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Question Number 148558 by puissant last updated on 29/Jul/21

$$\mathrm{Trouver}\mathrm{toutes}\mathrm{les}\mathrm{fonctions}\mathrm{continues}$$$$\mathrm{f}:\mathbb{R}\to \mathbb{R}\mathrm{verifiant}:$$$$\mathrm{\forall }(\mathrm{x},\mathrm{y})\in {\mathbb{R}}^2,\mathrm{f}(\mathrm{x}+\mathrm{y})\mathrm{f}(\mathrm{x}-\mathrm{y})={\mathrm{f}}^2\left(\mathrm{x}\right){\mathrm{f}}^2\left(\mathrm{y}\right)..$$$$\mathrm{monsieur}{\mathrm{j}}^{\prime }\mathrm{ai}\mathrm{suppos}\acute{\mathrm{e}}\mathrm{que}\mathrm{f}\mathrm{est}\mathrm{un}$$$$\mathrm{morphisme}\mathrm{mutiplicatif}\mathrm{de}\mathbb{R}..\mathrm{mais}\mathrm{ca}\mathrm{ne}$$$$\mathrm{sort}\mathrm{pas}...$$

Answered by Olaf_Thorendsen last updated on 29/Jul/21

$$\mathrm{\forall }(x,\mathrm{y})\in {\mathbb{R}}^2,f(x+y)f(x-y)=f^2\left(x\right)f^2\left(y\right)$$$$$$$$\bullet \mathrm{On}\mathrm{elimine}{\mathrm{d}}^{\prime }\mathrm{emblee}\mathrm{le}\mathrm{cas}\mathrm{trivial}$$$$\mathrm{de}\mathrm{la}\mathrm{fonction}\mathrm{identiquement}\mathrm{nulle}\left(1\right)$$$$$$$$\bullet \mathrm{En}\mathrm{particulier},\mathrm{pour}x=y=0:$$$$f(0+0)f(0-0)=f^2\left(0\right)f^2\left(0\right)$$$$\mathrm{Et}\mathrm{donc}f\left(0\right)=0\mathrm{ou}\pm 1.$$$$$$$$\bullet \mathrm{\forall }x\in \mathbb{R}\mathrm{et}\mathrm{pour}y=0:$$$$f(x+0)f(x-0)=f^2\left(x\right)f^2\left(0\right)$$$$f^2(x+0)=f^2\left(x\right)f^2\left(0\right)$$$$\mathrm{On}\mathrm{elimine}\mathrm{alors}\mathrm{le}\mathrm{cas}f\left(0\right)=0\mathrm{avec}\left(1\right).$$$$\mathrm{Et}\mathrm{donc}f\left(0\right)=\pm 1.$$$$$$$$\mathrm{\forall }(x,\mathrm{y})\in {\mathbb{R}}^2,f(x+y)f(x-y)=f^2\left(x\right)f^2\left(y\right)$$$$\mathrm{Vu}\mathrm{cette}\mathrm{relation},\mathrm{et}\mathrm{notamment}\mathrm{les}$$$$\mathrm{carres}\mathrm{a}\mathrm{droite},\mathrm{on}\mathrm{voit}\mathrm{que}f\mathrm{est}\mathrm{de}$$$$\mathrm{signe}\mathrm{constant}.\mathrm{Dans}\mathrm{toute}\mathrm{la}\mathrm{suite},$$$$\mathrm{on}\mathrm{cherche}\mathrm{des}\mathrm{fonctions}f\mathrm{strictement}$$$$\mathrm{positives}\mathrm{avec}f\left(0\right)=1.$$$$$$$$\mathrm{En}\mathrm{posant}g\left(x\right)=\ln f\left(x\right),\mathrm{il}\mathrm{vient}:$$$$\mathrm{\forall }(x,\mathrm{y})\in {\mathbb{R}}^2,g(x+y)+g(x-y)=2[g\left(x\right)+g\left(y\right)]$$$$$$$$\mathrm{Mon}\mathrm{intuition}\mathrm{est}\mathrm{alors}\mathrm{de}\mathrm{poser}$$$$g\left(x\right)={wx}^2\mathrm{et}{\mathrm{l}}^{\prime }\mathrm{on}\mathrm{voit}\mathrm{que}\mathrm{cette}$$$$\mathrm{famille}\mathrm{de}\mathrm{fonctions}\mathrm{verifie}{\mathrm{l}}^{\prime }\mathrm{equation}.$$$$$$$$\mathrm{La}\mathrm{famille}\mathrm{de}\mathrm{fonctions}\mathrm{definies}\mathrm{par}$$$$f\left(x\right)=e^{\omega x^2}\mathrm{verifie}\mathrm{donc}{\mathrm{l}}^{\prime }\mathrm{equation}$$$$\mathrm{initiale}.$$$$\mathrm{Apres},\mathrm{il}\mathrm{reste}\mathrm{a}\mathrm{demontrer}{\mathrm{qu}}^{\prime }\mathrm{il}{\mathrm{n}}^{\prime }\mathrm{y}\mathrm{en}$$$$\mathrm{a}\mathrm{pas}{\mathrm{d}}^{\prime }\mathrm{autres}.$$$$\mathrm{Mais}\mathrm{on}\mathrm{voit}\mathrm{que}\mathrm{si}f\mathrm{est}\mathrm{solution}\mathrm{alors}$$$$\mathrm{toute}\mathrm{fonction}h\mathrm{definie}\mathrm{par}$$$$h\left(x\right)=e^{\omega x^2}f\left(x\right)\mathrm{est}\mathrm{encore}\mathrm{solution}.$$$$\mathrm{La}\mathrm{famille}\mathrm{de}\mathrm{fonctions}\left\{e^{\omega x^2}\right\}\mathrm{est}\mathrm{donc}$$$$\mathrm{generatrice}\mathrm{de}{\mathrm{l}}^{\prime }\mathrm{espace}\mathrm{des}\mathrm{solutions}.$$

