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Question Number 636 by 123456 last updated on 17/Feb/15

$$iff,garefunctionsof\mathbb{R}\to \mathbb{R}$$$$notconstantsuchforall(x,y)\in {\mathbb{R}}^2$$$$\{\begin{array}{l}f(x+y)=f\left(x\right)f\left(y\right)-g\left(x\right)g\left(y\right)\\ g(x+y)=f\left(x\right)g\left(y\right)+g\left(x\right)f\left(y\right)\end{array}$$$$if{f}^{\prime }\left(0\right)=0thenproofosdisproof$$$$that\mathrm{\forall }x\in \mathbb{R},{\left[f\left(x\right)\right]}^2+{\left[g\left(x\right)\right]}^2=1$$

Commented by prakash jain last updated on 16/Feb/15

$$g(x+y)=f\left(x\right)g\left(y\right)-g\left(x\right)f\left(y\right)$$$$g(x+x)=f\left(x\right)g\left(x\right)-g\left(x\right)f\left(x\right)$$$$g\left(2x\right)=0?$$$$\mathrm{do}\mathrm{you}\mathrm{mean}$$$$g(x+y)=f\left(x\right)g\left(y\right)+g\left(x\right)f\left(y\right)?$$$$\mathrm{If}$$$$g(x+y)=f\left(x\right)g\left(y\right)-g\left(x\right)f\left(y\right)$$$$g\left(2x\right)=f\left(x\right)g\left(x\right)-g\left(x\right)f\left(y\right)=0$$$$\mathrm{Contradicts}\mathrm{assumption}f\left(x\right),g\left(x\right)\mathrm{are}$$$$\mathrm{not}\mathrm{constant}.$$

Commented by prakash jain last updated on 17/Feb/15

$$\mathrm{Squaring}\mathrm{and}\mathrm{adding}$$$${\left[f(x+y)\right]}^2+{\left[g(x+y)\right]}^2=({\left[f\left(x\right)\right]}^2+{\left[g\left(x\right)\right]}^2)({\left[f\left(y\right)\right]}^2+{\left[g\left(y\right)\right]}^2)$$$$\mathrm{If}u\left(x\right)={\left[f\left(x\right)\right]}^2+{\left[g\left(x\right)\right]}^2$$$$u(x+y)=u\left(x\right)u\left(y\right)\Rightarrow u\left(x\right)=e^{kx}$$$${\left[f\left(x\right)\right]}^2+{\left[g\left(x\right)\right]}^2=e^{kx}$$

Answered by prakash jain last updated on 17/Feb/15

$$\mathrm{From}\mathrm{comments}$$$${\left[f\left(x\right)\right]}^2+{\left[g\left(x\right)\right]}^2=e^{kx}$$$$f(x+x)={\left[f\left(x\right)\right]}^2-{\left[g\left(x\right)\right]}^2$$$$f\left(2x\right)=2{\left[f\left(x\right)\right]}^2-e^{kx}$$$$\mathrm{Differenting}\mathrm{both}\mathrm{sides}$$$$2f{}^{\prime }\left(2x\right)=4f\left(x\right)f{}^{\prime }\left(x\right)-{ke}^{kx}$$$$\mathrm{put}x=0$$$$2f{}^{\prime }\left(0\right)=4f\left(0\right)f{}^{\prime }\left(0\right)-k$$$$\mathrm{Given}f{}^{\prime }\left(0\right)=0$$$$0=0-k\Rightarrow k=0$$$$\mathrm{Hence}$$$${\left[f\left(x\right)\right]}^2+{\left[g\left(x\right)\right]}^2=1$$$$$$

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