Formelsammlung Mathematik: Kettenbrüche

Reguläre Kettenbrüche

Ein regulärer Kettenbruch hat die Form

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

,

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

Negativer Wert

-\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

\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

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

Eulersche Zahl

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

Beweis

Sei $\varrho \,$ der Kettenbruch $[a_{0};a_{1},a_{2},a_{3},...]\,$ mit $a_{0}=2,a_{3n-2}=1,a_{3n-1}=2n,a_{3n}=1\,$ für $n\in \mathbb {Z} ^{>0}$ .

Das heißt $\varrho =[2;1,2,1,1,4,1,1,6,1,1,8,1,...]=\left[2,{\overline {1,2n,1}}\right]_{n\geq 1}$ .

Für Näherungsbrüche ${\frac {p_{n}}{q_{n}}}$ gilt allgemein die Rekursion ${\begin{Bmatrix}p_{n}=a_{n}p_{n-1}+p_{n-2}\\q_{n}=a_{n}q_{n-1}+q_{n-2}\end{Bmatrix}}$ mit den Startwerten ${\begin{Bmatrix}p_{-2}=0\;,\;p_{-1}=1\\q_{-2}=1\;,\;q_{-1}=0\end{Bmatrix}}$ .

Insbesondere für die Näherungsbrüche $\left({\frac {p_{n}}{q_{n}}}\right)_{n\geq 0}=\left({\frac {2}{1}},{\frac {3}{1}},{\frac {8}{3}},{\frac {11}{4}},{\frac {19}{7}},...\right)$ von $\varrho \,$ gelten für $n\geq 1\,$ die Rekursionsformeln:

${\begin{Bmatrix}p_{3n-1}&=&2n&p_{3n-2}&+&p_{3n-3}\\p_{3n}&=&&p_{3n-1}&+&p_{3n-2}\\p_{3n+1}&=&&p_{3n}&+&p_{3n-1}\end{Bmatrix}}\quad \quad {\begin{Bmatrix}q_{3n-1}&=&2n&q_{3n-2}&+&q_{3n-3}\\q_{3n}&=&&q_{3n-1}&+&q_{3n-2}\\q_{3n+1}&=&&q_{3n}&+&q_{3n-1}\end{Bmatrix}}\qquad (*)$

Setze ${\begin{bmatrix}a_{n}(x)={\frac {x^{n+1}\,(x-1)^{n}}{n!}}\,e^{x}&,&b_{n}(x)={\frac {x^{n}\,(x-1)^{n+1}}{n!}}\,e^{x}&,&c_{n}(x)=-{\frac {x^{n+1}\,(x-1)^{n+1}}{(n+1)!}}\,e^{x}\\\\A_{n}=\int _{0}^{1}a_{n}(x)dx&,&B_{n}=\int _{0}^{1}b_{n}(x)dx&,&C_{n}=\int _{0}^{1}c_{n}(x)dx\end{bmatrix}}$

Behauptung: Für alle $n\geq 0\,$ gilt ${\begin{Bmatrix}A_{n}&=&p_{3n-1}-q_{3n-1}\cdot e\\B_{n}&=&p_{3n}-q_{3n}\cdot e\\C_{n}&=&p_{3n+1}-q_{3n+1}\cdot e\end{Bmatrix}}$

Der Induktionsanfang $A_{0}=1\;,\;B_{0}=2-e\;,\;C_{0}=3-e\,$ lässt sich leicht nachprüfen.

Für $n\geq 1\,$ gilt nun ${\begin{Bmatrix}a_{n}(x)&=&2n&c_{n-1}(x)&+&b_{n-1}(x)&+&b_{n}'(x)\\b_{n}(x)&=&&a_{n}(x)&+&c_{n-1}(x)\\c_{n}(x)&=&&b_{n}(x)&+&a_{n}(x)&+&c_{n}'(x)\end{Bmatrix}}$ , woraus

${\begin{Bmatrix}A_{n}&=&2n&C_{n-1}&+&B_{n-1}\\B_{n}&=&&A_{n}&+&C_{n-1}\\C_{n}&=&&B_{n}&+&A_{n}\end{Bmatrix}}$ folgt. Daher erfüllen $A_{n}\,,\;B_{n}\,,\;C_{n}$ die Rekursion $(*)\,$ , woraus die Behauptung folgt.