Answered by Kamel last updated on 29/Jul/21

$$f(x+y)f(x-y)={\left(f\left(x\right)f\left(y\right)\right)}^2...\left(E\right)$$$$Puty=x\Rightarrow f^4\left(x\right)=f\left(0\right)f\left(2x\right)...\left(1\right)$$$$x=y=0\Rightarrow f^4\left(0\right)=f^2\left(0\right)\Rightarrow f\left(0\right)=0\vee f\left(0\right)=\pm 1$$$$iff\left(0\right)=0thenf\left(x\right)=0.$$$$iff\left(0\right)=-1thenf\left(2x\right)<0\Rightarrow f\left(x\right)<0\mathrm{\forall }x\in \mathbb{R}.$$$$iff\left(0\right)=1thenf\left(2x\right)>0\Rightarrow f\left(x\right)>0\mathrm{\forall }x\in \mathbb{R}.$$$$Theniff\left(0\right)=-1putg\left(x\right)=-f\left(x\right)>0.$$$$Forf\left(0\right)=1:f\left(x\right)=f^{2^2}\left(\frac{x}{2}\right)=f^{2^{2n}}\left(\frac{x}{2^n}\right)$$$$f\left(x\right)existforexamplef\left(x\right)=0orf\left(x\right)=1.$$$$So:letn\to +\mathrm{\infty },then:f\left(x\right)={\underset{n\to +\mathrm{\infty }}{lime}}^{2^{2n}Ln\left(f\left(\frac{x}{2^n}\right)\right)}$$$$\therefore f\left(x\right)\stackrel{t=\frac{x}{2^n}}{=}{\underset{t\to 0}{lime}}^{x^2\frac{Ln\left(f\left(t\right)\right)}{t^2}}=e^{{ax}^2}/a=\underset{t\to 0}{lim}\frac{Ln\left(f\left(t\right)\right)}{t^2}$$$$aisexistbecausef\left(x\right)isexistandcontinouson\mathbb{R}.$$

Commented by puissant last updated on 29/Jul/21

$$\mathrm{thanks}..$$

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