Questions in category: 数论 (Number Theory)

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## 1. p 进数

Posted by haifeng on 2022-07-05 21:02:56 last update 2022-07-05 21:02:56 | Answers (0) | 收藏

>> decimal2binary(20)
in> decimal2binary(20)
out> 10100

------------------------

>> reverse(10100)
in> reverse(10100)
out> 00101

## 2. 代数整数(Algebraic Integer)

Posted by haifeng on 2015-06-05 10:08:18 last update 2015-06-05 10:08:18 | Answers (0) | 收藏

$x^n+a_{n-1}x^{n-1}+\cdots+a_1 x+a_0=0$

## 3. Minkowski 定理

Posted by haifeng on 2015-05-01 09:03:38 last update 2015-05-01 09:03:38 | Answers (0) | 收藏

$\text{vol}(S) > 2^n d(L),$

Reference:

http://en.wikipedia.org/wiki/Minkowski%27s_theorem

## 4. Kronecker-Чeбыщeв 定理

Posted by haifeng on 2015-04-30 12:19:09 last update 2015-04-30 13:21:38 | Answers (0) | 收藏

Kronecker-Чeбыщeв 定理

$|p\alpha-q+\beta| < \frac{3}{p}.$

$\lim_{k\rightarrow\infty}\sin n_k=\sin\beta.$

$2p_k\pi-q_k+\beta=o(1).$

References:

http://en.wikipedia.org/wiki/Dirichlet%27s_approximation_theorem

http://en.wikipedia.org/wiki/Kronecker%27s_theorem

## 5. 相邻 $k$-free 整数之间的间隙

Posted by haifeng on 2014-05-10 12:01:12 last update 2014-05-10 14:28:29 | Answers (0) | 收藏

On the gaps between consecutive $k$-free integers

H. Halberstam & K. F. Roth

http://jlms.oxfordjournals.org/content/s1-26/4/268.extract

1. J. London Math. Soc. s1-26 (4): 268-273. doi: 10.1112/jlms/s1-26.4.268

## 6. 从 $\mathbb{N}\setminus\{1\}$ 中删去 2 和 3 的倍数, 得到数列 $\{u_n\}_{n=1}^{\infty}$, 证明一些性质

Posted by haifeng on 2014-05-10 09:23:48 last update 2014-05-10 09:23:48 | Answers (0) | 收藏

Claim 1. 从 $\mathbb{N}\setminus\{1\}$ 中删去 2 和 3 的倍数, 得到数列 $\{u_n\}_{n=1}^{\infty}$, 证明

$\begin{cases} u_{2k+1}=u_{2k}+4\\ u_{2k}=u_{2k-1}+2\\ \end{cases}$

Question.  从 $\mathbb{N}\setminus\{1\}$ 中删去 2, 3, 5 的倍数, 得到数列 $\{u_n\}_{n=1}^{\infty}$, 有什么性质?

## 7. 有哪些已经被证明为超越数的数?

Posted by haifeng on 2014-01-03 15:59:59 last update 2021-09-22 09:01:02 | Answers (0) | 收藏

Def. $\alpha$ 是代数数(algebraic), 如果存在 $p\in\mathbb{Z}[x]$, $p\neq 0$, 使得 $p(\alpha)=0$. 否则称 $\alpha$ 是超越数(transcendental).

$e$, $\pi$, $\ln 2$

$e^{\pi}$, $\pi+e^\pi$, $\pi e^\pi$, $e^{\pi\sqrt{n}}$.

Hilbert 数 （与 Hilbert 的第七问题有关）

$2^{\sqrt{2}}$,

Liouville\'s number

$0.110001000000000000000001000\ldots$

Chaitin\'s "constant", the probability that a random algorithm halts. (Noam Elkies of Harvard notes that not only is this number transcendental but it is also incomputable.)

Chapernowne\'s number,

$0.12345678910111213141516171819202122232425\ldots$

Apéry\'s constant.  $\zeta(3)$.  (一般总可期望超越函数在某些有理点处能给出超越数值.)

Morse-Thue\'s number, 0.01101001 ...

$i^i=e^{i\text{Log}i}=e^{i[\log|i|+i\text{Arg}i]}=e^{-\frac{\pi}{2}}=0.207879576\ldots$

Euler 常数,

$\gamma=\lim_{n\rightarrow\infty}\biggl(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\ln(n)\biggr)= 0.577215 ...$

$\pi^e$

$\zeta(s)$ 在其他奇数处的值, 如 $\zeta(5),\zeta(7),\ldots$ 尚未证明是否是超越数.