Offenbar verschwinden $A_{n}\,,\;B_{n}\,,\;C_{n}$ für $n\to \infty \,$ , und daher ist $\varrho :=\lim _{n\to \infty }{\frac {p_{n}}{q_{n}}}=e$ .

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

Beweis

Für $m\in \mathbb {Z} ^{>1}$ sei $\varrho _{m}\,$ der Kettenbruch $[a_{0};a_{1},a_{2},a_{3},...]\,$ mit $a_{3n}=1\,,\,a_{3n+1}=(2n+1)m-1\,,\,a_{3n+2}=1$ für $n\in \mathbb {Z} ^{\geq 0}$ .

Das heißt $\varrho _{m}=[1,m-1,1,1,3m-1,1,1,5m-1,1,...]=\left[{\overline {1,(2n+1)m-1,1}}\right]_{n\geq 0}$ .

Für Näherungsbrüche ${\frac {p_{n}}{q_{n}}}$ gilt allgemein die Rekursion ${\begin{Bmatrix}p_{n}=a_{n}p_{n-1}+p_{n-2}\\q_{n}=a_{n}q_{n-1}+q_{n-2}\end{Bmatrix}}$ mit den Startwerten ${\begin{Bmatrix}p_{-2}=0\;,\;p_{-1}=1\\q_{-2}=1\;,\;q_{-1}=0\end{Bmatrix}}$ .

Insbesondere für die Näherungsbrüche $\left({\frac {p_{n}}{q_{n}}}\right)_{n\geq 0}=\left(1,{\frac {m}{m-1}},{\frac {m+1}{m}},{\frac {2m+1}{2m-1}},...\right)$ von $\varrho _{m}\,$ gelten für $n\geq 0\,$ die Rekursionsformeln:

${\begin{Bmatrix}p_{3n}&=&&p_{3n-1}&+&p_{3n-2}\\p_{3n+1}&=&{\big (}(2n+1)m-1{\big )}&p_{3n}&+&p_{3n-1}\\p_{3n+2}&=&&p_{3n+1}&+&p_{3n}\end{Bmatrix}}\quad {\begin{Bmatrix}q_{3n}&=&&q_{3n-1}&+&q_{3n-2}\\q_{3n+1}&=&{\big (}(2n+1)m-1{\big )}&q_{3n}&+&q_{3n-1}\\q_{3n+2}&=&&q_{3n+1}&+&q_{3n}\end{Bmatrix}}\quad (*)$

Setze ${\begin{bmatrix}a_{n}(x)=-{\frac {1}{m^{n+1}}}\,{\frac {x^{n}\,(x-1)^{n}}{n!}}\,e^{x/m}&,&b_{n}(x)={\frac {1}{m^{n+1}}}\,{\frac {x^{n+1}\,(x-1)^{n}}{n!}}\,e^{x/m}&,&c_{n}(x)={\frac {1}{m^{n+1}}}\,{\frac {x^{n}\,(x-1)^{n+1}}{n!}}\,e^{x/m}\\\\A_{n}=\int _{0}^{1}a_{n}(x)dx&,&B_{n}=\int _{0}^{1}b_{n}(x)dx&,&C_{n}=\int _{0}^{1}c_{n}(x)dx\end{bmatrix}}$

Behauptung: Für alle $n\geq 0\,$ gilt ${\begin{Bmatrix}A_{n}&=&p_{3n}-q_{3n}\cdot e^{1/m}\\B_{n}&=&p_{3n+1}-q_{3n+1}\cdot e^{1/m}\\C_{n}&=&p_{3n+2}-q_{3n+2}\cdot e^{1/m}\end{Bmatrix}}$

Der Induktionsanfang $A_{0}=1-e^{1/m}\;,\;B_{0}=m-(m-1)\cdot e^{1/m}\;,\;C_{0}=(m+1)-m\cdot e^{1/m}\,$ lässt sich leicht nachprüfen.

