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(Green's and Singmaster's theorem)
(Universal multistep, Goldbach’s and Foundation theorem)
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= Welcome to MWiki =
 
= Welcome to MWiki =
== Theorems of the month ==
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== Theorem of the month ==
=== Green's theorem ===
 
  
For some <math>h</math>-domain <math>\mathbb{D} \subseteq {}^{(\omega)}\mathbb{R}^{2}</math>, infinitesimal <math>h = |{\downarrow}x|= |{\downarrow}y| = |\overset{\rightharpoonup}{\gamma}(s) - \gamma(s)| = \mathcal{O}({\tilde{\omega}}^{m})</math>, sufficiently large <math>m \in \mathbb{N}^{*}, (x, y) \in \mathbb{D}, \mathbb{D}^{-} := \{(x, y) \in \mathbb{D} : (x + h, y + h) \in \mathbb{D}\}</math>, and a simply closed path <math>\gamma: [a, b[\rightarrow {\downarrow} \mathbb{D}</math> followed anticlockwise, choosing <math>\overset{\rightharpoonup}{\gamma}(s) = \gamma(\overset{\rightharpoonup}{s})</math> for <math>s \in [a, b[, A \subseteq {[a, b]}^{2}</math>, the following equation holds for sufficiently <math>\alpha</math>-continuous functions <math>u, v: \mathbb{D} \rightarrow \mathbb{R}</math> with not necessarily continuous <math>{\downarrow} u/{\downarrow} x, {\downarrow} u/{\downarrow} y, {\downarrow} v/{\downarrow} x</math> and <math>{\downarrow} v/{\downarrow} y</math><div style="text-align:center;"><math>{\uparrow}_{\gamma }{(u\,{\downarrow}x+v\,{\downarrow}y)}={\uparrow}_{(x,y)\in {\mathbb{D}^{-}}}{\left( \tfrac{{\downarrow} v}{{\downarrow} x}-\tfrac{{\downarrow} u}{{\downarrow} y} \right){\downarrow}(x,y)}.</math></div>
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=== Universal multistep theorem ===
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For <math>n \in {}^{\nu}\mathbb{N}_{\le p}, k, m, p \in {}^{\nu}\mathbb{N}^{*}, {\downarrow}\overset{\rightharpoonup}{x} \in\, ]0, 1[, x \in [a, b] \subseteq {}^{\omega}\mathbb{R}, y : [a, b] \rightarrow {}^{\omega}\mathbb{R}^q,</math> <math>f : [a, b]\times{}^{\omega}\mathbb{R}^{q \times n} \rightarrow {}^{\omega}\mathbb{R}^q, g_k(\overset{\rightharpoonup}{x}) := g_{\acute{k}}(x)</math>, and <math>g_0(a) = f(\overset{\leftharpoonup}{a}, y_0, ... , y_{\acute{n}})</math>, the <abbr title="Taylor series">TS</abbr> of the initial value problem <math>y^\prime(x) = f(x, y((\rightharpoonup)^0 x), ... , y((\rightharpoonup)^{\acute{n}} x))</math> of order <math>n</math> implies<div style="text-align:center;"><math>y(\overset{\rightharpoonup}{x}) = y(x) + {\downarrow}\overset{\rightharpoonup}{x}{\pm}_{k=1}^{p}{\left (g_{p-k}(\overset{\rightharpoonup}{x}){\LARGE{\textbf{+}}}_{m=k}^{p}{\widetilde{m!}\tbinom{\acute{m}}{\acute{k}}}\right )} + \mathcal{O}(({\downarrow}\overset{\rightharpoonup}{x})^{\grave{p}}).\square</math></div>
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=== Goldbach’s theorem ===
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Goldbach’s theorem: Every even whole number <math>&gt; 2</math> is the sum of two primes.
  
 
==== Proof: ====
 
==== Proof: ====
Only <math>\mathbb{D} := \{(x, y) : r \le x \le s, f(x) \le y \le g(x)\}, r, s \in {}^{(\omega)}\mathbb{R}, f, g : {\downarrow} \mathbb{D} \rightarrow {}^{(\omega)}\mathbb{R}</math> is proved, since the proof is analogous for each case rotated by <math>\check{\pi}</math>. Every <math>h</math>-domian is union of such sets. Simply showing <div style="text-align:center;"><math>{\uparrow}_{\gamma }{u\,{\downarrow}x}=-{\uparrow}_{(x,y)\in {\mathbb{D}^{-}}}{\tfrac{{\downarrow} u}{{\downarrow} y}{\downarrow}(x,y)}.</math></div> is sufficient because the other relation is given analogously. Neglecting the regions of <math>\gamma</math> with <math>{\downarrow}x = 0</math> and <math>s := h(u(r, g(r)) - u(t, g(t)))</math> shows <div style="text-align:center;"><math>-{\uparrow}_{\gamma }{u\,{\downarrow}x}-s={\uparrow}_{t}^{r}{u(x,g(x)){\downarrow}x}-{\uparrow}_{t}^{r}{u(x,f(x)){\downarrow}x}={\uparrow}_{t}^{r}{{\uparrow}_{f(x)}^{g(x)}{\tfrac{{\downarrow} u}{{\downarrow} y}}{\downarrow}y{\downarrow}x}={\uparrow}_{(x,y)\in {\mathbb{D}^{-}}}{\tfrac{{\downarrow} u}{{\downarrow} y}{\downarrow}(x,y)}.\square</math></div>
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For <math>\hat{m} + \hat{n} = p_{m+r,n-r} + q_{m+r,n-r} + r, r \in \{0, 2, , \max(g(n))\}</math>, it follows alike <math>\hat{m} + \hat{n} = p_{m+s,n-s} + q_{m+s,n-s} + s,</math> <math>s \in \{0, 2, … , \max(g(n)) + 2\}</math>. This implies <math>\hat{m} + \hat{n} + 2 = p_{\grave{m}+r,\grave{n}-r} + q_{\grave{m}+r,\grave{n}-r} + r, r \in \{0, 2, … , \max(g(\grave{n}))\}</math>. Induction yields then the claim by the previous theorem.<math>\square</math>
  
