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(Greatest-prime Criterion and Transcendence of Euler's Constant)
(Three-cube theorem and Fickett's theorem)
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= Welcome to MWiki =
 
= Welcome to MWiki =
 
== Theorems of the month ==
 
== Theorems of the month ==
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=== Greatest-prime Criterion ===
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=== Three-Cube Theorem ===
  
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If a real number may be represented as an irreducible fraction <math>\widehat{ap}b \pm \hat{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.
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By Fermat’s little theorem, <math>k \in {}^{\omega }{\mathbb{Z}}</math> is sum of three cubes if and only if
  
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==== Proof: ====
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<div style="text-align:center;"><math>k=(n - a)^3 + n^3 + (n + b)^3 = 3n^3 - a^3 + b^3+ 3c \ne \pm 4\mod 9</math></div>
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The denominator <math>\widehat{ap s} (bs \pm apt)</math> is <math>\ge 2p \ge 2\omega - \mathcal{O}({_e}\omega\sqrt{\omega}) &gt; \omega</math> by the prime number theorem.<math>\square</math>
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and <math>a, b, c, d, m, n \in {}^{\omega }{\mathbb{Z}}</math> implies both <math>(a^2 + b^2)n - (a - b)n^2 = c =: dn</math> and <math>m^2 = n^2 - 4(b^2 - bn + d)</math> for <math>2a_{1,2} = n \pm m.\square</math>
  
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=== Transcendence of Euler's Constant ===
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=== Fickett's Theorem ===
  
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For <math>x \in {}^{\omega }{\mathbb{R}}</math>, let be <math>s(x) := \sum\limits_{n=1}^{\omega}{\hat{n}{{x}^{n}}}</math> and <math>\gamma := s(1) - {_e}\omega = \int\limits_{1}^{\omega}{\left( \widehat{\left\lfloor x \right\rfloor} - \hat{x} \right)dx}</math> Euler's constant, where rearranging shows <math>\gamma \in \; ]0, 1[</math>.
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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>m := 2n - 1</math>, it can be stated for the exact standard measure <math>\mu</math>, where <math>\mu</math> for <math>n = 2</math> needs to be replaced by the Euclidean path length <math>L</math>, that:
  
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If <math>{_e}\omega = s(\hat{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}\hat{\omega}\;{_e}\omega)</math>.
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<div style="text-align:center;"><math>\hat{m} &lt; r := \mu(\partial Q \cap R)/\mu(\partial R \cap Q) &lt; m.</math></div>
  
 
==== Proof: ====
 
==== Proof: ====
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The (exact) integration of the geometric series yields <math>-{_e}(-\acute{x}) = s(x) + \mathcal{O}(\hat{\omega}{x}^{\grave{\omega}}/\acute{x}) + t(x)dx</math> for <math>x \in [-1, 1 - \hat{\nu}]</math> and <math>t(x) \in {}^{\omega }{\mathbb{R}}</math> such that <math>|t(x)| &lt; {\omega}</math>.
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Since the underlying extremal problem has its maximum for rectangles with the side lengths <math>s</math> and <math>s + 2d0</math>, min <math>r = s/(3s - 2d0) \le r \le</math> max <math>r = (3s - 2d0)/s</math> holds. The proof for <math>n &gt; 2</math> is analogous.<math>\square</math>
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After applying Fermat's little theorem to the numerator of <math>\hat{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 03:27, 1 November 2021

Welcome to MWiki

Theorems of the month

Three-Cube Theorem

By Fermat’s little theorem, [math]\displaystyle{ k \in {}^{\omega }{\mathbb{Z}} }[/math] is sum of three cubes if and only if

[math]\displaystyle{ k=(n - a)^3 + n^3 + (n + b)^3 = 3n^3 - a^3 + b^3+ 3c \ne \pm 4\mod 9 }[/math]

and [math]\displaystyle{ a, b, c, d, m, n \in {}^{\omega }{\mathbb{Z}} }[/math] implies both [math]\displaystyle{ (a^2 + b^2)n - (a - b)n^2 = c =: dn }[/math] and [math]\displaystyle{ m^2 = n^2 - 4(b^2 - bn + d) }[/math] for [math]\displaystyle{ 2a_{1,2} = n \pm m.\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{ m := 2n - 1 }[/math], it can be stated for the exact standard measure [math]\displaystyle{ \mu }[/math], where [math]\displaystyle{ \mu }[/math] for [math]\displaystyle{ n = 2 }[/math] needs to be replaced by the Euclidean path length [math]\displaystyle{ L }[/math], that:

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

Proof:

Since the underlying extremal problem has its maximum for rectangles with the side lengths [math]\displaystyle{ s }[/math] and [math]\displaystyle{ s + 2d0 }[/math], min [math]\displaystyle{ r = s/(3s - 2d0) \le r \le }[/math] max [math]\displaystyle{ r = (3s - 2d0)/s }[/math] holds. The proof for [math]\displaystyle{ n > 2 }[/math] is analogous.[math]\displaystyle{ \square }[/math]

Recommended reading

Nonstandard Mathematics

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