# Formelsammlung Mathematik: Kettenbrüche

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## Reguläre Kettenbrüche

Ein regulärer Kettenbruch hat die Form

${\displaystyle a_{0}+{\frac {1}{\displaystyle a_{1}+{\frac {1}{\displaystyle a_{2}+{\frac {1}{a_{3}+\cdots }}}}}}}$,

wobei ${\displaystyle a_{0}\in \mathbb {Z} }$ ist und ${\displaystyle a_{1},a_{2},a_{3},...\in \mathbb {Z} ^{>0}}$ sind. Man kürzt ihn mit ${\displaystyle [a_{0},a_{1},a_{2},a_{3},...]\,}$ ab.

### Negativer Wert

${\displaystyle -\left[a_{0},a_{1},a_{2},a_{3},a_{4},\ldots \right]=\left\{{\begin{matrix}\left[-(a_{0}+1),a_{2}+1,a_{3},a_{4},a_{5},\ldots \right]&,&a_{1}=1\\\\\left[-(a_{0}+1),1,a_{1}-1,a_{2},a_{3},a_{4},\ldots \right]&,&a_{1}>1\end{matrix}}\right.}$

### Kehrwert

${\displaystyle \left[a_{0},a_{1},a_{2},...\right]^{-1}=\left\{{\begin{matrix}\left[0,a_{0},a_{1},...\right]&,&a_{0}>0\\\\\left[a_{1},a_{2},a_{3}...\right]&,&a_{0}=0\end{matrix}}\right.}$

### Goldener Schnitt

${\displaystyle \varphi ={\frac {1+{\sqrt {5}}}{2}}=\left[{\overline {1}}\right]}$

### Eulersche Zahl

${\displaystyle e=\left[2,{\overline {1,2k,1}}\right]_{k\geq 1}}$

${\displaystyle e^{1/n}=\left[{\overline {1,(2k-1)n-1,1}}\right]_{k\geq 1}\qquad n\in \mathbb {Z} ^{>1}}$

${\displaystyle e^{2}=\left[7,{\overline {3k-1,1,1,3k,6(2k+1)}}\right]_{k\geq 1}}$

${\displaystyle e^{2/n}=\left[\,{\overline {1,{\frac {(6k-5)n-1}{2}},6(2k-1)n,{\frac {(6k-1)n-1}{2}},1}}\,\right]_{k\geq 1}\qquad n\in \mathbb {Z} ^{>1}\;\mathrm {ungerade} }$

${\displaystyle {\frac {e-1}{2}}=\left[0,1,{\overline {4k+2}}\right]_{k\geq 1}}$

${\displaystyle {\frac {1}{{\sqrt {e}}-1}}=\left[{\overline {4k+1,1,1}}\right]_{k\geq 0}}$

${\displaystyle {\frac {1}{{\sqrt {e}}+1}}=\left[0,2,{\overline {4k+1,1,1}}\right]_{k\geq 0}}$

### (Ko)tangens (Hyperbolicus)

${\displaystyle \tan(1)=\left[{\overline {1,2k-1}}\right]_{k\geq 1}\quad ,\quad \cot(1)=\left[0,{\overline {1,2k-1}}\right]_{k\geq 1}}$

${\displaystyle \cot \left({\frac {1}{n}}\right)=\left[n-1,{\overline {1,(2k+1)n-2}}\right]_{k\geq 1}\quad ,\quad \tan \left({\frac {1}{n}}\right)=\left[0,n-1,{\overline {1,(2k+1)n-2}}\right]_{k\geq 1}\qquad n\in \mathbb {Z} ^{>1}}$

${\displaystyle \coth \left({\frac {1}{n}}\right)=\left[{\overline {(2k-1)n}}\right]_{k\geq 1}\qquad ,\qquad \tanh \left({\frac {1}{n}}\right)=\left[0,{\overline {(2k-1)n}}\right]_{k\geq 1}\qquad n\in \mathbb {Z} ^{>0}}$

${\displaystyle {\sqrt {n}}\,\tan \left({\frac {1}{\sqrt {n}}}\right)=\left[{\overline {1,(4k-1)n-2,1,4k-1}}\right]_{k\geq 1}\qquad n\in \mathbb {Z} ^{>0}}$