Für $n\geq 1\,$ gilt nun ${\begin{Bmatrix}a_{n}(x)&=&&c_{n-1}(x)&+&b_{n-1}(x)&+&m\cdot a_{n}'(x)\\b_{n}(x)&=&{\big (}(2n+1)m-1{\big )}&a_{n}(x)&+&c_{n-1}(x)&+&m\cdot c_{n}'(x)\\c_{n}(x)&=&&b_{n}(x)&+&a_{n}(x)\end{Bmatrix}}$ , woraus

${\begin{Bmatrix}A_{n}&=&&C_{n-1}&+&B_{n-1}\\B_{n}&=&{\big (}(2n+1)m-1{\big )}&A_{n}&+&C_{n-1}\\C_{n}&=&&B_{n}&+&A_{n}\end{Bmatrix}}$ folgt. Daher erfüllen $A_{n}\,,\;B_{n}\,,\;C_{n}$ die Rekursion $(*)\,$ , woraus die Behauptung folgt.

Offenbar verschwinden $A_{n}\,,\;B_{n}\,,\;C_{n}$ für $n\to \infty \,$ , und daher ist $\varrho _{m}:=\lim _{n\to \infty }{\frac {p_{n}}{q_{n}}}=e^{1/m}$ .

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

Beweis

Sei $\varrho \,$ der Kettenbruch $[a_{0};a_{1},a_{2},a_{3},...]\,$ mit $a_{0}=7,\,$

$a_{5n-4}=3n-1,a_{5n-3}=1,a_{5n-2}=1,a_{5n-1}=3n,a_{5n}=6(2n+1)\,$ für $n\in \mathbb {Z} ^{>0}$ .

Das heißt $\varrho =[7;2,1,1,3,18,5,1,1,6,30,8,1,1,9,42,...]=\left[7,{\overline {3n-1,1,1,3n,6(2n+1)}}\right]_{n\geq 1}$ .

Für Näherungsbrüche ${\frac {p_{n}}{q_{n}}}$ gilt allgemein die Rekursion ${\begin{Bmatrix}p_{n}=a_{n}p_{n-1}+p_{n-2}\\q_{n}=a_{n}q_{n-1}+q_{n-2}\end{Bmatrix}}$ mit den Startwerten ${\begin{Bmatrix}p_{-2}=0\;,\;p_{-1}=1\\q_{-2}=1\;,\;q_{-1}=0\end{Bmatrix}}$ .

Insbesondere für die Näherungsbrüche $\left({\frac {p_{n}}{q_{n}}}\right)_{n\geq 0}=\left({\frac {7}{1}},{\frac {15}{2}},{\frac {22}{3}},{\frac {37}{5}},{\frac {133}{18}},{\frac {2431}{329}},...\right)$ von $\varrho \,$ gelten für $n\geq 1\,$ die Rekursionsformeln:

${\begin{Bmatrix}p_{5n-4}&=&(3n-1)&p_{5n-5}&+&p_{5n-6}\\p_{5n-3}&=&&p_{5n-4}&+&p_{5n-5}\\p_{5n-2}&=&&p_{5n-3}&+&p_{5n-4}\\p_{5n-1}&=&3n&p_{5n-2}&+&p_{5n-3}\\p_{5n}&=&6(2n+1)&p_{5n-1}&+&p_{5n-2}\end{Bmatrix}}\quad \quad {\begin{Bmatrix}q_{5n-4}&=&(3n-1)&q_{5n-5}&+&q_{5n-6}\\q_{5n-3}&=&&q_{5n-4}&+&q_{5n-5}\\q_{5n-2}&=&&q_{5n-3}&+&q_{5n-4}\\q_{5n-1}&=&3n&q_{5n-2}&+&q_{5n-3}\\q_{5n}&=&6(2n+1)&q_{5n-1}&+&q_{5n-2}\end{Bmatrix}}\qquad (*)$

