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(Three-cube theorem and Fickett's theorem)
(Prime number and Gelfond-Schneider theorem)
 
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= Welcome to MWiki =
 
= Welcome to MWiki =
 
== Theorems of the month ==
 
== Theorems of the month ==
=== Three-Cube Theorem ===
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=== Prime number theorem ===
  
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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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>.
  
<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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==== Proof: ====
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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>.
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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>
  
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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=== Gelfond-Schneider theorem ===
  
=== Fickett's Theorem ===
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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>.
  
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>:
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==== Proof: ====
  
<div style="text-align:center;"><math>\tilde{m} &lt; r := \mu(\partial Q \cap R)/\mu(\partial R \cap Q) &lt; m.</math></div>
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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>
  
==== Proof: ====
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== MWiki has moved ==
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>
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The new URL is: [https://en.hwiki.de/maths.html HWiki]
  
 
== Recommended reading ==
 
== Recommended reading ==

Latest revision as of 17:01, 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]

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Recommended reading

Nonstandard Mathematics