Difference between revisions of "Main Page"

From MWiki
Jump to: navigation, search
(Prime number and Gelfond-Schneider theorem)
 
(68 intermediate revisions by the same user not shown)
Line 1: Line 1:
 
__NOTOC__
 
__NOTOC__
 
= Welcome to MWiki =
 
= Welcome to MWiki =
−
== Theorem of the month ==
+
== Theorems of the month ==
−
=== RU-method ===
+
=== Prime number theorem ===
−
If the linear system (LS) <math>Ax = b \in  {}^{\nu}\mathbb{Q}^{n}</math> can be uniquely solved for <math>n \in {}^{\nu}\mathbb{N}^*</math>, the ''root of unity method (RU-method)'' computes <math>x \in {}^{\nu}\mathbb{Q}^{n}</math> for <math>A \in {}^{\nu}\mathbb{Q}^{n \times n}</math> in <math>\mathcal{O}(n^2)</math>.
 
  
−
=== Proof and algorithm ===
+
For <math>\pi(x) := |\{p \in {\mathbb{P}_{\le x}} : x \in {}^{\omega}{\mathbb{R}}\}|</math>, it holds that <math>\pi(\omega) = \widetilde{{_e}\omega}\omega + \mathcal{O}({_\epsilon}\omega\;{\omega}^{\tilde{2}})</math>.
−
Let <math>R_1 := (r_{1jk}) = (r_{1kj}) = R_1^T \in {}^{\nu}\mathbb{C}^{n \times n}, n \in {}^{\nu}2\mathbb{N}^*, r_{11k} := 1</math> and for <math>j &gt; 1</math> sowie <math>n_{jk} := j + k - 3</math> as well as <math>r_{1kj} := \hat{n}e^{i\tau n_{jk}/n}</math> both <math>n_{jk} &lt; n</math> and <math>r_{1kj} := \hat{n}e^{i\tau(n_{jk} - \acute{n})/n}</math> for <math>n_{jk} \ge n</math>. Interchanging the first and <math>j</math>-th row resp. column position and correspondingly interchanging the remaining row and column positions yields matrices <math>R_j = R_j^T</math> for <math>j &gt; 1</math>. That implies obviously rk<math>(R_j) = n</math>. If <math>A(x - x^\prime) = (1 - x_j, ..., 1 - x_j)^T</math> and <math>Ax^\prime = b</math> imply <math>x_j = 1</math> for all <math>j</math>, then most likely rk<math>(A) = n</math>.
 
  
−
If <math>a_{jk} \le 0</math> is given for at least one couple <math>(j, k)</math> and <math>A := (a_{jk})</math>, then compute the sums <math>s_0 := \sum\limits_{j=1}^m{b_j\varepsilon^j}</math> for an arbitrary transcendental number <math>\varepsilon</math> and <math>s_k := \sum\limits_{j=1}^m{a_{jk}\varepsilon^j} \ne 0</math> for all <math>k</math>. Replace <math>x_k</math> by <math>-x_k</math> for <math>s_k &lt; 0</math>. Then add a multiple of <math>s^Tx</math> resp. <math>s_0</math> to <math>Ax = b</math>, such that now <math>a_{jk} &gt; 0</math> holds for all <math>j, k</math>. From <math>D_j := (d_{jk}), d_{jk} = \delta_{jk}/\prod\limits_{m=1}^n{a_{jm}}</math> and <math>C_j := D_j R_j = (c_{jk})</math>, it follows that <math>x_j^\prime = (AC_jx^\prime)_j = (C_jb)_j</math>. If, however, <math>x_j^\prime = 0 \ne b_j</math> holds for one <math>j</math>, the LS cannot be solved. The transfer into complex numbers is easy.<math>\square</math>
+
==== Proof: ====
 +
From intervals of fix length <math>y \in {}^{\omega}{\mathbb{R}_{>0}}, \check{y}</math> set-2-tuples of prime numbers are formed such that the first interval has the unchanged representative prime number density and the second interval is empty, then the interval with the second most prime number density is followed by the second least one etc. The Stirling formula suggests the prime gap <math>n = {\epsilon}^{\sigma} = \mathcal{O}({_\epsilon}(n!))</math>.
 +
 
 +
For induction basis <math>n = 2</math> resp. 3, the hypothesis states the first interval to contain <math>x_n/{_\epsilon}x_n</math> primes for <math>n \in {}^{\omega}{\mathbb{N}_{\ge2}}</math> and <math>x_4 \in [2, 4[</math>. Stepping from <math>x_n</math> to <math>x_n^2</math> finds <math>\pi(x_n^2) = \pi(x_n) \check{x}_n</math> primes only from <math>\pi(x_n) = x_n/{_\epsilon}x_n</math>. The average prime gap is <math>{_\epsilon}x_n</math>, the maximal one <math>{_\epsilon}x_n^2</math> and the maximal <math>x_n^2</math> to <math>x_n</math> behaves like <math>\omega</math> to <math>{\omega}^{\tilde{2}}.\square</math>
 +
 
