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(Cube 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 ==
=== Cube Theorem ===
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=== Prime number theorem ===
  
A sum <math>m \in {}^{\omega }{\mathbb{Z}}</math> consists of three cubes for <math>a, b, c, n \in {}^{\omega }{\mathbb{Z}}</math> 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>m=n^3 + (n + a)^3 + (n - b)^3 = 3n^3 + a - b + 6c \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>.
  
is true. This implicitly quadratic equation yields the formula to be satisfied by <math>n.\square</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>
  
=== Fickett's Theorem ===
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=== Gelfond-Schneider theorem ===
  
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>, 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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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>.
  
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==== Proof: ====
  
<div style="text-align:center;"><math>1/(2n - 1) &lt; r := \mu(\partial Q \cap R)/\mu(\partial R \cap Q) &lt; 2n - 1.</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 ==
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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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]

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Recommended reading

Nonstandard Mathematics