${\displaystyle {\sqrt {n}}\,\cot \left({\frac {1}{\sqrt {n}}}\right)=\left[n-1,{\overline {1,4k-3,1,(4k+1)n-2}}\right]_{k\geq 1}\quad n\in \mathbb {Z} ^{>0}}$

${\displaystyle {\sqrt {n}}\,\tanh \left({\frac {1}{\sqrt {n}}}\right)=\left[0,{\overline {4k-3,(4k-1)n}}\right]_{k\geq 1}\qquad n\in \mathbb {Z} ^{>0}}$

${\displaystyle {\sqrt {n}}\,\coth \left({\frac {1}{\sqrt {n}}}\right)=\left[{\overline {(4k-3)n,4k-1}}\right]_{k\geq 1}\quad n\in \mathbb {Z} ^{>0}}$

${\displaystyle {\sqrt {2}}=\left[1,{\overline {2}}\right]}$
${\displaystyle {\sqrt {3}}=\left[1,{\overline {1,2}}\right]}$
${\displaystyle {\sqrt {4}}=\left[2\right]}$
${\displaystyle {\sqrt {5}}=\left[2,{\overline {4}}\right]}$
${\displaystyle {\sqrt {6}}=\left[2,{\overline {2,4}}\right]}$
${\displaystyle {\sqrt {7}}=\left[2,{\overline {1,1,1,4}}\right]}$
${\displaystyle {\sqrt {8}}=\left[2,{\overline {1,4}}\right]}$
${\displaystyle {\sqrt {9}}=\left[3\right]}$
${\displaystyle {\sqrt {10}}=\left[3,{\overline {6}}\right]}$
Ist ${\displaystyle d\in \mathbb {Z} ^{>0}\,}$ kein Quadrat, so lässt sich ${\displaystyle {\sqrt {d}}}$ schreiben in der Form ${\displaystyle \left[a_{0},{\overline {a_{1},...,a_{n},2a_{0}}}\right]}$

### Familien von Kettenbrüchen

Zum Beispiel

${\displaystyle {\sqrt {a^{2}+1}}=\left[a,{\overline {2a}}\right]}$
${\displaystyle {\sqrt {a^{2}+2}}=\left[a,{\overline {a,2a}}\right]}$
${\displaystyle {\sqrt {a^{2}+a}}=\left[a,{\overline {2,2a}}\right]}$
${\displaystyle {\sqrt {a^{2}+2a}}=\left[a,{\overline {1,2a}}\right]}$
${\displaystyle {\sqrt {9a^{2}+3}}=\left[3a,{\overline {2a,6a}}\right]}$

Eine ausführlichere Einteilung stellen die Bernstein Familien und die Levesque-Rhin Familien dar.

### Allgemeine Aussagen über reguläre Kettenbrüche

• Ein regulärer Kettenbruch ist genau dann irrational, wenn er nicht abbricht.
• Ein regulärer Kettenbruch ist genau dann periodisch, wenn er die Form ${\displaystyle {\frac {a+b{\sqrt {c}}}{d}}}$ besitzt, wobei ${\displaystyle a\in \mathbb {Z} ,b,c,d\in \mathbb {Z} ^{\neq 0}}$ ist und ${\displaystyle c\,}$ keine Quadratzahl ist.

### Satz von Galois

Sind ${\displaystyle a_{0},...,a_{n}\in \mathbb {Z} ^{>0}}$, dann lässt sich der reinperiodische Kettenbruch ${\displaystyle \alpha =[{\overline {a_{0},...,a_{n}}}]}$ schreiben in der Form ${\displaystyle {\frac {P+{\sqrt {D}}}{Q}}}$,
wobei ${\displaystyle P,Q,D\in \mathbb {Z} ^{>0}}$ sind und ${\displaystyle D\,}$ keine Quadratzahl ist.
Ist ${\displaystyle \beta =[{\overline {a_{n},...,a_{0}}}]}$ der zu ${\displaystyle \alpha \,}$ inverse Kettenbruch, so stimmt ${\displaystyle {\frac {-1}{\beta }}}$ mit der Wurzelkonjugierten ${\displaystyle {\frac {P-{\sqrt {D}}}{Q}}}$ überein.

