Question Number 1976 by 123456 last updated on 27/Oct/15
![f:R→R g:R→R f(x+y)=f(x)g(x)+f(y)g(y) g(x+y)=f(x)f(y)+g(x)g(y) [f(x)]^2 +[g(x)]^2 =? [f(x)+g(y)][g(x)+f(y)]=?? f(x)=??? g(x)=????](https://www.tinkutara.com/question/Q1976.png)
$${f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${g}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({x}+{y}\right)={f}\left({x}\right){g}\left({x}\right)+{f}\left({y}\right){g}\left({y}\right) \\ $$$${g}\left({x}+{y}\right)={f}\left({x}\right){f}\left({y}\right)+{g}\left({x}\right){g}\left({y}\right) \\ $$$$\left[{f}\left({x}\right)\right]^{\mathrm{2}} +\left[{g}\left({x}\right)\right]^{\mathrm{2}} =? \\ $$$$\left[{f}\left({x}\right)+{g}\left({y}\right)\right]\left[{g}\left({x}\right)+{f}\left({y}\right)\right]=?? \\ $$$${f}\left({x}\right)=??? \\ $$$${g}\left({x}\right)=???? \\ $$
Commented by prakash jain last updated on 27/Oct/15
![x=y=0 f(0)=2f(0)g(0)⇒f(0)=0 or g(0)=(1/2) g(0)=f^2 (0)+g^2 (0) Four solution for f(0), g(0) f(0)=0, g(0)=0 f(0)=0, g(0)=1.......(A) f(0)=(1/2), g(0)=(1/2), f(0)=−(1/2), g(0)=(1/2)......(B) y=0 f(x)=f(x)g(x)+f(0)g(0) f(x)[1−g(x)]=f(0)g(0) f(x)=((f(0)g(0))/(1−g(x)))........(1) g(x)=f(x)f(0)+g(x)g(0) subtitute f(x) from 1 above g(x)=((f(0)g(0))/(1−g(x)))+g(x)f(0) .....(2) ⇒g(x)=k_1 (constant) if g(x) is not constant then g(x)≠1 and (2) can only be solved with constant g(x) values. If g(x)=k_1 ⇒f(x)=k_2 continued in answer](https://www.tinkutara.com/question/Q1983.png)
$${x}={y}=\mathrm{0} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{2}{f}\left(\mathrm{0}\right){g}\left(\mathrm{0}\right)\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{0}\:{or}\:{g}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${g}\left(\mathrm{0}\right)={f}^{\mathrm{2}} \left(\mathrm{0}\right)+{g}^{\mathrm{2}} \left(\mathrm{0}\right) \\ $$$$\mathrm{Four}\:\mathrm{solution}\:\mathrm{for}\:{f}\left(\mathrm{0}\right),\:{g}\left(\mathrm{0}\right) \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0},\:{g}\left(\mathrm{0}\right)=\mathrm{0}\:\:\:{f}\left(\mathrm{0}\right)=\mathrm{0},\:{g}\left(\mathrm{0}\right)=\mathrm{1}…….\left({A}\right) \\ $$$${f}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{2}},\:{g}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{2}},\:{f}\left(\mathrm{0}\right)=−\frac{\mathrm{1}}{\mathrm{2}},\:{g}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{2}}……\left({B}\right) \\ $$$${y}=\mathrm{0} \\ $$$${f}\left({x}\right)={f}\left({x}\right){g}\left({x}\right)+{f}\left(\mathrm{0}\right){g}\left(\mathrm{0}\right) \\ $$$${f}\left({x}\right)\left[\mathrm{1}−{g}\left({x}\right)\right]={f}\left(\mathrm{0}\right){g}\left(\mathrm{0}\right) \\ $$$${f}\left({x}\right)=\frac{{f}\left(\mathrm{0}\right){g}\left(\mathrm{0}\right)}{\mathrm{1}−{g}\left({x}\right)}……..\left(\mathrm{1}\right) \\ $$$${g}\left({x}\right)={f}\left({x}\right){f}\left(\mathrm{0}\right)+{g}\left({x}\right){g}\left(\mathrm{0}\right) \\ $$$${subtitute}\:{f}\left({x}\right)\:{from}\:\mathrm{1}\:{above} \\ $$$${g}\left({x}\right)=\frac{{f}\left(\mathrm{0}\right){g}\left(\mathrm{0}\right)}{\mathrm{1}−{g}\left({x}\right)}+{g}\left({x}\right){f}\left(\mathrm{0}\right)\:\:\:\:…..