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Question Number 191960 by Rupesh123 last updated on 04/May/23

Answered by a.lgnaoui last updated on 05/May/23

f(f(x)−f(y))=f(f(x)−2x^2 f(y)+f(y^2 ) ⇒f^(−1) (f(x)−f(y))=f^(−1) (f(x))−2x^2 f^(−1) (y)+f^(−1) (y^2 ) ⇒f(x)−f(y)=f(x)−2x^2 y+y^2 ⇒f(y)=y(2x^2 −y)

$$\:\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)−\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{y}}\right)\right)=\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)−\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{y}}\right)+\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{y}}^{\mathrm{2}} \right)\right. \\ $$$$\:\:\Rightarrow\boldsymbol{\mathrm{f}}^{−\mathrm{1}} \left(\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)−\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{y}}\right)\right)=\boldsymbol{\mathrm{f}}^{−\mathrm{1}} \left(\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)\right)−\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{f}}^{−\mathrm{1}} \left(\boldsymbol{\mathrm{y}}\right)+\boldsymbol{\mathrm{f}}^{−\mathrm{1}} \left(\boldsymbol{\mathrm{y}}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)−\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{y}}\right)=\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)−\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}+\boldsymbol{\mathrm{y}}^{\mathrm{2}} \\ $$$$\Rightarrow\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{y}}\right)=\boldsymbol{\mathrm{y}}\left(\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{y}}\right) \\ $$$$\:\:\:\:\: \\ $$

Commented by mehdee42 last updated on 05/May/23

all operation have no mathemtical basis.??!!

$${all}\:{operation}\:{have}\:{no}\:{mathemtical}\:{basis}.??!! \\ $$

Answered by deleteduser1 last updated on 25/May/23

P(x,y): f(f(x)−f(y))=f(f(x))−2x^2 f(y)+f(y^2 ) f(f(0))=0; f(f(0)−f(1))=f(1) f(f(1)−f(0))=f(f(1))−f(0); f(0)=f(f(1))−f(1) f(f(0)−f(y))=f(y^2 ) f(0)=f(f(x))−2x^2 f(x)+f(x^2 ) f(f(0)^2 )=0⇒f(0)^2 =f(0)⇒f(0)=0 or 1 Case I: f(0)=0⇒f(−f(y))=f(y^2 ) f(−f(−y))=f(y^2 ) ⇒−f(y)=−f(−y)⇒f(y)=f(−y) f(f(1))=f(1); f(0)−f(1)=f(1)⇒f(1)=0 ⇒f(−f(y))=−2f(y)+f(y^2 ); f(−f(y))=f(y^2 ) ⇒f(y^2 )=−2f(y)+f(y^2 ) ⇒f(y)=0⇒f≡0 f is identically zero when f(0)=0 Case II: f(0)=1⇒f(f(0))=f(1)=0.....(a) But also observe that f(f(0)−f(1))=f(1)⇒f(1−f(1))=f(1)...(i) and thatf(f(1)−f(0))=f(f(1))−f(0) ⇒f(f(1)−1)=f(f(1))−1...(ii) We also have that f(0)+f(1)=1+f(1)=f(f(1))...(iii) Combining i,ii and iii ⇒f(f(1)−1)=f(1)=f(1−f(1)) ⇒f(1)−1=1−f(1)⇒f(1)=1 This contradicts (a) Hence,f(0) cannot be equal to 1