Setze ${\begin{bmatrix}a_{n}(x)=-2^{n+1}\,{\frac {x^{n}\,(x-1)^{n}}{n!}}\,e^{2x}&,&b_{n}(x)=2^{n+1}\,{\frac {x^{n+1}\,(x-1)^{n}}{n!}}\,e^{2x}&,&c_{n}(x)=2^{n+1}\,{\frac {x^{n}\,(x-1)^{n+1}}{n!}}\,e^{2x}\\\\A_{n}=\int _{0}^{1}a_{n}(x)dx&,&B_{n}=\int _{0}^{1}b_{n}(x)dx&,&C_{n}=\int _{0}^{1}c_{n}(x)dx\end{bmatrix}}$

Behauptung: Für alle $n\geq 1\,$ gilt ${\begin{Bmatrix}B_{3n-1}&=&p_{5n-4}-q_{5n-4}\cdot e^{2}\\C_{3n-1}&=&p_{5n-3}-q_{5n-3}\cdot e^{2}\\A_{3n}&=&p_{5n-2}-q_{5n-2}\cdot e^{2}\\{\frac {1}{2}}A_{3n+1}&=&p_{5n-1}-q_{5n-1}\cdot e^{2}\\A_{3n+2}&=&p_{5n}-q_{5n}\cdot e^{2}\\\end{Bmatrix}}$

Der Induktionsanfang $B_{2}=15-2e^{2}\;,\;C_{2}=22-3e^{2}\;,\;A_{3}=37-5e^{2}\;,\;{\frac {1}{2}}A_{4}=133-18e^{2}\;,\;A_{5}=2431-329e^{2}$

lässt sich leicht nachprüfen. Für $n\geq 1\,$ gilt nun

${\begin{Bmatrix}b_{3n-1}(x)&=&(3n-1)&a_{3n-1}(x)&+&{\frac {1}{2}}a_{3n-2}(x)&+&{\frac {1}{4}}a_{3n-1}'(x)+{\frac {1}{2}}b_{3n-1}'(x)\\c_{3n-1}(x)&=&&b_{3n-1}(x)&+&a_{3n-1}(x)\\a_{3n}(x)&=&&c_{3n-1}(x)&+&b_{3n-1}(x)&+&{\frac {1}{2}}a_{3n}'(x)\\{\frac {1}{2}}a_{3n+1}(x)&=&3n&a_{3n}(x)&+&c_{3n-1}(x)&+&{\frac {1}{4}}a_{3n+1}'(x)+{\frac {1}{2}}c_{3n}'(x)\\a_{3n+2}(x)&=&6(2n+1)&{\frac {1}{2}}a_{3n+1}(x)&+&a_{3n}(x)&+&{\frac {1}{2}}a_{3n+1}'(x)+{\frac {1}{2}}a_{3n+2}'(x)+b_{3n+1}'(x)\end{Bmatrix}}$ , woraus

${\begin{Bmatrix}B_{3n-1}&=&(3n-1)&A_{3n-1}&+&{\frac {1}{2}}A_{3n-2}\\C_{3n-1}&=&&B_{3n-1}&+&A_{3n-1}\\A_{3n}&=&&C_{3n-1}&+&B_{3n-1}\\{\frac {1}{2}}A_{3n+1}&=&3n&A_{3n}&+&C_{3n-1}\\A_{3n+2}&=&6(2n+1)&{\frac {1}{2}}A_{3n+1}&+&A_{3n}\end{Bmatrix}}$ folgt. Daher erfüllen $A_{n}\,,\;B_{n}\,,\;C_{n}$ die Rekursion $(*)\,$ , woraus die Behauptung folgt.

Offenbar verschwinden $A_{n}\,,\;B_{n}\,,\;C_{n}$ für $n\to \infty \,$ , und daher ist $\varrho :=\lim _{n\to \infty }{\frac {p_{n}}{q_{n}}}=e^{2}$ .

