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Suma uogolniona
: 20 lut 2010, o 09:19
autor: FAUSTVIII
Niech \(\displaystyle{ <x,y> \in R \times R}\)
\(\displaystyle{ Sn = \{ x^{2}+y^{2} \le n^{2}\}}\) gdzie\(\displaystyle{ n \in N}\)
Proszę wyznaczyć :
\(\displaystyle{ \bigcup_{n=1}^{\infty} (An+1/ An}\))
Suma uogolniona
: 20 lut 2010, o 14:39
autor: Crizz
Przecież \(\displaystyle{ A_{n} \subset A_{n+1}}\), zatem
\(\displaystyle{ (A_{n+1} \backslash A_{n}) \cup (A_{n+2} \backslash A_{n+1})=A_{n+2}\backslash A_{n}}\)
\(\displaystyle{ (A_{n+1} \backslash A_{n}) \cup (A_{n+2} \backslash A_{n+1}) \cup (A_{n+3} \backslash A_{n+2})=A_{n+3}\backslash A_{n}}\) itd.
Mamy więc
\(\displaystyle{ \bigcup_{n=1}^{p} (A_{n+1} \backslash A_{n}) =A_{p+1}/A_{1}}\)
\(\displaystyle{ \bigcup_{n=1}^{\infty}(A_{n+1} \backslash A_{n})=A_{\infty}/A_{1}}\)
\(\displaystyle{ \bigcup_{n=1}^{\infty}(A_{n+1} \backslash A_{n})=\{(x,y):x^{2}+y^{2} \le \infty\} \backslash \{(x,y):x^{2}+y^{2} \le 1\}= \\ =\{(x,y):x^{2}+y^{2}>1\}}\)