## Verallgemeinerte Kettenbrüche

Eine mögliche Verallgemeinerung ist es Kettenbrüche der Form

${\displaystyle b_{0}+{\frac {a_{1}}{\displaystyle b_{1}+{\frac {a_{2}}{\displaystyle b_{2}+{\frac {a_{3}}{b_{3}+\cdots }}}}}}}$

zu betrachten, wobei ${\displaystyle b_{0}\in \mathbb {Z} }$ ist und wobei ${\displaystyle b_{1},b_{2},...\,}$ und ${\displaystyle a_{1},a_{2}...\,}$ positive ganze Zahlen sind. Genauso könnte man für ${\displaystyle b_{0},b_{1},b_{2},...\,}$ und ${\displaystyle a_{1},a_{2},a_{3},...\,}$ auch reelle oder komplexe Zahlen zulassen. Ein verallgemeinerter Kettenbruch lässt sich nach Pringsheim abkürzend mit ${\displaystyle b_{0}+{\frac {a_{1}|}{|b_{1}}}+{\frac {a_{2}|}{|b_{2}}}+{\frac {a_{3}|}{|b_{3}}}+...}$ und nach Gauß abkürzend mit ${\displaystyle b_{0}+{\underset {k=1}{\overset {\infty }{K}}}{\frac {a_{k}}{b_{k}}}}$ notieren.

### Formel von Bombelli

Ist ${\displaystyle {\sqrt {z}}=x+y}$, so gilt ${\displaystyle z=x^{2}+2xy+y^{2}\Rightarrow {\frac {z-x^{2}}{2x+y}}=y\Rightarrow y={\underset {k=1}{\overset {\infty }{K}}}{\frac {z-x^{2}}{2x}}\Rightarrow {\sqrt {z}}=x+{\underset {k=1}{\overset {\infty }{K}}}{\frac {z-x^{2}}{2x}}}$

### Eulersche Zahl

${\displaystyle e=2+{\underset {k=2}{\overset {\infty }{K}}}{\frac {k}{k}}=2+{\frac {2|}{|2}}+{\frac {3|}{|3}}+{\frac {4|}{|4}}+...}$

${\displaystyle e=2+{\frac {1|}{|1}}+{\underset {k=1}{\overset {\infty }{K}}}{\frac {k}{k+1}}=2+{\frac {1|}{|1}}+{\frac {1|}{|2}}+{\frac {2|}{|3}}+{\frac {3|}{|4}}+...}$

${\displaystyle {\frac {1}{{\sqrt {e}}-1}}=1+{\underset {k=1}{\overset {\infty }{K}}}{\frac {2k}{2k+1}}=1+{\frac {2|}{|3}}+{\frac {4|}{|5}}+...}$

### Kreiszahl ${\displaystyle \pi }$

${\displaystyle {\frac {\pi }{2}}=1+{\underset {k=1}{\overset {\infty }{K}}}{\frac {1}{\frac {1}{k}}}=1+{\frac {1|}{|1}}+{\frac {1|}{|{\frac {1}{2}}}}+{\frac {1|}{|{\frac {1}{3}}}}+...}$

${\displaystyle {\frac {4}{\pi }}=1+{\underset {k=1}{\overset {\infty }{K}}}{\frac {k^{2}}{2k+1}}=1+{\frac {1|}{|3}}+{\frac {4|}{|5}}+{\frac {9|}{|7}}+...}$

${\displaystyle \pi =3+{\underset {k=1}{\overset {\infty }{K}}}{\frac {(2k-1)^{2}}{6}}=3+{\frac {1|}{|6}}+{\frac {9|}{|6}}+{\frac {25|}{|6}}+...}$

Formel von Brouncker
${\displaystyle {\frac {4}{\pi }}=1+{\underset {k=1}{\overset {\infty }{K}}}{\frac {(2k-1)^{2}}{2}}=1+{\frac {1|}{|2}}+{\frac {9|}{|2}}+{\frac {25|}{|2}}+...}$

