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(Universal multistep theorem)
(Counting theorem for algebraic numbers and Brocard's theorem)
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= Welcome to MWiki =
 
= Welcome to MWiki =
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== Theorem of the month ==
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== Theorems of the month ==
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=== Counting theorem for algebraic numbers ===
  
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=== Universal multistep 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>n \in {}^{\nu}\mathbb{N}_{\le p}, k, m, p \in {}^{\nu}\mathbb{N}^{*}, {\downarrow}_{{}^\curvearrowright B} x \in\, ]0, 1[, x \in [a, b] \subseteq {}^{\omega}\mathbb{R}, y : [a, b] \rightarrow {}^{\omega}\mathbb{R}^q, f : [a, b]\times{}^{\omega}\mathbb{R}^{q \times n} \rightarrow {}^{\omega}\mathbb{R}^q, g_k({}^\curvearrowright x) := g_{\acute{k}}(x)</math>, and <math>g_0(a) = f(({}^\curvearrowleft)a, y_0, ... , y_{\acute{n}})</math>, the Taylor series of the initial value problem <math>y^\prime(x) = f(x, y(({}^\curvearrowright)^0 x), ... , y(({}^\curvearrowright)^{\acute{n}} x))</math> of order <math>n</math> implies <div style="text-align:center;"><math>y({}^\curvearrowright x) = y(x) + {\downarrow}_{{}^\curvearrowright}x{\pm}_{k=1}^{p}{\left (g_{p-k}(({}^\curvearrowright) x){+}_{m=k}^{p}{\widetilde{m!}\tbinom{\acute{m}}{\acute{k}}}\right )} + \mathcal{O}(({\downarrow}_{{}^\curvearrowright} x)^{\grave{p}}).\square</math></div>
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==== 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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=== Goldbach’s theorem ===
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=== Brocard's theorem ===
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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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Every even whole number greater than 2 is the sum of two primes.
 
  
 
==== Proof: ====
 
==== Proof: ====
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Induction over all prime gaps until the maximally possible one each time.<math>\square</math>
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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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=== Foundation theorem ===
 
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Only the postulation of the axiom of foundation that every nonempty subset <math>X \subseteq Y</math> contains an element <math>x_0</math> such that <math>X</math> und <math>x_0</math> are disjoint guarantees cycle freedom.
 
  
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==== Proof: ====
 
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Set <math>X := \{x_m : x_0 := \{\emptyset\}, x_{\omega} := \{x_1\}</math> and <math>x_{\acute{n}} := \{x_n\}</math> for <math>m \in {}^{\omega}\mathbb{N}</math> and <math>n \in {}^{\omega}\mathbb{N}_{\ge 2}\}</math> .<math>\square</math>
 
 
== 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 23:31, 30 June 2023

Welcome to MWiki

Theorems of the month

Counting theorem for algebraic numbers

The number [math]\displaystyle{ \mathbb{A}(m, n) }[/math] of algebraic numbers of polynomial or series degree [math]\displaystyle{ m }[/math] and thus in general for the Riemann zeta function [math]\displaystyle{ \zeta }[/math] asymptotically satisfies the equation [math]\displaystyle{ \mathbb{A}(m, n) = \widetilde{\zeta(\grave{m})}\,z(m){{(2n+1)}^{m}}\left( n+\mathcal{O}({_e}n) \right) }[/math], where [math]\displaystyle{ z(m) }[/math] is the average number of zeros of a polynomial or series.

Proof:

The case [math]\displaystyle{ m = 1 }[/math] requires by [1] the error term [math]\displaystyle{ \mathcal{O}({_e}n n) }[/math] and represents the number [math]\displaystyle{ 4{+}_{k=1}^{n}{\varphi (k)}-1 }[/math] by the [math]\displaystyle{ \varphi }[/math]-function. For [math]\displaystyle{ m \gt 1 }[/math], the divisibility conditions neither change the error term [math]\displaystyle{ \mathcal{O}({_e}n) }[/math] nor the leading term. Polynomials or series such that [math]\displaystyle{ \text{gcd}({a}_{0}, {a}_{1}, ..., {a}_{m}) \ne 1 }[/math] are excluded by [math]\displaystyle{ 1/\zeta(\grave{m}) }[/math]: The latter is given by taking the product over the prime numbers [math]\displaystyle{ p }[/math] of all [math]\displaystyle{ (1 - {p}^{-\grave{m}}) }[/math] absorbing here multiples of [math]\displaystyle{ p }[/math] and representing sums of geometric series.[math]\displaystyle{ \square }[/math]

Brocard's theorem

It holds that [math]\displaystyle{ \{(m, n) \in {}^{\omega} \mathbb{N}^2 : n! + 1 = m^2\} = \{(5, 4), (11, 5), (71, 7)\}. }[/math]

Proof:

From [math]\displaystyle{ n! = \acute{m}\grave{m} }[/math], it follows that [math]\displaystyle{ m = \hat{r} \pm 1 }[/math] für [math]\displaystyle{ r \in {}^{\omega} \mathbb{N}^{*} }[/math] and [math]\displaystyle{ n \ge 3 }[/math]. Thus [math]\displaystyle{ n! = \hat{r}(\hat{r}\pm2) = 8s(\hat{s} \pm 1) }[/math] holds for [math]\displaystyle{ s \in {}^{\omega} \mathbb{N}^{*} }[/math]. Let [math]\displaystyle{ 2^q \mid n! }[/math] and [math]\displaystyle{ 2^{\grave{q}} \nmid n! }[/math] for maximal [math]\displaystyle{ q \in {}^{\omega} \mathbb{N}^{*} }[/math]. Therefore [math]\displaystyle{ n! = 2^q(\hat{u} + 1) }[/math] holds for [math]\displaystyle{ u \in {}^{\omega} \mathbb{N}^{*} }[/math] and necessarily [math]\displaystyle{ n! = 2^q(2^{q-2} \pm 1) }[/math]. Then the prime factorisation of [math]\displaystyle{ n! }[/math] requires [math]\displaystyle{ n \le 7 }[/math] giving the claim.[math]\displaystyle{ \square }[/math]

Recommended reading

Nonstandard Mathematics

References

  1. ↑ Scheid, Harald: Zahlentheorie : 1st Ed.; 1991; Bibliographisches Institut; Mannheim; ISBN 9783411148417, p. 323.
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