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Question Number 1976 by 123456 last updated on 27/Oct/15

$$f:\mathbb{R}\to \mathbb{R}$$$$g:\mathbb{R}\to \mathbb{R}$$$$f(x+y)=f\left(x\right)g\left(x\right)+f\left(y\right)g\left(y\right)$$$$g(x+y)=f\left(x\right)f\left(y\right)+g\left(x\right)g\left(y\right)$$$${\left[f\left(x\right)\right]}^2+{\left[g\left(x\right)\right]}^2=?$$$$[f\left(x\right)+g\left(y\right)][g\left(x\right)+f\left(y\right)]=??$$$$f\left(x\right)=???$$$$g\left(x\right)=????$$

Commented by prakash jain last updated on 27/Oct/15

$$x=y=0$$$$f\left(0\right)=2f\left(0\right)g\left(0\right)\Rightarrow f\left(0\right)=0org\left(0\right)=\frac{1}{2}$$$$g\left(0\right)=f^2\left(0\right)+g^2\left(0\right)$$$$\mathrm{Four}\mathrm{solution}\mathrm{for}f\left(0\right),g\left(0\right)$$$$f\left(0\right)=0,g\left(0\right)=0f\left(0\right)=0,g\left(0\right)=1.......\left(A\right)$$$$f\left(0\right)=\frac{1}{2},g\left(0\right)=\frac{1}{2},f\left(0\right)=-\frac{1}{2},g\left(0\right)=\frac{1}{2}......\left(B\right)$$$$y=0$$$$f\left(x\right)=f\left(x\right)g\left(x\right)+f\left(0\right)g\left(0\right)$$$$f\left(x\right)[1-g\left(x\right)]=f\left(0\right)g\left(0\right)$$$$f\left(x\right)=\frac{f\left(0\right)g\left(0\right)}{1-g\left(x\right)}........\left(1\right)$$$$g\left(x\right)=f\left(x\right)f\left(0\right)+g\left(x\right)g\left(0\right)$$$$subtitutef\left(x\right)from1above$$$$g\left(x\right)=\frac{f\left(0\right)g\left(0\right)}{1-g\left(x\right)}+g\left(x\right)f\left(0\right).....\left(2\right)$$$$\Rightarrow g\left(x\right)=k_1\left(constant\right)$$$$ifg\left(x\right)isnotconstanttheng\left(x\right)\ne 1and\left(2\right)$$$$canonlybesolvedwithconstantg\left(x\right)values.$$$$\mathrm{If}g\left(x\right)=k_1\Rightarrow f\left(x\right)=k_2$$$$continuedinanswer$$

Commented by Rasheed Soomro last updated on 28/Oct/15

$$VeryNice!$$

Answered by Rasheed Soomro last updated on 27/Oct/15

$$f:\mathbb{R}\to \mathbb{R}$$$$g:\mathbb{R}\to \mathbb{R}$$$$f(x+y)=f\left(x\right)g\left(x\right)+f\left(y\right)g\left(y\right).........\left(1\right)$$$$g(x+y)=f\left(x\right)f\left(y\right)+g\left(x\right)g\left(y\right)..........\left(2\right)$$$${\left[f\left(x\right)\right]}^2+{\left[g\left(x\right)\right]}^2=?$$$$[f\left(x\right)+g\left(y\right)][g\left(x\right)+f\left(y\right)]=??$$$$f\left(x\right)=???$$$$g\left(x\right)=????$$$$-------------------$$$$Lety=x,substitutingin\left(1\right)and\left(2\right)$$$$f\left(2x\right)=2f\left(x\right)g\left(x\right)........................\left(3\right)$$$$g\left(2x\right)={\left[f\left(x\right)\right]}^2+{\left[g\left(x\right)\right]}^2...................\left(4\right)$$$$From\left(3\right)g\left(x\right)=\frac{f\left(2x\right)}{2f\left(x\right)}$$$$\therefore g\left(2x\right)=\frac{f\left(4x\right)}{2f\left(2x\right)}$$$$Substitutingvaluesofg\left(x\right)andg\left(2x\right)in\left(4\right):$$$$\frac{f\left(4x\right)}{2f\left(2x\right)}={\left[f\left(x\right)\right]}^2+{\left[\frac{f\left(2x\right)}{2f\left(x\right)}\right]}^2$$$$\frac{f\left(4x\right)}{2f\left(2x\right)}={\left[f\left(x\right)\right]}^2+\frac{{\left[f\left(2x\right)\right]}^2}{4{\left[f\left(x\right)\right]}^2}$$$$Multiplyingby4f\left(2x\right){\left[f\left(x\right)\right]}^{2}tob.s.$$$$2f\left(4x\right){\left[f\left(x\right)\right]}^2=4f\left(2x\right){\left[f\left(x\right)\right]}^4+{\left[f\left(2x\right)\right]}^3$$$$Toocomplicatdd$$$$\ast \ast \ast$$$$Alsofrom\left(3\right)f\left(x\right)=\frac{f\left(2x\right)}{2g\left(x\right)}$$$$Substitutingin\left(4\right)$$$$g\left(2x\right)={\left[\frac{f\left(2x\right)}{2g\left(x\right)}\right]}^2+{\left[g\left(x\right)\right]}^2$$$$g\left(2x\right)=\frac{{\left[f\left(2x\right)\right]}^2}{4{\left[g\left(x\right)\right]}^2}+{\left[g\left(x\right)\right]}^2$$$$4g\left(2x\right){\left[g\left(x\right)\right]}^2={\left[f\left(2x\right)\right]}^2+4{\left[g\left(x\right)\right]}^4$$$$Toocomplicated.$$$$Continue$$$$$$

Answered by prakash jain last updated on 27/Oct/15

$$A\mathrm{s}\mathrm{shown}\mathrm{in}\mathrm{in}\mathrm{comments}$$$$f\left(x\right)=k_2\mathrm{an}dg\left(x\right)=k_1\mathrm{if}{k}_1\ne 1$$$$caseg\left(x\right)\ne 1$$$$sincef\left(x\right)andg\left(x\right)areconstantsit$$$$issufficienttofindvalueonlyforf\left(0\right)$$$$andg\left(0\right)$$$$(f,g)=(-\frac{1}{2},\frac{1}{2})\vee (\frac{1}{2},\frac{1}{2})\vee (0,0)$$$$caseg\left(x\right)=1$$$$f(x+y)=f\left(x\right)+f\left(y\right)$$$$f\left(x\right)f\left(y\right)=0$$$$onlysolutionwillbef\left(x\right)=0$$$$(f,g)=(-\frac{1}{2},\frac{1}{2})\vee (\frac{1}{2},\frac{1}{2})\vee (0,0)\vee (0,1)$$

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