${\displaystyle {\frac {2}{\pi }}=1-{\frac {1|}{|2}}+{\underset {k=1}{\overset {\infty }{K}}}{\frac {k\,(k+1)}{1}}=1-{\frac {1|}{|2}}+{\frac {1\cdot 2|}{|1}}+{\frac {2\cdot 3|}{|1}}+{\frac {3\cdot 4|}{|1}}+{\frac {4\cdot 5|}{|1}}+...}$

${\displaystyle {\frac {\pi }{2}}=1-{\frac {1|}{|3}}-{\frac {2\cdot 3|}{|1}}-{\frac {2\cdot 1|}{|3}}-{\frac {4\cdot 5|}{|1}}-{\frac {4\cdot 3|}{|3}}-{\frac {6\cdot 7|}{|1}}-{\frac {6\cdot 5|}{|3}}+...}$

${\displaystyle {\frac {12}{\pi ^{2}}}=1+{\underset {k=1}{\overset {\infty }{K}}}{\frac {k^{4}}{2k+1}}=1+{\frac {1|}{|3}}+{\frac {16|}{|5}}+{\frac {81|}{|7}}+...}$

${\displaystyle {\frac {6}{\pi ^{2}-6}}=1+{\frac {1\cdot 1|}{|1}}+{\frac {1\cdot 2|}{|1}}+{\frac {2\cdot 2|}{|1}}+{\frac {2\cdot 3|}{|1}}+{\frac {3\cdot 3|}{|1}}+{\frac {3\cdot 4|}{|1}}...}$

### Catalansche Konstante

${\displaystyle {\frac {1}{2K-1}}={\frac {1}{2}}+{\frac {1\cdot 1|}{|{\frac {1}{2}}}}+{\frac {1\cdot 2|}{|{\frac {1}{2}}}}+{\frac {2\cdot 2|}{|{\frac {1}{2}}}}+{\frac {2\cdot 3|}{|{\frac {1}{2}}}}+{\frac {3\cdot 3|}{|{\frac {1}{2}}}}+{\frac {3\cdot 4|}{|{\frac {1}{2}}}}...}$

### Exponentialfunktion

${\displaystyle e^{z}=1+{\frac {z|}{|1}}+{\underset {k=2}{\overset {\infty }{K}}}{\frac {-{\frac {z}{k}}}{1+{\frac {z}{k}}}}=1+{\frac {z|}{|1}}-{\frac {{\frac {z}{2}}|}{|1+{\frac {z}{2}}}}-{\frac {{\frac {z}{3}}|}{|1+{\frac {z}{3}}}}-{\frac {{\frac {z}{4}}|}{|1+{\frac {z}{4}}}}-...}$

### Sinus (Hyperbolicus)

${\displaystyle \sin z={\frac {z|}{|1}}+{\frac {z^{2}|}{|2\cdot 3-z^{2}}}+{\underset {k=2}{\overset {\infty }{K}}}{\frac {(2k-2)(2k-1)z^{2}}{2k\,(2k+1)-z^{2}}}={\frac {z|}{|1}}+{\frac {z^{2}|}{|2\cdot 3-z^{2}}}+{\frac {2\cdot 3\,z^{2}|}{|4\cdot 5-z^{2}}}+{\frac {4\cdot 5\,z^{2}|}{|6\cdot 7-z^{2}}}+...}$

${\displaystyle \sinh z={\frac {z|}{|1}}-{\frac {z^{2}|}{|2\cdot 3-z^{2}}}+{\underset {k=2}{\overset {\infty }{K}}}{\frac {-(2k-2)(2k-1)z^{2}}{2k\,(2k+1)+z^{2}}}={\frac {z|}{|1}}-{\frac {z^{2}|}{|2\cdot 3+z^{2}}}-{\frac {2\cdot 3\,z^{2}|}{|4\cdot 5+z^{2}}}-{\frac {4\cdot 5\,z^{2}|}{|6\cdot 7+z^{2}}}-...}$

