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Alta functie periodica
Posted: Tue Nov 25, 2008 9:10 pm
by alex2008
Se da functia \( f:\mathb{N}\rightarrow\mathb{Q} \) cu \( f(n+1)=\frac{2}{2-f(n)} \) . Sa se arate ca functia este periodica pe \( \mathb{N} \).
Posted: Tue Nov 25, 2008 9:51 pm
by Marius Mainea
\( f(n+2)=\frac{2}{2-f(n+1)}=\frac{2}{2-\frac{2}{2-f(n)}}=1+\frac{1}{1-f(n)} \) apoi
\( f(n+4)=1+\frac{1}{1-f(n+2)}=1+\frac{1}{1-1-\frac{1}{1-f(n)}}=1-1+f(n)=f(n) \)