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

Beweis

Für eine ungerade Zahl $m\geq 3$ sei $\varrho _{m}\,$ der Kettenbruch $[a_{0};a_{1},a_{2},a_{3},...]\,$ mit

$a_{5n}=1\,,\,a_{5n+1}={\frac {(6n+1)m-1}{2}}\,,\,a_{5n+2}=6(2n+1)m\,,\,a_{5n+3}={\frac {(6n+5)m-1}{2}}\,,\,a_{5n+4}=1$ für $n\in \mathbb {Z} ^{\geq 0}$ .

Das heißt $\varrho _{m}=\left[1,{\frac {m-1}{2}},6m,{\frac {5m-1}{2}},1,1,{\frac {7m-1}{2}},18m,{\frac {11m-1}{2}},1,...\right]$

$=\left[\,{\overline {1,{\frac {(6n+1)m-1}{2}},6(2n+1)m,{\frac {(6n+5)m-1}{2}},1}}\,\right]_{n\geq 0}$ .

Für Näherungsbrüche ${\frac {p_{n}}{q_{n}}}$ gilt allgemein die Rekursion ${\begin{Bmatrix}p_{n}=a_{n}p_{n-1}+p_{n-2}\\q_{n}=a_{n}q_{n-1}+q_{n-2}\end{Bmatrix}}$ mit den Startwerten ${\begin{Bmatrix}p_{-2}=0\;,\;p_{-1}=1\\q_{-2}=1\;,\;q_{-1}=0\end{Bmatrix}}$ .

Insbesondere für die Näherungsbrüche $\left({\frac {p_{n}}{q_{n}}}\right)_{n\geq 0}=\left({\frac {1}{1}},{\frac {(m+1)/2}{(m-1)/2}},{\frac {3m^{2}+3m+1}{3m^{2}-3m+1}},...\right)$ von $\varrho _{m}\,$ gelten für $n\geq 0\,$ die Rekursionsformeln:

${\begin{Bmatrix}p_{5n}&=&&p_{5n-1}&+&p_{5n-2}\\p_{5n+1}&=&{\frac {(6n+1)m-1}{2}}&p_{5n}&+&p_{5n-1}\\p_{5n+2}&=&6(2n+1)m&p_{5n+1}&+&p_{5n}\\p_{5n+3}&=&{\frac {(6n+5)m-1}{2}}&p_{5n+2}&+&p_{5n+1}\\p_{5n+4}&=&&p_{5n+3}&+&p_{5n+2}\end{Bmatrix}}\quad {\begin{Bmatrix}q_{5n}&=&&q_{5n-1}&+&q_{5n-2}\\q_{5n+1}&=&{\frac {(6n+1)m-1}{2}}&q_{5n}&+&q_{5n-1}\\q_{5n+2}&=&6(2n+1)m&q_{5n+1}&+&q_{5n}\\q_{5n+3}&=&{\frac {(6n+5)m-1}{2}}&q_{5n+2}&+&q_{5n+1}\\q_{5n+4}&=&&q_{5n+3}&+&q_{5n+2}\end{Bmatrix}}\quad (*)$

Setze ${\begin{bmatrix}a_{n}(x)=-{\frac {2^{n+1}}{m^{n+1}}}\,{\frac {x^{n}\,(x-1)^{n}}{n!}}\,e^{2x/m}&,&b_{n}(x)={\frac {2^{n+1}}{m^{n+1}}}\,{\frac {x^{n+1}\,(x-1)^{n}}{n!}}\,e^{2x/m}&,&c_{n}(x)={\frac {2^{n+1}}{m^{n+1}}}\,{\frac {x^{n}\,(x-1)^{n+1}}{n!}}\,e^{2x/m}\\\\A_{n}=\int _{0}^{1}a_{n}(x)dx&,&B_{n}=\int _{0}^{1}b_{n}(x)dx&,&C_{n}=\int _{0}^{1}c_{n}(x)dx\end{bmatrix}}$

Behauptung: Für alle $n\geq 0\,$ gilt ${\begin{Bmatrix}A_{3n}&=&p_{5n}-q_{5n}\cdot e^{2/m}\\{\frac {1}{2}}A_{3n+1}&=&p_{5n+1}-q_{5n+1}\cdot e^{2/m}\\A_{3n+2}&=&p_{5n+2}-q_{5n+2}\cdot e^{2/m}\\B_{3n+2}&=&p_{5n+3}-q_{5n+3}\cdot e^{2/m}\\C_{3n+2}&=&p_{5n+4}-q_{5n+4}\cdot e^{2/m}\end{Bmatrix}}$

