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Question Number 169842 by mathlove last updated on 10/May/22

Answered by Rasheed.Sindhi last updated on 10/May/22

f^( 2) (x).f(((1−x)/(1+x)))=x^2 .............(A) (A)^2 : f^( 4) (x).f^( 2) (((1−x)/(1+x)))=x^4 f^( 2) (((1−x)/(1+x)))=(x^4 /(f^4 (x))) •x→((1−x)/(1+x)) ((1−x)/(1+x))→((1−((1−x)/(1+x)))/(1+((1−x)/(1+x))))=((1+x−1+x)/(1+x+1−x))=((2x)/2)=x (A)⇒ f^2 (((1−x)/(1+x))).f(x)=(((1−x)/(1+x)))^2 (x^4 /(f^4 (x))).f(x)=(((1−x)/(1+x)))^2 (x^4 /(f^3 (x)))=(((1−x)/(1+x)))^2 f^3 (x)=x^4 .(((1+x)/(1−x)))^2

$${f}^{\:\mathrm{2}} \left({x}\right).{f}\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)={x}^{\mathrm{2}} .............\left(\mathrm{A}\right) \\ $$$$\left(\mathrm{A}\right)^{\mathrm{2}} :\:\:{f}^{\:\:\mathrm{4}} \left({x}\right).{f}^{\:\mathrm{2}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)={x}^{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{f}^{\:\mathrm{2}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)=\frac{{x}^{\mathrm{4}} }{{f}\:^{\mathrm{4}} \left({x}\right)} \\ $$$$\bullet{x}\rightarrow\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\: \\ $$$$\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\rightarrow\frac{\mathrm{1}−\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}{\mathrm{1}+\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}=\frac{\mathrm{1}+{x}−\mathrm{1}+{x}}{\mathrm{1}+{x}+\mathrm{1}−{x}}=\frac{\mathrm{2}{x}}{\mathrm{2}}={x} \\ $$$$\left(\mathrm{A}\right)\Rightarrow \\ $$$${f}^{\mathrm{2}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right).{f}\left({x}\right)=\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)^{\mathrm{2}} \\ $$$$\frac{{x}^{\mathrm{4}} }{{f}\:^{\mathrm{4}} \left({x}\right)}.{f}\left({x}\right)=\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)^{\mathrm{2}} \\ $$$$\frac{{x}^{\mathrm{4}} }{{f}\:^{\mathrm{3}} \left({x}\right)}=\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)^{\mathrm{2}} \\ $$$${f}\:^{\mathrm{3}} \left({x}\right)={x}^{\mathrm{4}} .\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)^{\mathrm{2}} \\ $$

Commented by mathlove last updated on 11/May/22

thanks sir

$${thanks}\:{sir} \\ $$

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