 +
=== Gelfond-Schneider theorem ===
 +
 
 +
It holds <math>a^b \notin {}_{\omega}^{\omega} \mathbb{A}_\mathbb{C}</math> where <math>a, c \in {}^{\omega} \mathbb{A}_\mathbb{C} \setminus \mathbb{B}</math> and infinitesimal <math>\varepsilon, b \in {}^{\omega}\mathbb{A}_\mathbb{C} \setminus {}_{\omega}^{\omega}\mathbb{R}</math>.
 +
 
 +
==== Proof: ====
 +
 
 +
The minimal polynomials <math>p</math> (and <math>q</math>) of <math>c^r</math> resp. <math>c^{r\pm\varepsilon} = a^b</math> for maximal <math>r \in {}_{\omega}^{\omega}\mathbb{R}_{>0}</math> and <math>f = p\;(q)</math> lead to the contradiction <math>{}^1f(c^{r(\pm\varepsilon)}) \ne 0 = (f(c^r) - f(c^{r\pm\varepsilon})) / (c^r - c^{r\pm\varepsilon}) = {}^1f(c^{r(\pm\varepsilon)}).\square</math>
 +
 
 +
== MWiki has moved ==
 +
The new URL is: [https://en.hwiki.de/maths.html HWiki]
  
 
== Recommended reading ==
 
== Recommended reading ==

Latest revision as of 18:01, 31 July 2024

Welcome to MWiki

Theorems of the month

Prime number theorem

For [math]\displaystyle{ \pi(x) := |\{p \in {\mathbb{P}_{\le x}} : x \in {}^{\omega}{\mathbb{R}}\}| }[/math], it holds that [math]\displaystyle{ \pi(\omega) = \widetilde{{_e}\omega}\omega + \mathcal{O}({_\epsilon}\omega\;{\omega}^{\tilde{2}}) }[/math].

Proof:

From intervals of fix length [math]\displaystyle{ y \in {}^{\omega}{\mathbb{R}_{\gt 0}}, \check{y} }[/math] set-2-tuples of prime numbers are formed such that the first interval has the unchanged representative prime number density and the second interval is empty, then the interval with the second most prime number density is followed by the second least one etc. The Stirling formula suggests the prime gap [math]\displaystyle{ n = {\epsilon}^{\sigma} = \mathcal{O}({_\epsilon}(n!)) }[/math].

For induction basis [math]\displaystyle{ n = 2 }[/math] resp. 3, the hypothesis states the first interval to contain [math]\displaystyle{ x_n/{_\epsilon}x_n }[/math] primes for [math]\displaystyle{ n \in {}^{\omega}{\mathbb{N}_{\ge2}} }[/math] and [math]\displaystyle{ x_4 \in [2, 4[ }[/math]. Stepping from [math]\displaystyle{ x_n }[/math] to [math]\displaystyle{ x_n^2 }[/math] finds [math]\displaystyle{ \pi(x_n^2) = \pi(x_n) \check{x}_n }[/math] primes only from [math]\displaystyle{ \pi(x_n) = x_n/{_\epsilon}x_n }[/math]. The average prime gap is [math]\displaystyle{ {_\epsilon}x_n }[/math], the maximal one [math]\displaystyle{ {_\epsilon}x_n^2 }[/math] and the maximal [math]\displaystyle{ x_n^2 }[/math] to [math]\displaystyle{ x_n }[/math] behaves like [math]\displaystyle{ \omega }[/math] to [math]\displaystyle{ {\omega}^{\tilde{2}}.\square }[/math]

Gelfond-Schneider theorem

It holds [math]\displaystyle{ a^b \notin {}_{\omega}^{\omega} \mathbb{A}_\mathbb{C} }[/math] where [math]\displaystyle{ a, c \in {}^{\omega} \mathbb{A}_\mathbb{C} \setminus \mathbb{B} }[/math] and infinitesimal [math]\displaystyle{ \varepsilon, b \in {}^{\omega}\mathbb{A}_\mathbb{C} \setminus {}_{\omega}^{\omega}\mathbb{R} }[/math].

Proof:

The minimal polynomials [math]\displaystyle{ p }[/math] (and [math]\displaystyle{ q }[/math]) of [math]\displaystyle{ c^r }[/math] resp. [math]\displaystyle{ c^{r\pm\varepsilon} = a^b }[/math] for maximal [math]\displaystyle{ r \in {}_{\omega}^{\omega}\mathbb{R}_{\gt 0} }[/math] and [math]\displaystyle{ f = p\;(q) }[/math] lead to the contradiction [math]\displaystyle{ {}^1f(c^{r(\pm\varepsilon)}) \ne 0 = (f(c^r) - f(c^{r\pm\varepsilon})) / (c^r - c^{r\pm\varepsilon}) = {}^1f(c^{r(\pm\varepsilon)}).\square }[/math]

MWiki has moved

The new URL is: HWiki

Recommended reading

Nonstandard Mathematics

Retrieved from "https://en.mwiki.de/index.php?title=Main_Page&oldid=322"