Question Number 191960 by Rupesh123 last updated on 04/May/23
Answered by a.lgnaoui last updated on 05/May/23
$$\:\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)−\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{y}}\right)\right)=\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)−\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{y}}\right)+\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{y}}^{\mathrm{2}} \right)\right. \\ $$$$\:\:\Rightarrow\boldsymbol{\mathrm{f}}^{−\mathrm{1}} \left(\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)−\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{y}}\right)\right)=\boldsymbol{\mathrm{f}}^{−\mathrm{1}} \left(\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)\right)−\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{f}}^{−\mathrm{1}} \left(\boldsymbol{\mathrm{y}}\right)+\boldsymbol{\mathrm{f}}^{−\mathrm{1}} \left(\boldsymbol{\mathrm{y}}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)−\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{y}}\right)=\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)−\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}+\boldsymbol{\mathrm{y}}^{\mathrm{2}} \\ $$$$\Rightarrow\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{y}}\right)=\boldsymbol{\mathrm{y}}\left(\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{y}}\right) \\ $$$$\:\:\:\:\: \\ $$
Commented by mehdee42 last updated on 05/May/23
$${all}\:{operation}\:{have}\:{no}\:{mathemtical}\:{basis}.??!! \\ $$
Answered by AST last updated on 25/May/23
$${P}\left({x},{y}\right):\:{f}\left({f}\left({x}\right)−{f}\left({y}\right)\right)={f}\left({f}\left({x}\right)\right)−\mathrm{2}{x}^{\mathrm{2}} {f}\left({y}\right)+{f}\left({y}^{\mathrm{2}} \right) \\ $$$${f}\left({f}\left(\mathrm{0}\right)\right)=\mathrm{0};\:{f}\left({f}\left(\mathrm{0}\right)−{f}\left(\mathrm{1}\right)\right)={f}\left(\mathrm{1}\right) \\ $$$${f}\left({f}\left(\mathrm{1}\right)−{f}\left(\mathrm{0}\right)\right)={f}\left({f}\left(\mathrm{1}\right)\right)−{f}\left(\mathrm{0}\right);\:{f}\left(\mathrm{0}\right)={f}\left({f}\left(\mathrm{1}\right)\right)−{f}\left(\mathrm{1}\right) \\ $$$${f}\left({f}\left(\mathrm{0}\right)−{f}\left({y}\right)\right)={f}\left({y}^{\mathrm{2}} \right) \\ $$$${f}\left(\mathrm{0}\right)={f}\left({f}\left({x}\right)\right)−\mathrm{2}{x}^{\mathrm{2}} {f}\left({x}\right)+{f}\left({x}^{\mathrm{2}} \right) \\ $$$${f}\left({f}\left(\mathrm{0}\right)^{\mathrm{2}} \right)=\mathrm{0}\Rightarrow{f}\left(\mathrm{0}\right)^{\mathrm{2}} ={f}\left(\mathrm{0}\right)\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{0}\:{or}\:\mathrm{1} \\ $$$${Case}\:{I}:\:{f}\left(\mathrm{0}\right)=\mathrm{0}\Rightarrow{f}\left(−{f}\left({y}\right)\right)={f}\left({y}^{\mathrm{2}} \right) \\ $$$${f}\left(−{f}\left(−{y}\right)\right)={f}\left({y}^{\mathrm{2}} \right) \\ $$$$\Rightarrow−{f}\left({y}\right)=−{f}\left(−{y}\right)\Rightarrow{f}\left({y}\right)={f}\left(−{y}\right) \\ $$$${f}\left({f}\left(\mathrm{1}\right)\right)={f}\left(\mathrm{1}\right);\:{f}\left(\mathrm{0}\right)−{f}\left(\mathrm{1}\right)={f}\left(\mathrm{1}\right)\Rightarrow{f}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{f}\left(−{f}\left({y}\right)\right)=−\mathrm{2}{f}\left({y}\right)+{f}\left({y}^{\mathrm{2}} \right);\:\:{f}\left(−{f}\left({y}\right)\right)={f}\left({y}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{f}\left({y}^{\mathrm{2}} \right)=−\mathrm{2}{f}\left({y}\right)+{f}\left({y}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{f}\left({y}\right)=\mathrm{0}\Rightarrow{f}\equiv\mathrm{0} \\ $$$${f}\:{is}\:{identically}\:{zero}\:{when}\:{f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${Case}\:{II}:\:{f}\left(\mathrm{0}\right)=\mathrm{1}\Rightarrow{f}\left({f}\left(\mathrm{0}\right)\right)={f}\left(\mathrm{1}\right)=\mathrm{0}…..\left({a}\right) \\ $$$${But}\:{also}\:{observe}\:{that} \\ $$$${f}\left({f}\left(\mathrm{0}\right)−{f}\left(\mathrm{1}\right)\right)={f}\left(\mathrm{1}\right)\Rightarrow{f}\left(\mathrm{1}−{f}\left(\mathrm{1}\right)\right)={f}\left(\mathrm{1}\right)…\left({i}\right) \\ $$$${and}\:{thatf}\left({f}\left(\mathrm{1}\right)−{f}\left(\mathrm{0}\right)\right)={f}\left({f}\left(\mathrm{1}\right)\right)−{f}\left(\mathrm{0}\right) \\ $$$$\Rightarrow{f}\left({f}\left(\mathrm{1}\right)−\mathrm{1}\right)={f}\left({f}\left(\mathrm{1}\right)\right)−\mathrm{1}…\left({ii}\right) \\ $$$${We}\:{also}\:{have}\:{that} \\ $$$${f}\left(\mathrm{0}\right)+{f}\left(\mathrm{1}\right)=\mathrm{1}+{f}\left(\mathrm{1}\right)={f}\left({f}\left(\mathrm{1}\right)\right)…\left({iii}\right) \\ $$$${Combining}\:{i},{ii}\:{and}\:{iii} \\ $$$$\Rightarrow{f}\left({f}\left(\mathrm{1}\right)−\mathrm{1}\right)={f}\left(\mathrm{1}\right)={f}\left(\mathrm{1}−{f}\left(\mathrm{1}\right)\right) \\ $$$$\Rightarrow{f}\left(\mathrm{1}\right)−\mathrm{1}=\mathrm{1}−{f}\left(\mathrm{1}\right)\Rightarrow{f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$${This}\:{contradicts}\:\left({a}\right) \\ $$$${Hence},{f}\left(\mathrm{0}\right)\:{cannot}\:{be}\:{equal}\:{to}\:\mathrm{1} \\ $$
Commented by witcher3 last updated on 06/May/23
$$\mathrm{5}\:\mathrm{ligne}\:\mathrm{this}\:\mathrm{is}\:\mathrm{true}\:\Leftrightarrow\mathrm{f}\:\mathrm{injective} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{2}} \:\mathrm{worcks}\:\mathrm{also} \\ $$$$ \\ $$
Commented by AST last updated on 06/May/23
$${f}\left({x}\right)={x}^{\mathrm{2}} \:{is}\:{not}\:{injective}. \\ $$