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O ecuatie functionala
Posted: Sat Feb 14, 2009 11:44 pm
by andy crisan
Sa se determine \( f:\mathbb{Z}\rightarrow \mathbb{Z} \) cu proprietatea ca \( 2f(f(x))-5f(x)=3x\ (\forall) x \in \mathbb{Z} \).
Posted: Sun Feb 15, 2009 12:21 am
by Marius Mainea
Notam \( x_n=f^{(n)}(x) \) si obtinem relatia de recurenta \( 2x_{n+2}-5x_{n+1}-3x_n=0 \) de unde
\( x_n=\frac{2f(x)+x}{7}3^n+\frac{6x-2f(x)}{7}(-\frac{1}{2})^n \).
Asadar \( f(x)=3x \).
Posted: Sun Feb 15, 2009 1:51 pm
by andy crisan
Am gasit o solutie destul de interesanta.
Daca notam \( f^{n}(x)-3f^{n-1}(x)=a_{n}(x) \) obtinem ca \( a_{n}(x)=\frac{-a_{n-1}(x)}{2}\forall x \in \mathbb{Z} \) si prin inductie se arata ca \( a_{n+1}(x)=\frac{(-1)^{n}\cdot a_{1}(x)}{2^{n}}\forall x \in \mathbb{Z}\forall n \in \mathbb{N}^* \).
\( \Rightarrow 2^{n}|a_{1}(x)\forall x \in \mathbb{Z}\forall n \in \mathbb{N}^* \) si cum \( a_{1}(x) \) nu depinde de \( n \)\( \Rightarrow a_{1}(x)=0 \forall x \in \mathbb{Z} \Leftrightarrow f(x)=3x \forall x \in \mathbb{Z} \)