=== Singmaster's theorem ===
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=== Foundation theorem ===
  
There are maximally 8 distinct binomial coefficients of the same value > 1.
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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.
  
 
==== Proof: ====
 
==== Proof: ====
The existence is clear due to <math>\tbinom{3003}{1} = \tbinom{78}{2} = \tbinom{15}{5} = \tbinom{14}{6}</math> and the structure of Pascal's triangle. With <math>p \in {}^{\omega }{\mathbb{P}}, a,b ,c, d \in {}^{\omega }{\mathbb{N^*}}, \hat{a} \le r := p - b, \hat{a} < \hat{c} \le n := p - d, b < d</math> and <math>s \notin \mathbb{P}</math> for every <math>s \in [\max(r - \acute{a},\grave{n}), r]</math>, Stirling's formula <math>{n!}^2\sim\pi(\hat{n}+\tilde{3}){(\tilde{\epsilon}n)}^{\hat{n}}</math> and the prime number theorem imply <math>\omega\tbinom{r}{a} \le {}_\epsilon\omega\tbinom{n}{c}</math> for <math>p \rightarrow \omega.\square</math>
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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 ==

Revision as of 19:08, 31 May 2024

Welcome to MWiki

Theorem of the month

Universal multistep theorem

For [math]\displaystyle{ n \in {}^{\nu}\mathbb{N}_{\le p}, k, m, p \in {}^{\nu}\mathbb{N}^{*}, {\downarrow}\overset{\rightharpoonup}{x} \in\, ]0, 1[, x \in [a, b] \subseteq {}^{\omega}\mathbb{R}, y : [a, b] \rightarrow {}^{\omega}\mathbb{R}^q, }[/math] [math]\displaystyle{ f : [a, b]\times{}^{\omega}\mathbb{R}^{q \times n} \rightarrow {}^{\omega}\mathbb{R}^q, g_k(\overset{\rightharpoonup}{x}) := g_{\acute{k}}(x) }[/math], and [math]\displaystyle{ g_0(a) = f(\overset{\leftharpoonup}{a}, y_0, ... , y_{\acute{n}}) }[/math], the TS of the initial value problem [math]\displaystyle{ y^\prime(x) = f(x, y((\rightharpoonup)^0 x), ... , y((\rightharpoonup)^{\acute{n}} x)) }[/math] of order [math]\displaystyle{ n }[/math] implies

[math]\displaystyle{ y(\overset{\rightharpoonup}{x}) = y(x) + {\downarrow}\overset{\rightharpoonup}{x}{\pm}_{k=1}^{p}{\left (g_{p-k}(\overset{\rightharpoonup}{x}){\LARGE{\textbf{+}}}_{m=k}^{p}{\widetilde{m!}\tbinom{\acute{m}}{\acute{k}}}\right )} + \mathcal{O}(({\downarrow}\overset{\rightharpoonup}{x})^{\grave{p}}).\square }[/math]

Goldbach’s theorem

Goldbach’s theorem: Every even whole number [math]\displaystyle{ > 2 }[/math] is the sum of two primes.

Proof:

For [math]\displaystyle{ \hat{m} + \hat{n} = p_{m+r,n-r} + q_{m+r,n-r} + r, r \in \{0, 2, … , \max(g(n))\} }[/math], it follows alike [math]\displaystyle{ \hat{m} + \hat{n} = p_{m+s,n-s} + q_{m+s,n-s} + s, }[/math] [math]\displaystyle{ s \in \{0, 2, … , \max(g(n)) + 2\} }[/math]. This implies [math]\displaystyle{ \hat{m} + \hat{n} + 2 = p_{\grave{m}+r,\grave{n}-r} + q_{\grave{m}+r,\grave{n}-r} + r, r \in \{0, 2, … , \max(g(\grave{n}))\} }[/math]. Induction yields then the claim by the previous theorem.[math]\displaystyle{ \square }[/math]

Foundation theorem

Only the postulation of the axiom of foundation that every nonempty subset [math]\displaystyle{ X \subseteq Y }[/math] contains an element [math]\displaystyle{ x_0 }[/math] such that [math]\displaystyle{ X }[/math] und [math]\displaystyle{ x_0 }[/math] are disjoint guarantees cycle freedom.

Proof:

Set [math]\displaystyle{ X := \{x_m : x_0 := \{\emptyset\}, x_{\omega} := \{x_1\} }[/math] and [math]\displaystyle{ x_{\acute{n}} := \{x_n\} }[/math] for [math]\displaystyle{ m \in {}^{\omega}\mathbb{N} }[/math] and [math]\displaystyle{ n \in {}^{\omega}\mathbb{N}_{\ge 2}\} }[/math] .[math]\displaystyle{ \square }[/math]

Recommended reading

Nonstandard Mathematics