[Proposed] Equation Fonctionnelle


  • Math&Maroc

    Trouver toutes les fonctions f:RR f: \mathbb R \to \mathbb R telles que:
    f(f(x)+y)=f(x2y)+4yf(y)     x,yR f(f(x)+y)=f(x^2-y)+4yf(y) \ \ \ \ \ \forall x,y \in \mathbb R

    Proposé par Amir Hossein Parvardi



  • Just some ideas :
    P(x,x2f(x)2)P(x, \frac{x^2-f( x)}{2} ) \implies f(x)=x2 f (x)=x^2 and f(x2f(x)2)=0f(\frac{x^2-f(x)}{2})=0
    First equation shows that f(x)=x2f (x)=x^2 for all x in R .
    Second equation needs to be used .
    I thought abt showing injectivity and surjectivity hence we can find an uinR such that f(u)=0f(u)=0 and hence x2f(x)=2ux^2-f(x)=2u so f(x)=x22uf (x)=x^2-2u .
    But we all see that f(x)=0f(x)=0 is a solution , so i suppose we should use the second equation to show that , for all x .

    Sauf erreur


  • Math&Maroc

    Ligne2: P(x,x2f(x)2)f(x)=x2P(x,\frac{x^2-f(x)}{2}) \rightarrow f(x) =x^2 OU f(x2f(x)2)=0f(\frac{x^2-f(x)}{2})=0 non pas ET
    Sinon tu es sur une bonne piste, continue.


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