\pi ^2}}}} \right)\left( {1 - \frac{{{x^2}}}{{4{\pi ^2}}}} \right)\left( {1 - \frac{{{x^2}}}{{9{\pi ^2}}}} \right) \cdots\hfill \\ \end{gathered}$

If we formally multiply out this product and collect all the $x^2$ terms, we see that the $x^2$ coefficient of $\frac{\sin x}{x}$ is

$\displaystyle -\left(\frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \cdots \right) = -\frac{1}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}$.

But from the original infinite series expansion of $\frac{\sin x}{x}$, the coefficient of $x^2$ is $-\frac{1}{3!} = -\frac{1}{6}$. These two coefficients must be equal; thus,

$\displaystyle -\frac{1}{6} = -\frac{1}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}$.

Multiplying through both sides of this equation by $-\pi^2$ gives the sum of the reciprocals of the positive square integers

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}$.

## September 11, 2008

### Một bài giới hạn khó về hàm lượng giác

Filed under: Các Bài Tập Nhỏ, Giải Tích 1 — Ngô Quốc Anh @ 22:02

Show that for any irrational $\alpha$ the limit

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \sin \left( {n\alpha \pi } \right)$

does not exist.

Solution. If the limit existed then we would get

$\displaystyle 0 = \mathop {\lim }\limits_{n \to \infty } \left( {\sin \left( {\left( {n + 2} \right)\alpha \pi } \right) - \sin \left( {n\alpha \pi } \right)} \right) = 2\sin \left( {\alpha \pi } \right)\mathop {\lim }\limits_{n \to \infty } \cos \left( {\left( {n + 1} \right)\alpha \pi } \right)$

and consequently, $\mathop {\lim }\limits_{n \to \infty } \cos \left( {n\alpha \pi } \right) = 0$. Similarly,

$\displaystyle0 = \mathop {\lim }\limits_{n \to \infty } \left( {\cos \left( {\left( {n + 2} \right)\alpha \pi } \right) - \cos \left( {n\alpha \pi } \right)} \right) = - 2\sin \left( {\alpha \pi } \right)\mathop {\lim }\limits_{n \to \infty } \cos \left( {\left( {n + 1} \right)\alpha \pi } \right)$

which is impossible because $\sin^2x+\cos^2x=1$ for all $x \in \mathbb R$. Therefore the limit does not exist.