\left(\mathrm{2}\right) \\ $$$$\Rightarrow{g}\left({x}\right)={k}_{\mathrm{1}} \:\left({constant}\right) \\ $$$${if}\:{g}\left({x}\right)\:{is}\:{not}\:{constant}\:{then}\:{g}\left({x}\right)\neq\mathrm{1}\:{and}\:\left(\mathrm{2}\right) \\ $$$${can}\:{only}\:{be}\:{solved}\:{with}\:{constant}\:{g}\left({x}\right)\:{values}. \\ $$$$\mathrm{If}\:{g}\left({x}\right)={k}_{\mathrm{1}} \Rightarrow{f}\left({x}\right)={k}_{\mathrm{2}} \\ $$$${continued}\:{in}\:{answer} \\ $$
Commented by Rasheed Soomro last updated on 28/Oct/15
$${Very}\:{Nice}! \\ $$
Answered by Rasheed Soomro last updated on 27/Oct/15
![f:R→R g:R→R f(x+y)=f(x)g(x)+f(y)g(y).........(1) g(x+y)=f(x)f(y)+g(x)g(y)..........(2) [f(x)]^2 +[g(x)]^2 =? [f(x)+g(y)][g(x)+f(y)]=?? f(x)=??? g(x)=???? −−−−−−−−−−−−−−−−−−− Let y=x,substituting in (1) and (2) f(2x)=2f(x) g(x)........................(3) g(2x)=[f(x)]^2 +[g(x)]^2 ...................(4) From (3) g(x)=((f(2x))/(2f(x))) ∴ g(2x)=((f(4x))/(2f(2x))) Substituting values of g(x) and g(2x) in (4): ((f(4x))/(2f(2x)))=[f(x)]^2 +[((f(2x))/(2f(x)))]^2 ((f(4x))/(2f(2x)))=[f(x)]^2 +(([f(2x)]^2 )/(4[f(x)]^2 )) Multiplying by 4f(2x)[f(x)]^2 to b. s. 2f(4x)[f(x)]^2 =4f(2x)[f(x)]^4 +[f(2x)]^3 Too complicatdd ∗∗∗ Also from (3) f(x)=((f(2x))/(2g(x))) Substituting in (4) g(2x)=[((f(2x))/(2g(x)))]^2 +[g(x)]^2 g(2x)=(([f(2x)]^2 )/(4[g(x)]^2 ))+[g(x)]^2 4g(2x)[g(x)]^2 =[f(2x)]^2 +4[g(x)]^4 Too complicated. Continue](https://www.tinkutara.com/question/Q1978.png)
$${f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${g}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({x}+{y}\right)={f}\left({x}\right){g}\left({x}\right)+{f}\left({y}\right){g}\left({y}\right)………\left(\mathrm{1}\right) \\ $$$${g}\left({x}+{y}\right)={f}\left({x}\right){f}\left({y}\right)+{g}\left({x}\right){g}\left({y}\right)……….\left(\mathrm{2}\right) \\ $$$$\left[{f}\left({x}\right)\right]^{\mathrm{2}} +\left[{g}\left({x}\right)\right]^{\mathrm{2}} =? \\ $$$$\left[{f}\left({x}\right)+{g}\left({y}\right)\right]\left[{g}\left({x}\right)+{f}\left({y}\right)\right]=?? \\ $$$${f}\left({x}\right)=??? \\ $$$${g}\left({x}\right)=???? \\ $$$$−−−−−−−−−−−−−−−−−−− \\ $$$${Let}\:{y}={x},{substituting}\:{in}\:\left(\mathrm{1}\right)\:\:{and}\:\left(\mathrm{2}\right) \\ $$$${f}\left(\mathrm{2}{x}\right)=\mathrm{2}{f}\left({x}\right)\:{g}\left({x}\right)……………………\left(\mathrm{3}\right) \\ $$$${g}\left(\mathrm{2}{x}\right)=\left[{f}\left({x}\right)\right]^{\mathrm{2}} +\left[{g}\left({x}\right)\right]^{\mathrm{2}} ……………….