Difference between revisions of "Main Page"

From MWiki
Jump to: navigation, search
(Greatest-prime Criterion and Transcendence of Euler's Constant)
(Gelfond-Schneider, Three-Cube and Fickett's Theorem)
Line 2: Line 2:
 
= Welcome to MWiki =
 
= Welcome to MWiki =
 
== Theorems of the month ==
 
== Theorems of the month ==
=== Greatest-prime Criterion ===
+
=== Gelfond-Schneider theorem ===
 +
It holds <math>a^b \in {}^{\omega} \mathbb{T}_\mathbb{C}</math> where <math>a, c \in {}^{\omega} \mathbb{A}_\mathbb{C}^{*} \setminus \{1\}, Q :=  {}^{\omega} \mathbb{R} \setminus {}^{\omega} \mathbb{T}_\mathbb{R}</math> and <math>b, \varepsilon \in {}^{\omega} \mathbb{A}_\mathbb{C} \setminus Q</math>.
  
If a real number may be represented as an irreducible fraction <math>\widetilde{ap}b \pm \tilde{s}t</math>, where <math>a, b, s</math>, and <math>t</math> are natural numbers, <math>abst \ne 0</math>, <math>a + s &gt; 2</math>, and the (second-)greatest prime number <math>p \in {}^{\omega }\mathbb{P}, p \nmid b</math> and <math>p \nmid s</math>, then <math>r</math> is <math>\omega</math>-transcendental.
+
==== Proof: ====
 +
Where <math>b \in Q</math> puts the minimal polynomial <math>p(a^b) = p(c^q) = 0</math>, assuming <math>a^b = c^{q+\varepsilon}</math> for maximum <math>q \in Q_{&gt;0}</math> leads to the contradiction <math>0 = (p(a^b) - p(c^q)) / (a^b - c^q) = p^\prime(a^b) = p^\prime(c^q) \ne 0.\square</math>
 +
 
 +
=== Three-Cube Theorem ===
  
==== Proof: ====
+
Three-cube theorem: It holds <math>S := \{n \in \mathbb{Z} : n \ne \pm 4\mod 9\} = \{n \in \mathbb{Z} : n = a^3 + b^3 + c^3 + 3(a + b)c(a - b + c) = (a + c)^3 + (b - c)^3 + c^3\} \subset a^3 + b^3 + c^3 + 6{\mathbb{Z}}</math>, since independent mathematical induction by equitable variables <math>a, b, c \in {\mathbb{Z}}</math> first shows <math>\{0, \pm 1, \pm 2, \pm 3\} \subset S</math>, and then the claim.<math>\square</math>
The denominator <math>\widetilde{ap s} (bs \pm apt)</math> is <math>\ge \hat{p} \ge \hat{\omega} - \mathcal{O}({_e}\omega{\omega}^{\tilde{2}}) &gt; \omega</math> by the prime number theorem.<math>\square</math>
 
  
=== Transcendence of Euler's Constant ===
+
=== Fickett's Theorem ===
  
For <math>x \in {}^{\omega }{\mathbb{R}}</math>, let be <math>s(x) := {+}_{n=1}^{\omega}{\tilde{n}{{x}^{n}}}</math> and <math>\gamma := s(1) - {_e}\omega = {\uparrow}_{1}^{\omega}{\left( \widetilde{\left\lfloor x \right\rfloor} - \tilde{x} \right)\downarrow x}</math> Euler's constant, where rearranging shows <math>\gamma \in \; ]0, 1[</math>.
+
For any relative positions of two overlapping congruent rectangular <math>n</math>-prisms <math>Q</math> and <math>R</math> with <math>n \in {}^{\omega }\mathbb{N}_{\ge 2}</math> and <math>\grave{m} := \hat{n}</math>, the exact standard measure <math>\mu</math> implies, where <math>\mu</math> for <math>n = 2</math> is the Euclidean path length <math>A</math>:
  
If <math>{_e}\omega = s(\tilde{2})\;{_2}\omega</math> is accepted, <math>\gamma \in {}^{\omega }\mathbb{T}_{\mathbb{R}}</math> is true with a precision of <math>\mathcal{O}({2}^{-\omega}\tilde{\omega}\;{_e}\omega)</math>.
+
<div style="text-align:center;"><math>\tilde{m} &lt; r := \mu(\partial Q \cap R)/\mu(\partial R \cap Q) &lt; m.</math></div>
  