### Tangens (Hyperbolicus)

${\displaystyle {\begin{matrix}\tanh \\\tan \end{matrix}}\left({\frac {x}{y}}\right)={\frac {x|}{|y}}+{\underset {k=1}{\overset {\infty }{K}}}{\frac {\pm x^{2}}{(2k+1)y}}={\frac {x|}{|y}}\pm {\frac {x^{2}|}{|3y}}\pm {\frac {x^{2}|}{|5y}}+...}$

${\displaystyle {\begin{matrix}\tanh \\\tan \end{matrix}}\left({\frac {\pi z}{4}}\right)={\frac {z|}{|1}}+{\underset {k=1}{\overset {\infty }{K}}}{\frac {(2k-1)^{2}\pm z^{2}}{2}}={\frac {z|}{|1}}+{\frac {1^{2}\pm z^{2}|}{|2}}+{\frac {3^{2}\pm z^{2}|}{|2}}+{\frac {5^{2}\pm z^{2}|}{|2}}+...}$

${\displaystyle {\begin{matrix}\tanh \\\tan \end{matrix}}{\Big (}nz{\Big )}={\frac {n\,\tan(z)|}{|1}}+{\underset {k=1}{\overset {n-1}{K}}}{\frac {(k^{2}\pm n^{2})\,\tan ^{2}(z)}{2k+1}}}$

${\displaystyle {\begin{matrix}\tanh \\\tan \end{matrix}}{\Big (}\alpha z{\Big )}={\frac {\alpha \,\tan(z)|}{|1}}+{\underset {k=1}{\overset {\infty }{K}}}{\frac {(k^{2}\pm \alpha ^{2})\,\tan ^{2}(z)}{2k+1}}}$

### Kotangens Hyperbolicus

${\displaystyle z\,\coth \left({\frac {z}{2}}\right)=2+{\underset {k=1}{\overset {\infty }{K}}}{\frac {z^{2}}{4k+2}}=2+{\frac {z^{2}|}{|6}}+{\frac {z^{2}|}{|10}}+{\frac {z^{2}|}{|14}}+...}$

${\displaystyle z\,\coth z=1+{\underset {k=1}{\overset {\infty }{K}}}{\frac {z^{2}|}{|2k+1}}=1+{\frac {z^{2}|}{|3}}+{\frac {z^{2}|}{|5}}+{\frac {z^{2}|}{|7}}+...}$

### Arkussinus und Areasinus Hyperbolicus

${\displaystyle {\text{arsinh}}\,z={\frac {z{\sqrt {1+z^{2}}}|}{|1}}+{\frac {1\cdot 2\,z^{2}|}{|3}}+{\frac {1\cdot 2\,z^{2}|}{|5}}+{\frac {3\cdot 4\,z^{2}|}{|7}}+{\frac {3\cdot 4\,z^{2}|}{|9}}+{\frac {5\cdot 6\,z^{2}|}{|11}}+{\frac {5\cdot 6\,z^{2}|}{|13}}+...}$

${\displaystyle \arcsin z={\frac {z{\sqrt {1-z^{2}}}|}{|1}}-{\frac {1\cdot 2\,z^{2}|}{|3}}-{\frac {1\cdot 2\,z^{2}|}{|5}}-{\frac {3\cdot 4\,z^{2}|}{|7}}-{\frac {3\cdot 4\,z^{2}|}{|9}}-{\frac {5\cdot 6\,z^{2}|}{|11}}-{\frac {5\cdot 6\,z^{2}|}{|13}}-...}$

### Arkustangens und Areatangens Hyperbolicus

${\displaystyle \arctan z={\frac {z|}{|1}}+{\underset {k=1}{\overset {\infty }{K}}}{\frac {k^{2}z^{2}}{2k+1}}={\frac {z|}{|1}}+{\frac {z^{2}|}{|3}}+{\frac {4z^{2}|}{|5}}+{\frac {9z^{2}|}{|7}}+...}$

${\displaystyle {\text{artanh}}\,z={\frac {z|}{|1}}+{\underset {k=1}{\overset {\infty }{K}}}{\frac {-k^{2}z^{2}}{2k+1}}={\frac {z|}{|1}}-{\frac {z^{2}|}{|3}}-{\frac {4z^{2}|}{|5}}-{\frac {9z^{2}|}{|7}}-...}$