Der Induktionsanfang $A_{0}=1-e^{2/m}\;,\;{\frac {1}{2}}A_{1}={\frac {m+1}{2}}-{\frac {m-1}{2}}\cdot e^{2/m}\;,\;A_{2}=(3m^{2}+3m+1)-(3m^{2}-3m+1)\cdot e^{2/m}\;,$

$B_{2}={\frac {15m^{3}+12m^{2}+3m}{2}}-{\frac {15m^{3}-18m^{2}+9m-2}{2}}\cdot e^{2/m}\;,\;C_{2}={\frac {15m^{3}+18m^{2}+9m+2}{2}}-{\frac {15m^{3}-18m^{2}+3m}{2}}\cdot e^{2/m}$

lässt sich leicht nachprüfen. Für $n\geq 1\,$ gilt nun

${\begin{Bmatrix}a_{3n}(x)&=&&c_{3n-1}(x)&+&b_{3n-1}(x)&+&{\frac {m}{2}}\cdot a_{3n}'(x)\\{\frac {1}{2}}a_{3n+1}(x)&=&{\frac {(6n+1)m-1}{2}}&a_{3n}(x)&+&c_{3n-1}(x)&+&{\frac {m}{4}}\cdot a_{3n+1}'(x)+{\frac {m}{2}}\cdot c_{3n}'(x)\\a_{3n+2}(x)&=&6(2n+1)m&{\frac {1}{2}}a_{3n+1}(x)&+&a_{3n}(x)&+&{\frac {m}{2}}\cdot a_{3n+1}'(x)+{\frac {m}{2}}\cdot a_{3n+2}'(x)+m\cdot b_{3n+1}'(x)\\b_{3n+2}(x)&=&{\frac {(6n+1)m-1}{2}}&a_{3n+2}(x)&+&{\frac {1}{2}}a_{3n+1}(x)&+&{\frac {m}{2}}\cdot c_{3n+2}'(x)-{\frac {m}{4}}\cdot a_{3n+2}'(x)\\c_{3n+2}(x)&=&&b_{3n+2}(x)&+&a_{3n+2}(x)\end{Bmatrix}}$ , woraus

${\begin{Bmatrix}A_{3n}&=&&C_{3n-1}&+&B_{3n-1}\\{\frac {1}{2}}A_{3n+1}&=&{\frac {(6n+1)m-1}{2}}&A_{3n}&+&C_{3n-1}\\A_{3n+2}&=&6(2n+1)m&{\frac {1}{2}}A_{3n+1}&+&A_{3n}\\B_{3n+2}&=&{\frac {(6n+5)m-1}{2}}&A_{3n+2}&+&{\frac {1}{2}}A_{3n+1}\\C_{3n+2}&=&&B_{3n+2}&+&A_{3n+2}\end{Bmatrix}}$ folgt. Daher erfüllen $A_{n}\,,\;B_{n}\,,\;C_{n}$ die Rekursion $(*)\,$ , woraus die Behauptung folgt.

Offenbar verschwinden $A_{n}\,,\;B_{n}\,,\;C_{n}$ für $n\to \infty \,$ , und daher ist $\varrho _{m}:=\lim _{n\to \infty }{\frac {p_{n}}{q_{n}}}=e^{2/m}$ .

