Question Number 1295 by prakash jain last updated on 20/Jul/15
![Solve for f(x) f(x+1)=[f(x)]^3](https://www.tinkutara.com/question/Q1295.png)
$$\mathrm{Solve}\:\mathrm{for}\:{f}\left({x}\right) \\ $$$${f}\left({x}+\mathrm{1}\right)=\left[{f}\left({x}\right)\right]^{\mathrm{3}} \\ $$
Commented by Rasheed Ahmad last updated on 20/Jul/15
![(Rasheed Soomro) f(x)=±1 Verification: 1) f(x)=1⇒f(x+1)=1 ∧ [f(x)]^3 =(1)^3 =1 2) f(x)=−1⇒f(x+1)=−1 ∧ [f(x)]^3 =(−1)^3 =−1](https://www.tinkutara.com/question/Q1296.png)
$$\left({Rasheed}\:{Soomro}\right) \\ $$$${f}\left({x}\right)=\pm\mathrm{1} \\ $$$${Verification}: \\ $$$$\left.\mathrm{1}\right)\:{f}\left({x}\right)=\mathrm{1}\Rightarrow{f}\left({x}+\mathrm{1}\right)=\mathrm{1}\:\:\wedge\:\left[{f}\left({x}\right)\right]^{\mathrm{3}} =\left(\mathrm{1}\right)^{\mathrm{3}} =\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{f}\left({x}\right)=−\mathrm{1}\Rightarrow{f}\left({x}+\mathrm{1}\right)=−\mathrm{1}\:\wedge\:\:\left[{f}\left({x}\right)\right]^{\mathrm{3}} =\left(−\mathrm{1}\right)^{\mathrm{3}} =−\mathrm{1} \\ $$
Commented by Rasheed Soomro last updated on 20/Jul/15
![Let′s first consider for f(x) to be polynomial. Then f(x)=a or f(x)=ax+b or ax^2 +bx+c ........... 1) f(x)=a=ax^0 f(x+1)=[f(x)]^3 ................(i) f(x+1)=a(x+1)^0 =a.........(ii) and [f(x)]^3 =a^3 ...................(iii) from(i),(ii) and (iii) a=a^3 ⇒a^2 =1 ⇒a^3 −a=0⇒a=0 ∨a^2 =1 ⇒a=0,1, −1 Hence f(x)=0,±1 f(x)=ax+b and f(x)=ax^2 +bx+c both will give same result though calculation will be complicated. I think for f(x) to be poynomial there exist only these three solutions.](https://www.tinkutara.com/question/Q1297.png)
$${Let}'{s}\:{first}\:{consider}\:{for}\:{f}\left({x}\right)\:{to}\:{be}\:{polynomial}. \\ $$$${Then} \\ $$$${f}\left({x}\right)={a}\:{or}\:{f}\left({x}\right)={ax}+{b}\:{or}\:{ax}^{\mathrm{2}} +{bx}+{c}\:……….. \\ $$$$\left.\mathrm{1}\right)\:\:{f}\left({x}\right)={a}={ax}^{\mathrm{0}} \\ $$$$\:\:{f}\left({x}+\mathrm{1}\right)=\left[{f}\left({x}\right)\right]^{\mathrm{3}} …………….\left({i}\right) \\ $$$${f}\left({x}+\mathrm{1}\right)={a}\left({x}+\mathrm{1}\right)^{\mathrm{0}} ={a}………\left({ii}\right) \\ $$$${and}\:\left[{f}\left({x}\right)\right]^{\mathrm{3}} ={a}^{\mathrm{3}} ……………….\left({iii}\right) \\ $$$${from}\left({i}\right),\left({ii}\right)\:{and}\:\left({iii}\right) \\ $$$${a}={a}^{\mathrm{3}} \:\Rightarrow{a}^{\mathrm{2}} =\mathrm{1}\:\Rightarrow{a}^{\mathrm{3}} −{a}=\mathrm{0}\Rightarrow{a}=\mathrm{0}\:\vee{a}^{\mathrm{2}} =\mathrm{1}\:\Rightarrow{a}=\mathrm{0},\mathrm{1},\:−\mathrm{1} \\ $$$${Hence}\:{f}\left({x}\right)=\mathrm{0},\pm\mathrm{1} \\ $$$$\:{f}\left({x}\right)={ax}+{b}\:{and}\:{f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c}\:\:{both}\:{will}\:{give}\:{same}\: \\ $$$${result}\:{though}\:{calculation}\:{will}\:{be}\:{complicated}. \\ $$$${I}\:{think}\:{for}\:{f}\left({x}\right)\:{to}\:{be}\:{poynomial}\:{there}\:{exist}\:{only}\:{these}\:{three} \\ $$$${solutions}. \\ $$
Answered by prakash jain last updated on 20/Jul/15
$$\mathrm{Use}\:\mathrm{another}\:\mathrm{technique}\:\mathrm{to}\:\mathrm{remove}\:\mathrm{power}. \\ $$$$\mathrm{F}\left({x}\right)=\mathrm{log}_{{q}} {f}\left({x}\right) \\ $$$$\mathrm{F}\left({x}+\mathrm{1}\right)=\mathrm{3F}\left({x}\right) \\ $$$$\mathrm{F}\left({x}\right)=\mathrm{3}^{{x}} \\ $$$$\mathrm{log}_{{q}} {f}\left({x}\right)=\mathrm{3}^{{x}} \\ $$$${f}\left({x}\right)={q}^{\mathrm{3}^{{x}} } \\ $$$${f}\left({x}+\mathrm{1}\right)={q}^{\mathrm{3}^{{x}+\mathrm{1}} } ={q}^{\mathrm{3}^{{x}} } \centerdot{q}^{\mathrm{3}^{{x}} } \centerdot{q}^{\mathrm{3}^{{x}} } =\left({q}^{\mathrm{3}^{{x}} } \right)^{\mathrm{3}} =\left({f}\left({x}\right)\right)^{\mathrm{3}} \\ $$
Commented by Rasheed Ahmad last updated on 20/Jul/15
$${Marvellous}!\:{I}\:\:{couldn}'{t}\:{even}\:{think} \\ $$$${of}\:{it}.\:{However}\:{I}\:{want}\:{to}\:{know}\:{how} \\ $$$${you}\:{determine}\:{F}\left({x}\right)=\mathrm{3}^{{x}} ? \\ $$