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(Prime number and Gelfond-Schneider theorem)
 
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= Welcome to MWiki =
 
= Welcome to MWiki =
 
== Theorems of the month ==
 
== Theorems of the month ==
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=== Leibniz' differentiation rule ===
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=== Prime number theorem ===
  
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For <math>f: {}^{(\omega)}\mathbb{K}^{\grave{n}} \rightarrow {}^{(\omega)}\mathbb{K}, a, b: {}^{(\omega)}\mathbb{K}^{n} \rightarrow {}^{(\omega)}\mathbb{K}, \curvearrowright x := {(s, {x}_{2}, ..., {x}_{n})}^{T}</math> and <math>s \in {}^{(\omega)}\mathbb{K} \setminus \{{x}_{1}\}</math>, choosing <math>\curvearrowright a(x) = a(\curvearrowright x)</math> and <math>\curvearrowright b(x) = b(\curvearrowright x)</math>, it holds that<div style="text-align:center;"><math>\tfrac{{\downarrow} }{{\downarrow} {{x}_{1}}}\left( {\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}t} \right)={\uparrow}_{a(x)}^{b(x)}{\tfrac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}{\downarrow}t}+\tfrac{{\downarrow} b(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright x,b(x))-\tfrac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright x,a(x)).</math></div>
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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>.
  
 
==== Proof: ====
 
==== Proof: ====
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<div style="text-align:center;"><math>\begin{aligned}\tfrac{{\downarrow} }{{\downarrow} {{x}_{1}}}\left( {\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}t} \right) &={\left( {\uparrow}_{a(\curvearrowright x)}^{b(\curvearrowright x)}{f(\curvearrowright x,t){\downarrow}t}-{\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}t} \right)}/{{\downarrow} {{x}_{1}}}\; \\ &={\left( {\uparrow}_{a(x)}^{b(x)}{(f(\curvearrowright x,t)-f(x,t)){\downarrow}t}+{\uparrow}_{b(x)}^{b(\curvearrowright x)}{f(\curvearrowright x,t){\downarrow}t}-{\uparrow}_{a(x)}^{a(\curvearrowright x)}{f(\curvearrowright x,t){\downarrow}t} \right)}/{{\downarrow} {{x}_{1}}}\; \\ &={\uparrow}_{a(x)}^{b(x)}{\tfrac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}{\downarrow}t}+\tfrac{{\downarrow} b(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright x,b(x))-\tfrac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright x,a(x)).\square\end{aligned}</math></div>
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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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=== Beal's theorem ===
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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>
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Equation <math>a^m + b^n = c^k</math> for <math>a, b, c \in \mathbb{N}^{*}</math> and <math>k, m, n \in \mathbb{N}_{\ge 3}</math> implies gcd<math>(a, b, c) > 1.</math>
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=== Gelfond-Schneider 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>.
  
 
==== Proof: ====
 
==== Proof: ====
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For <math>b^n = (c^{kq}-a^{mr})\left(\tilde{c}^{k\acute{q}} + \tilde{a}^{m\acute{r}}\right) = c^k - a^m + c^{kq} \tilde{a}^{m\acute{r}} - \tilde{c}^{k\acute{q}} a^{mr}</math>, the function <math>f(q,r) := c^{k(\hat{q}-1)} - a^{m(\hat{r}-1)} = 0</math> is continuous in <math>q, r \in {}^{\omega} \mathbb{R}_{>0}</math> and <math>(q, r) = \left(\check{1}, \check{1}\right)</math> solves the equation. Every further solution in fractions yields after exponentiation gcd<math>(a, c) > 1</math> and thus proves the claim.<math>\square</math>
 
  
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=== Conclusion: ===
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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>
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The Fermat-Catalan conjecture can be proven analogously and an infinite descent implies because of gcd<math>(a, b, c) > 1</math> that no <math>n \in {}^{\omega}\mathbb{N}_{\ge 3}</math> satisfies <math>a^n + b^n = c^n</math> for arbitrary <math>a, b, c \in {}^{\omega}\mathbb{N}^{*}.\square</math>
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== MWiki has moved ==
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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]

MWiki has moved

The new URL is: HWiki

Recommended reading

Nonstandard Mathematics

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