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

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

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

(Ko)tangens (Hyperbolicus)

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

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

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

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

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

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

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

Quadratwurzeln

${\sqrt {2}}=\left[1,{\overline {2}}\right]$

${\sqrt {3}}=\left[1,{\overline {1,2}}\right]$

${\sqrt {4}}=\left[2\right]$

${\sqrt {5}}=\left[2,{\overline {4}}\right]$

${\sqrt {6}}=\left[2,{\overline {2,4}}\right]$

${\sqrt {7}}=\left[2,{\overline {1,1,1,4}}\right]$

${\sqrt {8}}=\left[2,{\overline {1,4}}\right]$

${\sqrt {9}}=\left[3\right]$

${\sqrt {10}}=\left[3,{\overline {6}}\right]$

Ist $d\in \mathbb {Z} ^{>0}\,$ kein Quadrat, so lässt sich ${\sqrt {d}}$ schreiben in der Form $\left[a_{0},{\overline {a_{1},...,a_{n},2a_{0}}}\right]$

Familien von Kettenbrüchen

Zum Beispiel

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

${\sqrt {a^{2}+2}}=\left[a,{\overline {a,2a}}\right]$

${\sqrt {a^{2}+a}}=\left[a,{\overline {2,2a}}\right]$

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

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

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 ${\frac {a+b{\sqrt {c}}}{d}}$ besitzt, wobei $a\in \mathbb {Z} ,b,c,d\in \mathbb {Z} ^{\neq 0}$ ist und $c\,$ keine Quadratzahl ist.

Satz von Galois

Sind $a_{0},...,a_{n}\in \mathbb {Z} ^{>0}$ , dann lässt sich der reinperiodische Kettenbruch $\alpha =[{\overline {a_{0},...,a_{n}}}]$ schreiben in der Form ${\frac {P+{\sqrt {D}}}{Q}}$ ,

wobei $P,Q,D\in \mathbb {Z} ^{>0}$ sind und $D\,$ keine Quadratzahl ist.

Ist $\beta =[{\overline {a_{n},...,a_{0}}}]$ der zu $\alpha \,$ inverse Kettenbruch, so stimmt ${\frac {-1}{\beta }}$ mit der Wurzelkonjugierten ${\frac {P-{\sqrt {D}}}{Q}}$ überein.

Verallgemeinerte Kettenbrüche

Eine mögliche Verallgemeinerung ist es Kettenbrüche der Form

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

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

Formel von Bombelli

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

Eulersche Zahl

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

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

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

Kreiszahl $\pi$

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

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

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

Formel von Brouncker

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

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

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

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

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

Catalansche Konstante

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

Exponentialfunktion

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

Sinus (Hyperbolicus)

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

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

Tangens (Hyperbolicus)

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

Beweis

Zu vorgegebenem $x,y$ erfüllt die Folge $a_{n}:={\frac {J_{n+1/2}\left({\frac {x}{y}}\right)}{x^{n+1/2}}}$

die Rekursion $a_{n-1}=(2n+1)y\cdot a_{n}-x^{2}\cdot a_{n+1}$ .

$\Rightarrow \,{\frac {a_{n-1}}{a_{n}}}=(2n+1)y-x^{2}\cdot {\frac {a_{n+1}}{a_{n}}}\Rightarrow \,{\frac {a_{n}}{a_{n-1}}}={\frac {1}{(2n+1)y-x^{2}\cdot {\frac {a_{n+1}}{a_{n}}}}}$

Die Folge $b_{n}:={\frac {a_{n}}{a_{n-1}}}={\frac {1}{x}}\,{\frac {J_{n+1/2}\left({\frac {x}{y}}\right)}{J_{n-1/2}\left({\frac {x}{y}}\right)}}$

erfüllt daher die Rekursion $b_{n}={\frac {1}{(2n+1)y-x^{2}\cdot b_{n+1}}}$ , wobei $b_{0}={\frac {\tan \left({\frac {x}{y}}\right)}{x}}$ ist.

Also ist ${\frac {\tan \left({\frac {x}{y}}\right)}{x}}={\frac {1}{y-{\frac {x^{2}}{3y-{\frac {x^{2}}{5y-{\frac {x^{2}}{7y-...}}}}}}}}$ .

