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(Counting theorem for algebraic numbers and Brocard'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 ==
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=== Counting theorem for algebraic numbers ===
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=== Prime number theorem ===
  
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The number <math>\mathbb{A}(m, n)</math> of algebraic numbers of polynomial or series degree <math>m</math> and thus in general for the Riemann zeta function <math>\zeta</math> asymptotically satisfies the equation <math>\mathbb{A}(m, n) = \widetilde{\zeta(\grave{m})}\,z(m){{(2n+1)}^{m}}\left( n+\mathcal{O}({_e}n) \right)</math>, where <math>z(m)</math> is the average number of zeros of a polynomial or series.
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For <math>\pi(x) := |\{p \in {}^{\omega}{\mathbb{P}} : p \le x \in {}^{\omega}{\mathbb{R}}\}|</math> holds <math>\pi(\omega) = \widetilde{{_e}\omega}\omega + \mathcal{O}({_e}\omega{\omega}^{\tilde{2}})</math>.
  
 
==== Proof: ====
 
==== Proof: ====
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The case <math>m = 1</math> requires by <ref name="Scheid">[[w:Harald Scheid|<span class="wikipedia">Scheid, Harald</span>]]: ''Zahlentheorie'' : 1st Ed.; 1991; Bibliographisches Institut; Mannheim; ISBN 9783411148417, p. 323.</ref> the error term <math>\mathcal{O}({_e}n n)</math> and represents the number <math>4{+}_{k=1}^{n}{\varphi (k)}-1</math> by the <math>\varphi</math>-function. For <math>m > 1</math>, the divisibility conditions neither change the error term <math>\mathcal{O}({_e}n)</math> nor the leading term. Polynomials or series such that <math>\text{gcd}({a}_{0}, {a}_{1}, ..., {a}_{m}) \ne 1</math> are excluded by <math>1/\zeta(\grave{m})</math>: The latter is given by taking the product over the prime numbers <math>p</math> of all <math>(1 - {p}^{-\grave{m}})</math> absorbing here multiples of <math>p</math> and representing sums of geometric series.<math>\square</math>
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In the sieve of Eratosthenes, the number of prime numbers decreases almost regularly. From intervals of fix length <math>y \in {}^{\omega}{\mathbb{R}_{&gt;0}}, \hat{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.
  
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=== Brocard's theorem ===
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For induction basis <math>n = 2</math> resp. 3, the induction hypothesis is that the first interval contains <math>x_n/{_e}x_n</math> prime numbers for <math>n \in {}^{\omega}{\mathbb{N}_{\ge2}}</math> and arbitrary <math>x_4 \in [2, 4[</math>. Then the induction step from <math>x_n</math> to <math>x_n^2</math> by considering the prime gaps of prime <math>p\# /q + 1</math> for <math>p, q \in {}^{\omega}\mathbb{P}</math> proves that there are <math>\pi(x_n^2) = \pi(x_n) \check{x}_n</math> prime numbers only from <math>\pi(x_n) = x_n/{_e}x_n</math>. The average distance between the prime numbers is <math>{_e}x_n</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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It holds that <math>\{(m, n) \in {}^{\omega} \mathbb{N}^2 : n! + 1 = m^2\} = \{(5, 4), (11, 5), (71, 7)\}.</math>
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=== Gelfond-Schneider theorem ===
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It holds <math>a^b \in {}^{\omega} \mathbb{T}_\mathbb{C}</math> where <math>a, c \in {}^{\omega} \mathbb{A}_\mathbb{C}^{*} \setminus \{1\}, Q := {}^{\omega} \mathbb{R} \setminus {}^{\omega} \mathbb{T}_\mathbb{R}</math> and <math>b, \varepsilon \in {}^{\omega} \mathbb{A}_\mathbb{C} \setminus Q</math>.
  
