Formelsammlung Mathematik: Kettenbrüche

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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

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

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

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