Nach
Aufgabe
ist
-
![{\displaystyle {}{\binom {n}{k+1}}={\frac {n-k}{k+1}}\cdot {\binom {n}{k}}\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d7a04df4070914d9293f04cf6fc90029ccff02e)
Somit besteht zwischen
und
der Zusammenhang
![{\displaystyle {}{\begin{aligned}{\binom {n}{\frac {n}{2}}}&={\frac {{\frac {n}{2}}+1}{\frac {n}{2}}}\cdot {\binom {n}{{\frac {n}{2}}-1}}\\&={\frac {{\frac {n}{2}}+1}{\frac {n}{2}}}\cdot {\frac {{\frac {n}{2}}+2}{{\frac {n}{2}}-1}}\cdot {\binom {n}{{\frac {n}{2}}-2}}\\&={\frac {{\frac {n}{2}}+1}{\frac {n}{2}}}\cdot {\frac {{\frac {n}{2}}+2}{{\frac {n}{2}}-1}}\cdots {\frac {{\frac {n}{2}}+\left\lfloor \beta n\right\rfloor +1}{{\frac {n}{2}}-\left\lfloor \beta n\right\rfloor }}\cdot {\binom {n}{{\frac {n}{2}}-\left\lfloor \beta n\right\rfloor }}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6889cafb694b6d97f37ba47caf9cc19d8f6e0fea)
Dies bedeutet umgekehrt
![{\displaystyle {}{\begin{aligned}{\binom {n}{{\frac {n}{2}}-\left\lfloor \beta n\right\rfloor }}&={\frac {\frac {n}{2}}{{\frac {n}{2}}+1}}\cdot {\frac {{\frac {n}{2}}-1}{{\frac {n}{2}}+2}}\cdots {\frac {{\frac {n}{2}}-\left\lfloor \beta n\right\rfloor }{{\frac {n}{2}}+\left\lfloor \beta n\right\rfloor +1}}\cdot {\binom {n}{\frac {n}{2}}}\\&={\frac {{\frac {n}{2}}-\left\lfloor \beta n\right\rfloor }{{\frac {n}{2}}+1}}\cdot {\frac {{\frac {n}{2}}-\left\lfloor \beta n\right\rfloor +1}{{\frac {n}{2}}+2}}\cdots {\frac {\frac {n}{2}}{{\frac {n}{2}}+\left\lfloor \beta n\right\rfloor +1}}\cdot {\binom {n}{\frac {n}{2}}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a784e6f8dd9443b2021ea2aecadf2726ca3e68ce)
Die Faktoren sind alle von der Form
-
mit
.
Sie sind alle
und für das maximale
, also für
, am größten. Da es
viele Faktoren gibt, kann man das Produkt unter Verwendung von
Fakt (1),
Fakt (6)
und
Fakt (8)
durch
![{\displaystyle {}{\begin{aligned}{\left({\frac {\frac {n}{2}}{{\frac {n}{2}}+\left\lfloor \beta n\right\rfloor +1}}\right)}^{\left\lfloor \beta n\right\rfloor +1}&\leq {\left({\frac {\frac {n}{2}}{{\frac {n}{2}}+\beta n}}\right)}^{\left\lfloor \beta n\right\rfloor +1}\\&\leq {\left({\frac {\frac {n}{2}}{{\frac {n}{2}}+\beta n}}\right)}^{\beta n}\\&={\left({\frac {\frac {1}{2}}{{\frac {1}{2}}+\beta }}\right)}^{\beta n}\\&={\left({\left({\frac {1}{1+2\beta }}\right)}^{\beta }\right)}^{n}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7920b7a0d57544d97d09f36f62debaebf7607e7b)
nach oben abschätzen. Also ist
-
![{\displaystyle {}{\binom {n}{{\frac {n}{2}}-\left\lfloor \beta n\right\rfloor }}\leq {\left({\left({\frac {1}{1+2\beta }}\right)}^{\beta }\right)}^{n}\cdot {\binom {n}{\frac {n}{2}}}\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/041e1ecd6d04d80c44e3d0eceb0d76fc63410810)