 
==== Proof: ====
 
==== Proof: ====
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From <math>n! = \acute{m}\grave{m}</math>, it follows that <math>m = \hat{r} \pm 1</math> für <math>r \in {}^{\omega} \mathbb{N}^{*}</math> and <math>n \ge 3</math>. Thus <math>n! = \hat{r}(\hat{r}\pm2) = 8s(\hat{s} \pm 1)</math> holds for <math>s \in {}^{\omega} \mathbb{N}^{*}</math>. Let <math>2^q \mid n!</math> and <math>2^{\grave{q}} \nmid n!</math> for maximal <math>q \in {}^{\omega} \mathbb{N}^{*}</math>. Therefore <math>n! = 2^q(\hat{u} + 1)</math> holds for <math>u \in {}^{\omega} \mathbb{N}^{*}</math> and necessarily <math>n! = 2^q(2^{q-2} \pm 1)</math>. Then the prime factorisation of <math>n!</math> requires <math>n \le 7</math> giving the claim.<math>\square</math>
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Where <math>b \in Q</math> puts the minimal polynomial <math>p(a^b) = p(c^q) = 0</math>, assuming <math>a^b = c^{q+\varepsilon}</math> for maximum <math>q \in Q_{>0}</math> leads to the contradiction <math>0 = (p(a^b) - p(c^q)) / (a^b - c^q) = p^\prime(a^b) = p^\prime(c^q) \ne 0.\square</math>
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== Recommended reading ==
 
== Recommended reading ==
  
 
[https://en.calameo.com/books/003777977258f7b4aa332 Nonstandard Mathematics]
 
[https://en.calameo.com/books/003777977258f7b4aa332 Nonstandard Mathematics]
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== References ==
 
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<references />
 
  
 
[[de:Hauptseite]]
 
[[de:Hauptseite]]

Revision as of 19:42, 31 July 2023

Welcome to MWiki

Theorems of the month

Prime number theorem

For [math]\displaystyle{ \pi(x) := |\{p \in {}^{\omega}{\mathbb{P}} : p \le x \in {}^{\omega}{\mathbb{R}}\}| }[/math] holds [math]\displaystyle{ \pi(\omega) = \widetilde{{_e}\omega}\omega + \mathcal{O}({_e}\omega{\omega}^{\tilde{2}}) }[/math].

Proof:

In the sieve of Eratosthenes, the number of prime numbers decreases almost regularly. From intervals of fix length [math]\displaystyle{ y \in {}^{\omega}{\mathbb{R}_{>0}}, \hat{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.

For induction basis [math]\displaystyle{ n = 2 }[/math] resp. 3, the induction hypothesis is that the first interval contains [math]\displaystyle{ x_n/{_e}x_n }[/math] prime numbers for [math]\displaystyle{ n \in {}^{\omega}{\mathbb{N}_{\ge2}} }[/math] and arbitrary [math]\displaystyle{ x_4 \in [2, 4[ }[/math]. Then the induction step from [math]\displaystyle{ x_n }[/math] to [math]\displaystyle{ x_n^2 }[/math] by considering the prime gaps of prime [math]\displaystyle{ p\# /q + 1 }[/math] for [math]\displaystyle{ p, q \in {}^{\omega}\mathbb{P} }[/math] proves that there are [math]\displaystyle{ \pi(x_n^2) = \pi(x_n) \check{x}_n }[/math] prime numbers only from [math]\displaystyle{ \pi(x_n) = x_n/{_e}x_n }[/math]. The average distance between the prime numbers is [math]\displaystyle{ {_e}x_n }[/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 \in {}^{\omega} \mathbb{T}_\mathbb{C} }[/math] where [math]\displaystyle{ a, c \in {}^{\omega} \mathbb{A}_\mathbb{C}^{*} \setminus \{1\}, Q := {}^{\omega} \mathbb{R} \setminus {}^{\omega} \mathbb{T}_\mathbb{R} }[/math] and [math]\displaystyle{ b, \varepsilon \in {}^{\omega} \mathbb{A}_\mathbb{C} \setminus Q }[/math].

Proof:

Where [math]\displaystyle{ b \in Q }[/math] puts the minimal polynomial [math]\displaystyle{ p(a^b) = p(c^q) = 0 }[/math], assuming [math]\displaystyle{ a^b = c^{q+\varepsilon} }[/math] for maximum [math]\displaystyle{ q \in Q_{\gt 0} }[/math] leads to the contradiction [math]\displaystyle{ 0 = (p(a^b) - p(c^q)) / (a^b - c^q) = p^\prime(a^b) = p^\prime(c^q) \ne 0.\square }[/math]

Recommended reading

Nonstandard Mathematics

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