$${P}\left({x},{y}\right):\:{f}\left({f}\left({x}\right)−{f}\left({y}\right)\right)={f}\left({f}\left({x}\right)\right)−\mathrm{2}{x}^{\mathrm{2}} {f}\left({y}\right)+{f}\left({y}^{\mathrm{2}} \right) \\ $$$${f}\left({f}\left(\mathrm{0}\right)\right)=\mathrm{0};\:{f}\left({f}\left(\mathrm{0}\right)−{f}\left(\mathrm{1}\right)\right)={f}\left(\mathrm{1}\right) \\ $$$${f}\left({f}\left(\mathrm{1}\right)−{f}\left(\mathrm{0}\right)\right)={f}\left({f}\left(\mathrm{1}\right)\right)−{f}\left(\mathrm{0}\right);\:{f}\left(\mathrm{0}\right)={f}\left({f}\left(\mathrm{1}\right)\right)−{f}\left(\mathrm{1}\right) \\ $$$${f}\left({f}\left(\mathrm{0}\right)−{f}\left({y}\right)\right)={f}\left({y}^{\mathrm{2}} \right) \\ $$$${f}\left(\mathrm{0}\right)={f}\left({f}\left({x}\right)\right)−\mathrm{2}{x}^{\mathrm{2}} {f}\left({x}\right)+{f}\left({x}^{\mathrm{2}} \right) \\ $$$${f}\left({f}\left(\mathrm{0}\right)^{\mathrm{2}} \right)=\mathrm{0}\Rightarrow{f}\left(\mathrm{0}\right)^{\mathrm{2}} ={f}\left(\mathrm{0}\right)\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{0}\:{or}\:\mathrm{1} \\ $$$${Case}\:{I}:\:{f}\left(\mathrm{0}\right)=\mathrm{0}\Rightarrow{f}\left(−{f}\left({y}\right)\right)={f}\left({y}^{\mathrm{2}} \right) \\ $$$${f}\left(−{f}\left(−{y}\right)\right)={f}\left({y}^{\mathrm{2}} \right) \\ $$$$\Rightarrow−{f}\left({y}\right)=−{f}\left(−{y}\right)\Rightarrow{f}\left({y}\right)={f}\left(−{y}\right) \\ $$$${f}\left({f}\left(\mathrm{1}\right)\right)={f}\left(\mathrm{1}\right);\:{f}\left(\mathrm{0}\right)−{f}\left(\mathrm{1}\right)={f}\left(\mathrm{1}\right)\Rightarrow{f}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{f}\left(−{f}\left({y}\right)\right)=−\mathrm{2}{f}\left({y}\right)+{f}\left({y}^{\mathrm{2}} \right);\:\:{f}\left(−{f}\left({y}\right)\right)={f}\left({y}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{f}\left({y}^{\mathrm{2}} \right)=−\mathrm{2}{f}\left({y}\right)+{f}\left({y}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{f}\left({y}\right)=\mathrm{0}\Rightarrow{f}\equiv\mathrm{0} \\ $$$${f}\:{is}\:{identically}\:{zero}\:{when}\:{f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${Case}\:{II}:\:{f}\left(\mathrm{0}\right)=\mathrm{1}\Rightarrow{f}\left({f}\left(\mathrm{0}\right)\right)={f}\left(\mathrm{1}\right)=\mathrm{0}.....\left({a}\right) \\ $$$${But}\:{also}\:{observe}\:{that} \\ $$$${f}\left({f}\left(\mathrm{0}\right)−{f}\left(\mathrm{1}\right)\right)={f}\left(\mathrm{1}\right)\Rightarrow{f}\left(\mathrm{1}−{f}\left(\mathrm{1}\right)\right)={f}\left(\mathrm{1}\right)...\left({i}\right) \\ $$$${and}\:{thatf}\left({f}\left(\mathrm{1}\right)−{f}\left(\mathrm{0}\right)\right)={f}\left({f}\left(\mathrm{1}\right)\right)−{f}\left(\mathrm{0}\right) \\ $$$$\Rightarrow{f}\left({f}\left(\mathrm{1}\right)−\mathrm{1}\right)={f}\left({f}\left(\mathrm{1}\right)\right)−\mathrm{1}...\left({ii}\right) \\ $$$${We}\:{also}\:{have}\:{that} \\ $$$${f}\left(\mathrm{0}\right)+{f}\left(\mathrm{1}\right)=\mathrm{1}+{f}\left(\mathrm{1}\right)={f}\left({f}\left(\mathrm{1}\right)\right)...\left({iii}\right) \\ $$$${Combining}\:{i},{ii}\:{and}\:{iii} \\ $$$$\Rightarrow{f}\left({f}\left(\mathrm{1}\right)−\mathrm{1}\right)={f}\left(\mathrm{1}\right)={f}\left(\mathrm{1}−{f}\left(\mathrm{1}\right)\right) \\ $$$$\Rightarrow{f}\left(\mathrm{1}\right)−\mathrm{1}=\mathrm{1}−{f}\left(\mathrm{1}\right)\Rightarrow{f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$${This}\:{contradicts}\:\left({a}\right) \\ $$$${Hence},{f}\left(\mathrm{0}\right)\:{cannot}\:{be}\:{equal}\:{to}\:\mathrm{1} \\ $$

Commented by witcher3 last updated on 06/May/23

5 ligne this is true ⇔f injective f(x)=x^2 worcks also

$$\mathrm{5}\:\mathrm{ligne}\:\mathrm{this}\:\mathrm{is}\:\mathrm{true}\:\Leftrightarrow\mathrm{f}\:\mathrm{injective} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{2}} \:\mathrm{worcks}\:\mathrm{also} \\ $$$$ \\ $$

Commented by deleteduser1 last updated on 06/May/23

f(x)=x^2 is not injective.

$${f}\left({x}\right)={x}^{\mathrm{2}} \:{is}\:{not}\:{injective}. \\ $$

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