\left(\mathrm{4}\right) \\ $$$${From}\:\left(\mathrm{3}\right)\:\:\:\:{g}\left({x}\right)=\frac{{f}\left(\mathrm{2}{x}\right)}{\mathrm{2}{f}\left({x}\right)}\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\therefore\:\:\:\:\:\:\:\:\:{g}\left(\mathrm{2}{x}\right)=\frac{{f}\left(\mathrm{4}{x}\right)}{\mathrm{2}{f}\left(\mathrm{2}{x}\right)} \\ $$$${Substituting}\:{values}\:{of}\:\:{g}\left({x}\right)\:\:{and}\:\:\:{g}\left(\mathrm{2}{x}\right)\:{in}\:\left(\mathrm{4}\right): \\ $$$$\frac{{f}\left(\mathrm{4}{x}\right)}{\mathrm{2}{f}\left(\mathrm{2}{x}\right)}=\left[{f}\left({x}\right)\right]^{\mathrm{2}} +\left[\frac{{f}\left(\mathrm{2}{x}\right)}{\mathrm{2}{f}\left({x}\right)}\right]^{\mathrm{2}} \\ $$$$\frac{{f}\left(\mathrm{4}{x}\right)}{\mathrm{2}{f}\left(\mathrm{2}{x}\right)}=\left[{f}\left({x}\right)\right]^{\mathrm{2}} +\frac{\left[{f}\left(\mathrm{2}{x}\right)\right]^{\mathrm{2}} }{\mathrm{4}\left[{f}\left({x}\right)\right]^{\mathrm{2}} } \\ $$$${Multiplying}\:{by}\:\mathrm{4}{f}\left(\mathrm{2}{x}\right)\left[{f}\left({x}\right)\right]^{\mathrm{2}} \:{to}\:{b}.\:{s}. \\ $$$$\mathrm{2}{f}\left(\mathrm{4}{x}\right)\left[{f}\left({x}\right)\right]^{\mathrm{2}} =\mathrm{4}{f}\left(\mathrm{2}{x}\right)\left[{f}\left({x}\right)\right]^{\mathrm{4}} +\left[{f}\left(\mathrm{2}{x}\right)\right]^{\mathrm{3}} \\ $$$${Too}\:{complicatdd} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ast\ast\ast \\ $$$${Also}\:{from}\:\left(\mathrm{3}\right)\:\:\:\:{f}\left({x}\right)=\frac{{f}\left(\mathrm{2}{x}\right)}{\mathrm{2}{g}\left({x}\right)} \\ $$$${Substituting}\:{in}\:\left(\mathrm{4}\right)\: \\ $$$${g}\left(\mathrm{2}{x}\right)=\left[\frac{{f}\left(\mathrm{2}{x}\right)}{\mathrm{2}{g}\left({x}\right)}\right]^{\mathrm{2}} +\left[{g}\left({x}\right)\right]^{\mathrm{2}} \\ $$$${g}\left(\mathrm{2}{x}\right)=\frac{\left[{f}\left(\mathrm{2}{x}\right)\right]^{\mathrm{2}} }{\mathrm{4}\left[{g}\left({x}\right)\right]^{\mathrm{2}} }+\left[{g}\left({x}\right)\right]^{\mathrm{2}} \\ $$$$\mathrm{4}{g}\left(\mathrm{2}{x}\right)\left[{g}\left({x}\right)\right]^{\mathrm{2}} =\left[{f}\left(\mathrm{2}{x}\right)\right]^{\mathrm{2}} +\mathrm{4}\left[{g}\left({x}\right)\right]^{\mathrm{4}} \\ $$$${Too}\:{complicated}. \\ $$$${Continue} \\ $$$$ \\ $$
Answered by prakash jain last updated on 27/Oct/15
$${A}\mathrm{s}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{in}\:\mathrm{comments} \\ $$$${f}\left({x}\right)={k}_{\mathrm{2}} \:\mathrm{an}{d}\:{g}\left({x}\right)={k}_{\mathrm{1}} \:\mathrm{if}\:{k}_{\mathrm{1}} \neq\mathrm{1} \\ $$$${case}\:{g}\left({x}\right)\neq\mathrm{1} \\ $$$${since}\:{f}\left({x}\right)\:{and}\:{g}\left({x}\right)\:{are}\:{constants}\:{it} \\ $$$${is}\:{sufficient}\:{to}\:{find}\:{value}\:{only}\:{for}\:{f}\left(\mathrm{0}\right) \\ $$$${and}\:{g}\left(\mathrm{0}\right) \\ $$$$\left({f},{g}\right)=\left(−\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)\vee\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)\vee\left(\mathrm{0},\mathrm{0}\right) \\ $$$${case}\:{g}\left({x}\right)=\mathrm{1} \\ $$$${f}\left({x}+{y}\right)={f}\left({x}\right)+{f}\left({y}\right) \\ $$$${f}\left({x}\right){f}\left({y}\right)=\mathrm{0} \\ $$$${only}\:{solution}\:{will}\:{be}\:{f}\left({x}\right)=\mathrm{0} \\ $$$$\left({f},{g}\right)=\left(−\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)\vee\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)\vee\left(\mathrm{0},\mathrm{0}\right)\vee\left(\mathrm{0},\mathrm{1}\right) \\ $$