${\displaystyle {\begin{matrix}\arctan \\\operatorname {artanh} \end{matrix}}(z)={\frac {z|}{|1}}+{\underset {k=1}{\overset {\infty }{K}}}{\frac {\pm (2k-1)^{2}z^{2}}{2k+1\mp (2k-1)z^{2}}}={\frac {z|}{|1}}\pm {\frac {z^{2}|}{|3\mp z^{2}}}\pm {\frac {9z^{2}|}{|5\mp 3z^{2}}}\pm {\frac {25z^{2}|}{|7\mp 5z^{2}}}\pm ...}$

${\displaystyle \operatorname {artanh} (z)=\pm {\frac {i\pi }{2}}+{\frac {1|}{|z}}+{\underset {k=2}{\overset {\infty }{K}}}{\frac {-{\frac {k-1}{k}}}{{\frac {2k-1}{k}}z}}=\pm {\frac {i\pi }{2}}+{\frac {1|}{|z}}-{\frac {{\frac {1}{2}}|}{|{\frac {3}{2}}z}}-{\frac {{\frac {2}{3}}|}{|{\frac {5}{3}}z}}-{\frac {{\frac {3}{4}}|}{|{\frac {7}{4}}z}}-...\qquad {\begin{matrix}{\textrm {Im}}(z)>0\\{\textrm {Im}}(z)<0\end{matrix}}}$

${\displaystyle e^{2\mu \arctan \left({\frac {1}{z}}\right)}=1+{\frac {2\mu |}{|z-\mu }}+{\underset {k=1}{\overset {\infty }{K}}}{\frac {\mu ^{2}+k^{2}}{(2k+1)z}}=1+{\frac {2\mu |}{|z-\mu }}+{\frac {\mu ^{2}+1|}{|3z}}+{\frac {\mu ^{2}+4|}{|5z}}+{\frac {\mu ^{2}+9|}{|7z}}+...}$

### Logarithmus

${\displaystyle \log(1+z)={\frac {z|}{|1}}+{\frac {1^{2}z|}{|2}}+{\frac {1^{2}z|}{|3}}+{\frac {2^{2}z|}{|4}}+{\frac {2^{2}z|}{|5}}+{\frac {3^{2}z|}{|6}}+{\frac {3^{2}z|}{|7}}+{\frac {4^{2}z|}{|8}}+{\frac {4^{2}z|}{|9}}+...}$

${\displaystyle \log(1+z)={\frac {1|}{|{\frac {1}{z}}}}+{\frac {1|}{|{\frac {2}{1}}}}+{\frac {1|}{|{\frac {3}{z}}}}+{\frac {1|}{|{\frac {2}{2}}}}+{\frac {1|}{|{\frac {5}{z}}}}+{\frac {1|}{|{\frac {2}{3}}}}+{\frac {1|}{|{\frac {7}{z}}}}+{\frac {1|}{|{\frac {2}{4}}}}+{\frac {1|}{|{\frac {9}{z}}}}+{\frac {1|}{|{\frac {2}{5}}}}+...}$

${\displaystyle \log(1+z)={\frac {z|}{|1}}+{\underset {k=1}{\overset {\infty }{K}}}{\frac {k^{2}z}{k+1-kz}}={\frac {z|}{|1}}+{\frac {z|}{|2-z}}+{\frac {4z|}{|3-2z}}+{\frac {9z|}{|4-3z}}+{\frac {16z|}{|5-4z}}+...}$

### Fehlerfunktion

${\displaystyle \operatorname {erf} (z)=\pm 1-{\frac {{\frac {1}{\sqrt {\pi }}}e^{-z^{2}}|}{|z}}+{\frac {1|}{|2z}}+{\frac {2|}{|z}}+{\frac {3|}{|2z}}+{\frac {4|}{|z}}+{\frac {5|}{|2z}}+{\frac {6|}{|z}}+{\frac {7|}{|2z}}+{\frac {8|}{|z}}+...\qquad {\begin{matrix}\operatorname {Re} (z)>0\\\operatorname {Re} (z)<0\end{matrix}}}$

