Determinati toate functiile f : Z-->Z cu proprietatea ca
\( f(f(n+1)+3)=n \) pentru orice numar intreg.
Mihai Opincariu G.M.B. 1999
Ecuatie functionala
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opincariumihai
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Ecuatie functionala
Last edited by opincariumihai on Sat May 23, 2009 12:31 pm, edited 1 time in total.
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Mateescu Constantin
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In \( f(f(n+1)+3)=n \ (1) \) inlocuim pe \( n \) cu \( n+2 \) si obtinem:
\( f(f(n+3)+3)=n+2. \)
Inlocuind acum \( n \) cu \( f(n+1) \), obtinem:
\( f(f(f(n+1)+3)+3)=f(n+1)+2 \)
\( \Longleftrightarrow f(n+3)=f(n+1)+2 \)
\( \Longleftrightarrow f(n+2)=f(n)+2. \)
Sa notam \( f(0)=a \)
\( \Longrightarrow f(2)=a+2,\ f(4)=a+4 \) si prin inductie
\( \Longrightarrow f(2n)=a+2n \)
Apoi avem \( f(-2)=a-2,\ f(-4)=a-4 \) si prin inductie
\( \Longrightarrow f(-2n)=a-2n \)
Atunci \( f(2n)=a+2n,\ \forall n\in \mathbb{Z}. \)
Punand \( f(1)=b \) avem \( f(3)=b+2 \) si inductiv \( f(2n+1)=b+2n \) si apoi \( f(-2n+1)=b-2n \), de unde
\( f(2n+1)=b+2n,\ \forall n\in \mathbb{Z}. \)
In relatia \( (1) \) luam \( k=0 \)
\( \Longrightarrow f(f(1)+3)=0 \)
\( \Longleftrightarrow f(b+3)=0 \ (2) \)
Daca \( b=2k \), atunci ultima relatie se scrie
\( f(2k+3)=0 \Longleftrightarrow b+2k+2=0 \Longleftrightarrow 4k+2=0 \), imposibil.
Deci \( b=2k+1 \) si \( (2) \) devine:
\( f(2k+4)=0 \Longleftrightarrow a+2k+4=0 \) si atunci \( a=2p \), unde \( k=-p-2. \)
Avem atunci \( f(n)=\left\{ \begin 2p+n,\ \mbox{daca n este par}\\-2p-n-4,\ \mbox{daca n este impar} \), unde \( p\in \mathbb{Z} \), functie care verifica enuntul.
\( f(f(n+3)+3)=n+2. \)
Inlocuind acum \( n \) cu \( f(n+1) \), obtinem:
\( f(f(f(n+1)+3)+3)=f(n+1)+2 \)
\( \Longleftrightarrow f(n+3)=f(n+1)+2 \)
\( \Longleftrightarrow f(n+2)=f(n)+2. \)
Sa notam \( f(0)=a \)
\( \Longrightarrow f(2)=a+2,\ f(4)=a+4 \) si prin inductie
\( \Longrightarrow f(2n)=a+2n \)
Apoi avem \( f(-2)=a-2,\ f(-4)=a-4 \) si prin inductie
\( \Longrightarrow f(-2n)=a-2n \)
Atunci \( f(2n)=a+2n,\ \forall n\in \mathbb{Z}. \)
Punand \( f(1)=b \) avem \( f(3)=b+2 \) si inductiv \( f(2n+1)=b+2n \) si apoi \( f(-2n+1)=b-2n \), de unde
\( f(2n+1)=b+2n,\ \forall n\in \mathbb{Z}. \)
In relatia \( (1) \) luam \( k=0 \)
\( \Longrightarrow f(f(1)+3)=0 \)
\( \Longleftrightarrow f(b+3)=0 \ (2) \)
Daca \( b=2k \), atunci ultima relatie se scrie
\( f(2k+3)=0 \Longleftrightarrow b+2k+2=0 \Longleftrightarrow 4k+2=0 \), imposibil.
Deci \( b=2k+1 \) si \( (2) \) devine:
\( f(2k+4)=0 \Longleftrightarrow a+2k+4=0 \) si atunci \( a=2p \), unde \( k=-p-2. \)
Avem atunci \( f(n)=\left\{ \begin 2p+n,\ \mbox{daca n este par}\\-2p-n-4,\ \mbox{daca n este impar} \), unde \( p\in \mathbb{Z} \), functie care verifica enuntul.