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Ecuatie functionala cu functii rationale
Posted: Sun Dec 02, 2007 6:54 pm
by Andrei Velicu
Sa se determine functiile \( f:\mathbb{Q}\to \mathbb{Q} \) cu proprietatea:
\( f(x+y)+f(x-y)=f(x)+f(y)+f(f(x)-f(y)), \forall x, y \in \mathbb{Q} \).
Mihai Onucu Drimbe, "Nicolae Coculescu" 2007
Re: determinarea functiilor rationale
Posted: Fri Dec 07, 2007 3:19 pm
by turcas
Andrei Velicu wrote:Sa se determine functiile \( f:\mathbb{Q}\to \mathbb{Q} \) cu proprietatea:
\( f(x+y)+f(x-y)=f(x)+f(y)+f(f(x)-f(y)), \forall x, y \in \mathbb{Q} \).
Mihai Onucu Drimbe, "Nicolae Coculescu" 2007
\( f(x+y)+f(x-y)=f(x)+f(y)+f(f(x)-f(y)) \);
(1)
Daca inlocuim
\( x=y=0 \) in
(1), obtinem
\( f(0)=0 \) si apoi daca punem
\( y=0 \) in (1), obtinem
\( f(f(x))=f(x) \) (2);
Punem acum
\( x=0 \) in
(1) si obtinem
\( f(-f(y))=f(-y) \) (3);
Observam acum, luand cateva exemple
\( x=y \) si inlocuind in ipoteza ca
\( f(2x)=2f(x) \).
Se arata usor prin inductie ca
\( f(nx)=n \cdot f(x), \forall n \in \mathbb{N}^{*}, x \in \mathbb{Q} \Longleftrightarrow f(\frac{x}{n})=\frac{1}{n}f(x). \)
Daca avem
\( p,q \in \mathbb{N}^{*} \), atunci
\( f(\frac{p}{q})=\frac{p}{q}f(1) \).
De asemenea
\( f(-\frac{p}{q})=\frac{p}{q}f(-1) \).
Din cele de mai sus avem:
\( f(x)=\left\{\begin{array}{c} xf(1), \ x > 0 \\ -xf(-1), \ x <0 \end{array} \).
Se analizeaza cazurile
\( f(1)>0, \ f(-1)>0, \ f(1)<0, \ f(-1)<0 \) si se obtin solutiile:
\( f(x)=|x|, \ f(x)=x, \ f(x)=-|x| \) si
\( f(x)=0 \), care verifica ipoteza.