Aratati ca exista o singura functie \( f\ :\ \mathbb{R}\rightarrow \mathbb{R} \) care satisface egalitatea:
\( f^{2}(x+y)+f^{2}(x-y)=f^{2}(x)+y^{2}f^{2}\left\(\frac{x}{y}\right\)\ ,\ \forall\ x\in \mathbb{R}\ ,\ \forall\ y\in \mathbb{R}^{*}\ . \)
Ecuatie functionala
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Mateescu Constantin
- Newton
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Ecuatie functionala
Last edited by Mateescu Constantin on Wed May 19, 2010 10:58 am, edited 2 times in total.
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DrAGos Calinescu
- Thales
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Din substitutiile \( y=1 \) si \( y=-1 \) obtinem \( f^2(x)=f^2(-x) \)
Pentru \( x=0 \) avem \( f^2(y)+f^2(-y)=f^2(0)+y^2f^2(0)\forall y\in\mathbb{R}^*\Longrightarrow 2f^2(y)=f^2(0)(y^2+1) \)
Notand \( f(0)=a \) obtinem \( f(x)=a^2(x^2+1)\forall x\in\mathbb{R}^* \)
Prin inlocuire in ecuatie obtinem \( a^2(x^2+1)=0\Longrightarrow a=0\Longrightarrow f(x)=0\forall x\in\mathbb{R} \)
Deci singura functie care satisface conditiile problemei este functia nula.
Pentru \( x=0 \) avem \( f^2(y)+f^2(-y)=f^2(0)+y^2f^2(0)\forall y\in\mathbb{R}^*\Longrightarrow 2f^2(y)=f^2(0)(y^2+1) \)
Notand \( f(0)=a \) obtinem \( f(x)=a^2(x^2+1)\forall x\in\mathbb{R}^* \)
Prin inlocuire in ecuatie obtinem \( a^2(x^2+1)=0\Longrightarrow a=0\Longrightarrow f(x)=0\forall x\in\mathbb{R} \)
Deci singura functie care satisface conditiile problemei este functia nula.