«K12/K13» 8. Hausaufgabe «PDF», «POD»

0.0.1 ↑ 8. Hausaufgabe

0.0.1.1 ↑ Analysis-Buch Seite 36, Aufgabe 17a

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F := \left\{ (x,y) \,|\, 0 \leq x \leq 1 \wedge 0 \leq y \leq 1 - x^2 \right\};$F:=\left\{\left(x,y\right)\mid 0\le x\le 1\wedge 0\le y\le 1-x^2\right\};$

[\sum\limits_{i = 1}^n i = \frac{n\left(n + 1\right)}{2}; \quad$\sum _{i=1}^{n}i=\frac{n\left(n+1\right)}{2};$ \sum\limits_{i = 1}^n i^2 = \frac{n\left(n + 1\right)\left(2n + 1\right)}{6};$\sum _{i=1}^n{i}^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6};$]

\mathrm{f}(x) = 1 - x^2;$\text{f}\left(x\right)=1-x^2;$

{} \renewcommand{\arraystretch}{3.0} \begin{array}{rcl} {} S_n &=& \sum\limits_{i = 1}^n \frac{1}{n} \mathrm{f}\!\left(\frac{i - 1}{n}\right) = {} \frac{1}{n} \sum\limits_{i = 1}^n \left[1 - \frac{i^2 - 2i + 1}{n^2}\right] = {} \frac{n \cdot 1}{n} + \frac{1}{n}\sum\limits_{i = 1}^n - \frac{i^2 - 2i + 1}{n^2} = \\ {} &=& 1 - \frac{1}{n^3}\sum\limits_{i = 1}^n \left[i^2 - 2i + 1\right] = {} 1 - \frac{1}{n^3}\left(\,\sum\limits_{i = 1}^n i^2 + 2 \sum\limits_{i = 1}^n i + n\right) = \\ {} &=& 1 - \frac{1}{n^3}\left[\frac{n\left(n + 1\right)\left(2n + 1\right)}{6} + n\left(n + 1\right) + n\right] = 1 - \frac{2n^2 + 9n + 13}{6n^2}; {} \end{array}$\begin{array}{ccc}\hfill S_n& \hfill =\hfill & \sum _{i=1}^n\frac{1}{n}\text{f}\left(\frac{i-1}{n}\right)=\frac{1}{n}\sum _{i=1}^n\left[1-\frac{i^2-2i+1}{n^2}\right]=\frac{n\cdot 1}{n}+\frac{1}{n}\sum _{i=1}^n-\frac{i^2-2i+1}{n^2}=\hfill \\ \hfill & \hfill =\hfill & 1-\frac{1}{n^3}\sum _{i=1}^n\left[i^2-2i+1\right]=1-\frac{1}{n^3}\left(\sum _{i=1}^n{i}^2+2\sum _{i=1}^{n}i+n\right)=\hfill \\ \hfill & \hfill =\hfill & 1-\frac{1}{n^3}\left[\frac{n\left(n+1\right)\left(2n+1\right)}{6}+n\left(n+1\right)+n\right]=1-\frac{2n^2+9n+13}{6n^2};\hfill \end{array}$

{} \renewcommand{\arraystretch}{3.0} \begin{array}{rcl} {} s_n &=& \sum\limits_{i = 1}^n \frac{1}{n} \mathrm{f}\!\left(\frac{i}{n}\right) = {} \frac{1}{n} \sum\limits_{i = 1}^n \left[1 - \frac{i^2}{n^2}\right] = {} \frac{n \cdot 1}{n} - \frac{1}{n^3} \sum\limits_{i = 1}^n i^2 = \\ {} &=& 1 - \frac{n\left(n + 1\right)\left(2n + 1\right)}{6n^3} = {} 1 - \frac{2n^2 + n + 2n + 1}{6n^2} = {} 1 - \frac{2n^2 + 3n + 1}{6n^2}; {} \end{array}$\begin{array}{ccc}\hfill s_n& \hfill =\hfill & \sum _{i=1}^n\frac{1}{n}\text{f}\left(\frac{i}{n}\right)=\frac{1}{n}\sum _{i=1}^n\left[1-\frac{i^2}{n^2}\right]=\frac{n\cdot 1}{n}-\frac{1}{n^3}\sum _{i=1}^n{i}^2=\hfill \\ \hfill & \hfill =\hfill & 1-\frac{n\left(n+1\right)\left(2n+1\right)}{6n^3}=1-\frac{2n^2+n+2n+1}{6n^2}=1-\frac{2n^2+3n+1}{6n^2};\hfill \end{array}$

\left.\begin{array}{l} {} \lim\limits_{n \to \infty} S_n = 1 - \frac{1}{3} = \frac{2}{3}; \\ {} \lim\limits_{n \to \infty} s_n = 1 - \frac{1}{3} = \frac{2}{3}; \end{array}\right\} \Rightarrow \lim\limits_{n \to \infty} S_n = \lim\limits_{n \to \infty} s_n = \frac{2}{3} =: A;$\left.\begin{array}{c}\underset{n\to \infty }{lim}{S}_n=1-\frac{1}{3}=\frac{2}{3};\hfill \\ \underset{n\to \infty }{lim}{s}_n=1-\frac{1}{3}=\frac{2}{3};\hfill \end{array}\right\}\Rightarrow \underset{n\to \infty }{lim}{S}_n=\underset{n\to \infty }{lim}{s}_n=\frac{2}{3}=:A;$