 
==== Proof: ====
 
==== Proof: ====
The (exact) integration of the geometric series yields <math>-{_e}(-\acute{x}) = s(x) + \mathcal{O}(\tilde{\omega}{x}^{\grave{\omega}}/\acute{x}) + t(x)dx</math> for <math>x \in [-1, 1 - \tilde{\nu}]</math> and <math>t(x) \in {}^{\omega }{\mathbb{R}}</math> such that <math>|t(x)| &lt; {\omega}</math>.
+
The underlying extremal problem has its maximum for rectangles with side lengths <math>s</math> and <math>s + \hat{\iota}</math>. Putting <math>q := 3 - \hat{\iota}\tilde{s}</math> implies min <math>r = \tilde{q} \le r \le</math> max <math>r = q</math>. The proof for <math>n &gt; 2</math> works analogously.<math>\square</math>
 
 
After applying Fermat's little theorem to the numerator of <math>\tilde{p}(1 - 2^{-p}\,{_2}\omega)</math> for <math>p = \max\, {}^{\omega}\mathbb{P}</math>, the greatest-prime criterion yields the claim.<math>\square</math>
 
  
 
== Recommended reading ==
 
== Recommended reading ==

Revision as of 05:49, 1 November 2023

Welcome to MWiki

Theorems of the month

Gelfond-Schneider theorem

It holds [math]\displaystyle{ a^b \in {}^{\omega} \mathbb{T}_\mathbb{C} }[/math] where [math]\displaystyle{ a, c \in {}^{\omega} \mathbb{A}_\mathbb{C}^{*} \setminus \{1\}, Q := {}^{\omega} \mathbb{R} \setminus {}^{\omega} \mathbb{T}_\mathbb{R} }[/math] and [math]\displaystyle{ b, \varepsilon \in {}^{\omega} \mathbb{A}_\mathbb{C} \setminus Q }[/math].

Proof:

Where [math]\displaystyle{ b \in Q }[/math] puts the minimal polynomial [math]\displaystyle{ p(a^b) = p(c^q) = 0 }[/math], assuming [math]\displaystyle{ a^b = c^{q+\varepsilon} }[/math] for maximum [math]\displaystyle{ q \in Q_{>0} }[/math] leads to the contradiction [math]\displaystyle{ 0 = (p(a^b) - p(c^q)) / (a^b - c^q) = p^\prime(a^b) = p^\prime(c^q) \ne 0.\square }[/math]

Three-Cube Theorem

Three-cube theorem: It holds [math]\displaystyle{ S := \{n \in \mathbb{Z} : n \ne \pm 4\mod 9\} = \{n \in \mathbb{Z} : n = a^3 + b^3 + c^3 + 3(a + b)c(a - b + c) = (a + c)^3 + (b - c)^3 + c^3\} \subset a^3 + b^3 + c^3 + 6{\mathbb{Z}} }[/math], since independent mathematical induction by equitable variables [math]\displaystyle{ a, b, c \in {\mathbb{Z}} }[/math] first shows [math]\displaystyle{ \{0, \pm 1, \pm 2, \pm 3\} \subset S }[/math], and then the claim.[math]\displaystyle{ \square }[/math]

Fickett's Theorem

For any relative positions of two overlapping congruent rectangular [math]\displaystyle{ n }[/math]-prisms [math]\displaystyle{ Q }[/math] and [math]\displaystyle{ R }[/math] with [math]\displaystyle{ n \in {}^{\omega }\mathbb{N}_{\ge 2} }[/math] and [math]\displaystyle{ \grave{m} := \hat{n} }[/math], the exact standard measure [math]\displaystyle{ \mu }[/math] implies, where [math]\displaystyle{ \mu }[/math] for [math]\displaystyle{ n = 2 }[/math] is the Euclidean path length [math]\displaystyle{ A }[/math]:

[math]\displaystyle{ \tilde{m} < r := \mu(\partial Q \cap R)/\mu(\partial R \cap Q) < m. }[/math]

Proof:

The underlying extremal problem has its maximum for rectangles with side lengths [math]\displaystyle{ s }[/math] and [math]\displaystyle{ s + \hat{\iota} }[/math]. Putting [math]\displaystyle{ q := 3 - \hat{\iota}\tilde{s} }[/math] implies min [math]\displaystyle{ r = \tilde{q} \le r \le }[/math] max [math]\displaystyle{ r = q }[/math]. The proof for [math]\displaystyle{ n > 2 }[/math] works analogously.[math]\displaystyle{ \square }[/math]

Recommended reading

Nonstandard Mathematics