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\[ \definecolor{formcolor}{RGB}{0,0,0} \color{formcolor} { \mathbb{N}\;\;\;≔\;\;\;\mathbb{N_{0}} \setminus {0} } \]

(SN.Ein.Gr.1)

So lässt sich die Riemann'sche Zeta-Funktion schreiben als:

\[ \definecolor{formcolor}{RGB}{0,0,0} \color{formcolor} { (\forall s \in \mathbb{C}) \left[\; \zeta(s)\;\;\;≔\;\;\;\sum_{\forall n \in \mathbb{N}} \frac{ 1 }{ n^{s} } \\ \qquad\qquad\quad\;\;\;\;\;\;=\;\;\;\prod_{\forall p \in \mathbb{P}} \frac{ p^{s} }{ p^{s} - 1 } \;\right] } \]

(SN.Ein.Gr.2)

Für den Spezialfall s = 2 ergibt sich:

\[ \definecolor{formcolor}{RGB}{0,0,0} \color{formcolor} { \Rightarrow\hspace{10mm} \zeta(2)\;\;\;=\;\;\;\sum_{\forall n \in \mathbb{N}} \frac{ 1 }{ n^{2} } } \]	(SN.Ein.Gr.3)
\[ \definecolor{formcolor}{RGB}{0,0,0} \color{formcolor} { \Leftrightarrow\hspace{10mm} \zeta(2)\;\;\;=\;\;\;\frac{ 1 }{ 1 } + \frac{ 1 }{ 4 } + \frac{ 1 }{ 9 } + \frac{ 1 }{ 16 } + \frac{ 1 }{ 25 } + \cdots \\ \qquad\qquad\;\;\;\;=\;\;\;\frac{ \pi^{2} }{ 6 } } \]	(SN.Ein.Gr.4)

Und damit auch:

\[ \definecolor{formcolor}{RGB}{0,0,0} \color{formcolor} { \Leftrightarrow\hspace{10mm} \zeta(2)\;\;\;=\;\;\;\prod_{\forall p \in \mathbb{P}} \frac{ p^{2} }{ p^{2} - 1 } \\ \qquad\qquad\;\;\;\;=\;\;\;\prod_{\forall p \in \mathbb{P}} \frac{ p^{2} }{ (p - 1) \cdot (p + 1) } \\ \qquad\qquad\;\;\;\;=\;\;\;\frac{ \pi^{2} }{ 6 } } \]	(SN.Ein.Gr.5)
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\[ \definecolor{formcolor}{RGB}{0,0,0} \color{formcolor} { \Leftrightarrow\hspace{10mm} \frac{ \pi^{2} }{ 6 }\;\;\;=\;\;\;\prod_{\forall p \in \mathbb{P}} \frac{ p^{2} }{ (p - 1) \cdot (p + 1) } } \]	(SN.Ein.Gr.6)
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\[ \definecolor{formcolor}{RGB}{0,0,0} \color{formcolor} { \Leftrightarrow\hspace{10mm} \pi^{2}\;\;\;=\;\;\;6 \cdot \prod_{\forall p \in \mathbb{P}} \frac{ p^{2} }{ (p - 1) \cdot (p + 1) } } \]	(SN.Ein.Gr.7)
\[ \definecolor{formcolor}{RGB}{0,0,0} \color{formcolor} { \Leftrightarrow\hspace{10mm} \pi^{2}\;\;\;=\;\;\;3 \cdot \prod_{\forall p \in \mathbb{P}} \frac{ p^{2} }{ (p - 1) \cdot (p + 1) } \cdot 2 } \]	(SN.Ein.Gr.8)
\[ \definecolor{formcolor}{RGB}{0,0,0} \color{formcolor} { \Leftrightarrow\hspace{10mm} \pi^{2}\;\;\;=\;\;\;2 \cdot \prod_{\forall p \in \mathbb{P}} \frac{ p^{2} }{ (p - 1) \cdot (p + 1) } \\ \qquad\qquad\quad\;\; + 2 \cdot \prod_{\forall p \in \mathbb{P}} \frac{ p^{2} }{ (p - 1) \cdot (p + 1) } \\ \qquad\qquad\quad\;\; + 2 \cdot \prod_{\forall p \in \mathbb{P}} \frac{ p^{2} }{ (p - 1) \cdot (p + 1) } } \]	(SN.Ein.Gr.9)

