# Ngô Quốc Anh

## September 17, 2008

### 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

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

## August 9, 2008

### Tính tích phân suy rộng loại I

Filed under: Các Bài Tập Nhỏ, Giải Tích 2, Giải Tích 4 — Ngô Quốc Anh @ 13:16

Tính

$\displaystyle \int_{0}^{+\infty}{\frac{e^{-x^2}}{\left(x^2+\frac{1}{2}\right)^2}\; dx}$.

Lời giải. Đặt

$\displaystyle I(\alpha) = - \int_{0}^{\infty} \frac {e^{ - x^2}}{x^2 + \alpha} \, dx$.

Khi đó ta có

$\displaystyle\begin{gathered} \int_0^\infty {\frac{{{e^{ - {x^2}}}}}{{{x^2} + \alpha }}} \,dx = \int_0^\infty {{e^{ - {x^2}}}} \int_0^\infty {{e^{ - ({x^2} + \alpha )t}}} \,dtdx \hfill \\\qquad \qquad \qquad \; \,= \int_0^\infty {{e^{ - \alpha t}}} \int_0^\infty {{e^{ - (1 + t){x^2}}}} \,dxdt \hfill \\ \qquad \qquad \qquad \; \,= \int_0^\infty {\frac{{\sqrt \pi }}{2}} \frac{{{e^{ - \alpha t}}}}{{\sqrt {1 + t} }}\,dt, \hfill \\ \end{gathered}$

ta thấy

$\displaystyle\int_{0}^{\infty} \frac {e^{ - x^2}}{(x^2 + \frac {1}{2})^2} \, dx = I'(\tfrac{1}{2}) = \frac {\sqrt {\pi}}{2} \int_{0}^{\infty} \frac {t}{\sqrt {1 + t}} \, e^{ - \frac {t}{2}} \, dt$.

Nhưng

$\displaystyle \begin{gathered} \int_0^\infty {\frac{t}{{\sqrt {1 + t} }}} \,{e^{ - \frac{t}{2}}}\,dt = \int_0^\infty {\sqrt {1 + t} } \,{e^{ - \frac{t}{2}}}\,dt - \int_0^\infty {\frac{1}{{\sqrt {1 + t} }}} \,{e^{ - \frac{t}{2}}}\,dt \hfill \\ \qquad \qquad\qquad \qquad= \left[ { - 2\sqrt {1 + t} \,{e^{ - \frac{t}{2}}}} \right]_0^\infty + \int_0^\infty {\frac{1}{{\sqrt {1 + t} }}} \,{e^{ - \frac{t}{2}}}\,dt - \int_0^\infty {\frac{1}{{\sqrt {1 + t} }}} \,{e^{ - \frac{t}{2}}}\,dt \hfill \\ \qquad \qquad\qquad \qquad= 2. \hfill \\\end{gathered}$

Vậy

$\displaystyle\int_{0}^{\infty} \frac {e^{ - x^2}}{(x^2 + \frac {1}{2})^2} \, dx = \sqrt {\pi}$.

## March 30, 2008

### Một số bài tập Giải tích của Khoa Toán đại học Florida

Filed under: Giải Tích 2, Giải Tích 3, Giải Tích 4, Giải Tích 5 — Ngô Quốc Anh @ 12:04

## March 22, 2008

### Khảo sát sự hội cụ của chuỗi số liên quan đến hàm \sin

Filed under: Các Bài Tập Nhỏ, Giải Tích 4 — Ngô Quốc Anh @ 1:46

Đây là một bài tập khá hay. Khảo sát sự hội tụ của chuỗi số $\sum\limits_{n = 1}^{\infty} \sin(\pi(2 + \sqrt {3})^{n})$.

Lời giải.

## January 31, 2008

### GT4 – Bài Tập 05

Filed under: Giải Tích 4 — Ngô Quốc Anh @ 21:43

Lấy bài tập ở đây: gt4-05.pdf

## January 7, 2008

### GT4 – Bài Tập 06

Filed under: Giải Tích 4 — Ngô Quốc Anh @ 0:17

Lấy bài tập ở đây: gt4-06.pdf

## January 6, 2008

### GT4 – Bài Tập 04

Filed under: Giải Tích 4 — Ngô Quốc Anh @ 23:39

Lấy bài tập ở đây: gt4-04.pdf

### GT4 – Bài Tập 03

Filed under: Giải Tích 4 — Ngô Quốc Anh @ 21:31

Lấy bài tập ở đây: gt4-03.pdf

## January 5, 2008

### GT4 – Bài Tập 02

Filed under: Giải Tích 4 — Ngô Quốc Anh @ 23:44

Lấy bài tập ở đây: gt4-02.pdf

### GT4 – Bài Tập 01

Filed under: Giải Tích 4 — Ngô Quốc Anh @ 21:04

Lấy bài tập ở đây: gt4-01.pdf

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