Catalan\'s constant, $G$

$G=\sum_{k=0}^{+\infty}\frac{(-1)^k}{(2k+1)^2}=1-\frac{1}{9}+\frac{1}{25}-\frac{1}{49}+\cdots$

Feigenbaum numbers, e.g. 4.669 ... . (These are related to properties of dynamical systems with period-doubling. The ratio of successive differences between period-doubling bifurcation parameters approaches the number 4.669 ... , and it has been discovered in many physical systems before they enter the chaotic regime. It has not been proven to be transcendental, but is generally believed to be.)

Keith Briggs from the Mathematics Department of the University of Melbourne in Australia computed what he believes to be the world-record for the number of digits for the Feigenbaum number:

4. 669201609102990671853203820466201617258185577475768632745651 343004134330211314737138689744023948013817165984855189815134 408627142027932522312442988890890859944935463236713411532481 714219947455644365823793202009561058330575458617652222070385 410646749494284981453391726200568755665952339875603825637225

Briggs carried out the computation using special-purpose software designed by David Bailey of NASA Ames running on an IBM RISC System/6000. The computation required a few hours of computation time.

1844 年, 数学天才约瑟夫.柳维尔(Joseph Liouville (1809-1882)) 首先证明了超越数的存在性. (更确切的说法是, 他证明了某个特定的数是超越数.) 柳维尔给出了一个数是代数数的必要条件, 从而也就给出了一个数成为超越数的充分条件. 于是证明了超越数的存在.

Charles Hermite 在 1873 年证明了 $e$ 的超越性.

Ferdinand von Lindemann 于 1882 年证明了 $\pi$ 的超越性.

References:

http://en.wikipedia.org/wiki/Alexander_Gelfond

A. K. 苏凯什维奇 著, 叶乃膺 译 《数论初等教程》

The 15 Most Famous Transcendental Numbers

http://mathworld.wolfram.com/TranscendentalNumber.html

http://en.wikipedia.org/wiki/Transcendental_number

## 8. 关于 $\pi$ 和 $e$ 的公式

Posted by haifeng on 2014-01-03 11:50:49 last update 2014-01-03 12:14:23 | Answers (0) | 收藏

$e^{i\pi}=-1$

$\sqrt{\frac{\pi e}{2}}=1+\frac{1}{1\cdot 3}+\frac{1}{1\cdot 3\cdot 5}+\frac{1}{1\cdot 3\cdot 5\cdot 7}+\cdots$

References:

http://www.math.niu.edu/~rusin/known-math/98/pi_e

## 9. $\pi+re,r\in\mathbb{Q}$ 和 $\pi e$, $\pi/e$ 这些数中至多只能有一个是有理数

Posted by haifeng on 2014-01-01 16:38:30 last update 2014-01-03 20:53:46 | Answers (0) | 收藏

$x^2-(\pi+e)x+\pi e=0$

$\pi+ne$ 和 $\pi e$ 至少有一个是无理数.
$\pi+\frac{p}{q}e$ 和 $\pi e$ 至少有一个是无理数, 其中 $p,q$ 是互素的整数.

$\lim_{n\rightarrow\infty}(\pi+\frac{p_n}{q_n}e)=\pi e.$

$e^n$ 是无理数.

$(r_1-r_2)e=r_3-r_4,$

Cor. $\{\pi+re\mid r\in\mathbb{Q}\}$ 中至多只有一个是有理数.

• 若 $\pi e\in\mathbb{Q}$, 则 $\pi/e\in\mathbb{Q}^c$.
• 若 $\pi/e\in\mathbb{Q}$, 则 $\pi e\in\mathbb{Q}^c$.

$\pi e$ 和 $\pi/e$ 至少有一个是无理数.

$\{\pi+re,\quad\pi e,\quad\frac{\pi}{e}\mid r\in\mathbb{Q}\}$

中至多只有一个有理数.

Remark:

Hermite proved that e is transcendental in 1873, and Lindemann proved that pi is transcendental in 1882. In fact,
Lindemann\'s proof was similar to Hermite\'s proof and was based on the fact that e is also transcendental.

References:

http://en.wikipedia.org/wiki/Irrational_number

http://math.stackexchange.com/questions/159350/why-is-it-hard-to-prove-whether-pie-is-an-irrational-number

http://mathforum.org/library/drmath/view/51617.html

## 10. [Zhuo (Aubrey) Yang]Primes in Arbitrarily Long Arithmetic Progression

Posted by haifeng on 2013-01-09 17:10:59 last update 2013-01-09 17:10:59 | Answers (0) | 收藏

Primes in Arbitrarily Long Arithmetic Progression
By Zhuo (Aubrey) Yang
Advisor: Yum-Tong Siu and Leslie G. Valiant
March 29, 2012

## 1. 摘要

$43142746595714191+23681770223092870n,\quad n = 0, 1,\ldots, 25.$

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