${\displaystyle \operatorname {erf} (z)={\frac {{\frac {z}{\sqrt {\pi }}}e^{-z^{2}}|}{|{\frac {1}{2}}}}+{\underset {k=1}{\overset {\infty }{K}}}{\frac {(-1)^{k}{\frac {k}{2}}z^{2}}{\frac {2k+1}{2}}}={\frac {{\frac {z}{\sqrt {\pi }}}e^{-z^{2}}|}{|{\frac {1}{2}}}}-{\frac {{\frac {1}{2}}z^{2}|}{|{\frac {3}{2}}}}+{\frac {{\frac {2}{2}}z^{2}|}{|{\frac {5}{2}}}}-{\frac {{\frac {3}{2}}z^{2}|}{|{\frac {7}{2}}}}+{\frac {{\frac {4}{2}}z^{2}|}{|{\frac {9}{2}}}}-{\frac {{\frac {5}{2}}z^{2}|}{|{\frac {11}{2}}}}+{\frac {{\frac {6}{2}}z^{2}|}{|{\frac {13}{2}}}}-...}$

### Gammafunktion

${\displaystyle {\frac {\Gamma \left({\frac {z+\alpha +1}{4}}\right)\,\Gamma \left({\frac {z-\alpha +1}{4}}\right)}{\Gamma \left({\frac {z+\alpha +3}{4}}\right)\,\Gamma \left({\frac {z-\alpha +3}{4}}\right)}}={\frac {4|}{|z}}+{\underset {k=1}{\overset {\infty }{K}}}{\frac {(2k-1)^{2}-\alpha ^{2}}{2z}}={\frac {4|}{|z}}+{\frac {1^{2}-\alpha ^{2}|}{|2z}}+{\frac {3^{2}-\alpha ^{2}|}{|2z}}+{\frac {5^{2}-\alpha ^{2}|}{|2z}}+{\frac {7^{2}-\alpha ^{2}|}{|2z}}+...}$

### Besselfunktion

${\displaystyle {\frac {{\text{BesselI}}\left(1+{\frac {a}{d}},{\frac {2x}{d}}\right)}{{\text{BesselI}}\left({\frac {a}{d}},{\frac {2x}{d}}\right)}}\cdot x={\underset {k=1}{\overset {\infty }{K}}}{\frac {x^{2}}{a+kd}}}$

### Ramanujan-Kettenbrüche

${\displaystyle \left({\sqrt {\frac {5+{\sqrt {5}}}{5}}}-{\frac {{\sqrt {5}}+1}{2}}\right)\,e^{2\pi /5}={\underset {k=0}{\overset {\infty }{K}}}{\frac {e^{-2k\pi }}{1}}={\frac {1|}{|1}}+{\frac {e^{-2\pi }|}{|1}}+{\frac {e^{-4\pi }|}{|1}}+{\frac {e^{-6\pi }|}{|1}}+{\frac {e^{-8\pi }|}{|1}}+...}$

${\displaystyle \left({\frac {\sqrt {5}}{1+{\sqrt[{5}]{{\sqrt[{4}]{5}}^{\,3}\cdot {\sqrt {\frac {{\sqrt {5}}-1}{2}}}^{\,5}-1}}}}-{\frac {{\sqrt {5}}+1}{2}}\right)\,e^{2\pi /{\sqrt {5}}}={\underset {k=0}{\overset {\infty }{K}}}{\frac {e^{-2k\pi {\sqrt {5}}}}{1}}={\frac {1|}{|1}}+{\frac {e^{-2\pi {\sqrt {5}}}|}{|1}}+{\frac {e^{-4\pi {\sqrt {5}}}|}{|1}}+{\frac {e^{-6\pi {\sqrt {5}}}|}{|1}}+{\frac {e^{-8\pi {\sqrt {5}}}|}{|1}}+...}$

${\displaystyle {\sqrt {\frac {\pi \,e^{x}}{2x}}}=\sum _{k=0}^{\infty }{\frac {k!\,(2x)^{k}}{(2k+1)!}}+{\frac {1|}{|x}}+{\frac {1|}{|1}}+{\frac {2|}{|x}}+{\frac {3|}{|1}}+{\frac {4|}{|x}}+{\frac {5|}{|1}}+{\frac {6|}{|x}}+...}$