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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 ==
=== Greatest-prime Criterion ===
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=== Three-Cube Theorem ===
  
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.
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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
  
==== 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>
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>
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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>
  
=== Transcendence of Euler's Constant ===
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=== 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>.
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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>\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>L</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>.
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<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>.
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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 04:35, 1 November 2022

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{ \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{ L }[/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