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{\displaystyle \mathrm {q} (x)=\sum _{i=0}^{\mathrm {Int} \log x}\left(\mathrm {Int} {\frac {x}{10^{i}}}\ -\ 10\mathrm {Int} {\frac {x}{10^{i+1}}}\right)}
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{\displaystyle \mu _{0}\,\varepsilon _{0}\,c^{2}=-e^{i\pi }}
Satz : Seien
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{\displaystyle \left(x_{n}\right)_{n\in \mathbb {N} }}
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{\displaystyle \left(y_{n}\right)_{n\in \mathbb {N} }}
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{\displaystyle \mathbb {R} }
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{\displaystyle x_{n}\to x}
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{\displaystyle y_{n}\to y}
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{\displaystyle \left(x_{n}y_{n}\right)_{n\in \mathbb {N} }\to \left(xy\right)}
Beweis :
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{\displaystyle \left|x_{n}y_{n}-xy\right|\leq \varepsilon }
Dazu suchen wir eine Konstante k, für die gilt:
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{\displaystyle \forall n\in \mathbb {N} :\left|x_{n}\right|\leq k\land \left|y_{n}\right|\leq k}
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{\displaystyle x_{n}}
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{\displaystyle y_{n}}
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{\displaystyle \varepsilon _{\text{neu}}:={\frac {\varepsilon }{2k}}}
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{\displaystyle x_{n}}
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{\displaystyle y_{n}}
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{\displaystyle {\begin{aligned}\forall n\leq N_{\varepsilon _{\text{neu}}}:\\\left|x_{n}-x\right|\leq {\frac {\varepsilon }{2k}}\\\left|y_{n}-y\right|\leq {\frac {\varepsilon }{2k}}\end{aligned}}}
Erinnern wir uns an das, was wir eigentlich zeigen wollten:
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{\displaystyle {\begin{aligned}\left|x_{n}y_{n}-xy\right|\leq \varepsilon \\\left|x_{n}y_{n}-xy\right|&=\left|x_{n}y_{n}-x_{n}y+x_{n}y-xy\right|\\&\leq \left|x_{n}y_{n}-x_{n}y\right|+\left|x_{n}y-xy\right|\\&=\left|x_{n}\right|\left|y_{n}-y\right|+\left|y\right|\left|x_{n}-x\right|\\&\leq k{\frac {\varepsilon }{2k}}+k{\frac {\varepsilon }{2k}}\\&={\frac {\varepsilon }{2}}+{\frac {\varepsilon }{2}}\\\left|x_{n}y_{n}-xy\right|&\leq \varepsilon \quad {\ddot {\smile }}\end{aligned}}}