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

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

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

Kotangens Hyperbolicus

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

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

Arkussinus und Areasinus Hyperbolicus

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

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

Arkustangens und Areatangens Hyperbolicus

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

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

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

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

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

Logarithmus

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

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

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

Fehlerfunktion

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

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

Gammafunktion

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

Besselfunktion

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

Beweis

Setze $c_{k}={\text{BesselI}}\left(k+{\frac {a}{d}},{\frac {2x}{d}}\right)x^{k}=\sum _{n=0}^{\infty }{\frac {x^{2n+2k+{\frac {a}{d}}}}{n!\,d^{n}\,\left(a+k+{\frac {a}{d}}\right)!\,d^{\,n+k+{\frac {a}{d}}}}}$ .

Es gilt

${\frac {\left(n+k+{\frac {a}{d}}\right)\,x^{2n+2k+{\frac {a}{d}}}}{n!\,d^{n}\,\left(n+k+{\frac {a}{d}}\right)!\,d^{\,n+k+{\frac {a}{d}}}}}-{\frac {\left(k+{\frac {a}{d}}\right)\,x^{2n+2k+{\frac {a}{d}}}}{n!\,d^{n}\,\left(n+k+{\frac {a}{d}}\right)!\,d^{\,n+k+{\frac {a}{d}}}}}={\frac {n\,x^{2n+2k+{\frac {a}{d}}}}{n!\,d^{n}\,\left(n+k+{\frac {a}{d}}\right)!\,d^{\,n+k+{\frac {a}{d}}}}}$ ,

gleichbedeutend mit

${\frac {x^{2}\,{\frac {1}{d}}\,x^{2n+2k-2+{\frac {a}{d}}}}{n!\,d^{n}\,\left(n+k-1+{\frac {a}{d}}\right)!\,d^{\,n+k-1+{\frac {a}{d}}}}}-(a+kd)\cdot {\frac {{\frac {1}{d}}\,x^{2n+2k+{\frac {a}{d}}}}{n!\,d^{n}\,\left(n+k+{\frac {a}{d}}\right)!\,d^{\,n+k+{\frac {a}{d}}}}}={\frac {{\frac {1}{d}}\,x^{2n+2k+{\frac {a}{d}}}}{(n-1)!\,d^{n-1}\,\left(n+k+{\frac {a}{d}}\right)!\,d^{\,n+k+{\frac {a}{d}}}}}$ .

Multipliziert man mit $d\,$ durch und summiert über $n\,$ , so führt dies zur Gleichung $x^{2}\,c_{k-1}-(a+kd)c_{k}=c_{k+1}$ .

Daher ist $x^{2}\,{\frac {c_{k-1}}{c_{k}}}=(a+kd)+{\frac {c_{k+1}}{c_{k}}}$ und somit ${\frac {c_{k}}{c_{k-1}}}={\frac {x^{2}}{(a+kd)+{\frac {c_{k+1}}{c_{k}}}}}$ .

Also ist ${\frac {c_{1}}{c_{0}}}={\frac {x^{2}}{a+d+{\frac {c_{2}}{c_{1}}}}}={\frac {x^{2}}{a+d+{\frac {x^{2}}{a+2d+{\frac {c_{3}}{c_{2}}}}}}}={\frac {x^{2}}{a+d+{\frac {x^{2}}{a+2d+{\frac {x^{2}}{a+3d+{\frac {c_{4}}{c_{3}}}}}}}}}=...$

Ramanujan-Kettenbrüche

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

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

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

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Reguläre Kettenbrüche

Negativer Wert

Kehrwert

Goldener Schnitt

Eulersche Zahl

(Ko)tangens (Hyperbolicus)

Quadratwurzeln

Familien von Kettenbrüchen

Allgemeine Aussagen über reguläre Kettenbrüche

Satz von Galois

Verallgemeinerte Kettenbrüche

Formel von Bombelli

Eulersche Zahl

Kreiszahl π {\displaystyle \pi }

Catalansche Konstante

Exponentialfunktion

Sinus (Hyperbolicus)

Tangens (Hyperbolicus)

Kotangens Hyperbolicus

Arkussinus und Areasinus Hyperbolicus

Arkustangens und Areatangens Hyperbolicus

Logarithmus

Fehlerfunktion

Gammafunktion

Besselfunktion

Ramanujan-Kettenbrüche

Kreiszahl $\pi$