Kleiner Exkurs

Wir substituieren:

\[ \definecolor{formcolor}{RGB}{0,0,0} \color{formcolor} { a^{2}\;\;\;=\;\;\;\prod_{\forall p \in \mathbb{P}} \frac{ p^{2} }{ (p - 1) \cdot (p + 1) } } \]	(SN.Ein.Gr.10)
\[ \definecolor{formcolor}{RGB}{0,0,0} \color{formcolor} { \Rightarrow\hspace{10mm} \pi^{2}\;\;\;=\;\;\;2 \cdot a^{2} \cdot 3 } \]	(SN.Ein.Gr.11)
\[ \definecolor{formcolor}{RGB}{0,0,0} \color{formcolor} { \Leftrightarrow\hspace{10mm} \pi^{2}\;\;\;=\;\;\;2 \cdot a^{2} + 2 \cdot a^{2} + 2 \cdot a^{2} } \]	(SN.Ein.Gr.12)
\[ \definecolor{formcolor}{RGB}{0,0,0} \color{formcolor} { r^{2}\;\;\;=\;\;\;x^{2} + y^{2} + z^{2} } \]	(SN.Ein.Gr.13)

Vektoriell betrachtet:

\[ \definecolor{formcolor}{RGB}{0,0,0} \color{formcolor} { \vec{r}\;\;\;=\;\;\;\left( \begin{array}{c} x \\ y \\ z \end{array} \right) } \]	(SN.Ein.Gr.14)
\[ \definecolor{formcolor}{RGB}{0,0,0} \color{formcolor} { \vec{\pi}\;\;\;=\;\;\;\left( \begin{array}{c} \sqrt{2} \cdot a \\ \sqrt{2} \cdot a \\ \sqrt{2} \cdot a \end{array} \right) } \]	(SN.Ein.Gr.15)
\[ \definecolor{formcolor}{RGB}{0,0,0} \color{formcolor} { b^{2}\;\;\;=\;\;\;\frac{ a }{ \sqrt{2} } } \]	(SN.Ein.Gr.16)
\[ \definecolor{formcolor}{RGB}{0,0,0} \color{formcolor} { \vec{\pi}\;\;\;=\;\;\;\left( \begin{array}{c} 2 \cdot b^{2} \\ 2 \cdot b^{2} \\ 2 \cdot b^{2} \end{array} \right)\;\;\;=\;\;\;\left( \begin{array}{c} b^{2} + b^{2} \\ b^{2} + b^{2} \\ b^{2} + b^{2} \end{array} \right) } \]	(SN.Ein.Gr.17)
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\[ \definecolor{formcolor}{RGB}{0,0,0} \color{formcolor} { \pi^{2}\;\;\;=\;\;\;\left( 2 \cdot b^{2} \right)^{2} + \left( 2 \cdot b^{2} \right)^{2} + \left( 2 \cdot b^{2} \right)^{2} } \]	(SN.Ein.Gr.18)
\[ \definecolor{formcolor}{RGB}{0,0,0} \color{formcolor} { \Leftrightarrow\hspace{10mm} \pi^{2}\;\;\;=\;\;\;\left( 2 \cdot \frac{ a }{ \sqrt{2} } \right)^{2} + \left( 2 \cdot \frac{ a }{ \sqrt{2} } \right)^{2} + \left( 2 \cdot \frac{ a }{ \sqrt{2} } \right)^{2} } \]	(SN.Ein.Gr.19)
\[ \definecolor{formcolor}{RGB}{0,0,0} \color{formcolor} { \Leftrightarrow\hspace{10mm} \pi^{2}\;\;\;=\;\;\;\left( \sqrt{2} \cdot a \right)^{2} + \left( \sqrt{2} \cdot a \right)^{2} + \left( \sqrt{2} \cdot a \right)^{2} } \]	(SN.Ein.Gr.20)
\[ \definecolor{formcolor}{RGB}{0,0,0} \color{formcolor} { \Leftrightarrow\hspace{10mm} \pi^{2}\;\;\;=\;\;\;2 \cdot a^{2} + 2 \cdot a^{2} + 2 \cdot a^{2} } \]	(SN.Ein.Gr.21)
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\[ \definecolor{formcolor}{RGB}{0,0,0} \color{formcolor} { \vec{\pi}\;\;\;=\;\;\;\left( \begin{array}{c} \left| \left( \begin{array}{c} b \\ b \end{array} \right) \right|^{2} \\ \left| \left( \begin{array}{c} b \\ b \end{array} \right) \right|^{2} \\ \left| \left( \begin{array}{c} b \\ b \end{array} \right) \right|^{2} \end{array} \right) } \]	(SN.Ein.Gr.22)

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Stand 28. August 2022, 09:00 CET.

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