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Question Number 208164 by Fridunatjan08 last updated on 06/Jun/24

Solve for x: x^2 +x^2 (1−x^2 )+x^2 (1−x^2 )^2 +x^2 (1−x^2 )^3 +...+x^2 (1−x^2 )^(100) =1

$${Solve}\:{for}\:{x}: \\ $$$${x}^{\mathrm{2}} +{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)+{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} +{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{3}} +...+{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{100}} =\mathrm{1} \\ $$

Answered by A5T last updated on 06/Jun/24

x^2 +x^2 (1−x^2 )+x^2 (1−x^2 )^2 +...+x^2 (1−x^2 )^(100) =1 ⇒x^2 (1−x^2 )+x^2 (1−x^2 )^2 +...+x^2 (1−x^2 )^(100) =1−x^2 Dividing through by 1−x^2 [x≠+_− 1] ⇒x^2 +x^2 (1−x^2 )+x^2 (1−x^2 )^2 +...+x^2 (1−x^2 )^(99) =1 Subtracting x^2 from both sides, dividing by (1−x^2 ) and iterating ⇒x^2 (1−x^2 )=1−x^2 ⇒x^2 =1⇒x=+_− 1 But by assumption that x≠+_− 1 initially ⇒x=+_− 1

$${x}^{\mathrm{2}} +{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)+{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} +...+{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{100}} =\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)+{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} +...+{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{100}} =\mathrm{1}−{x}^{\mathrm{2}} \\ $$$${Dividing}\:{through}\:{by}\:\mathrm{1}−{x}^{\mathrm{2}} \:\left[{x}\neq\underset{−} {+}\mathrm{1}\right] \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)+{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} +...+{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{99}} =\mathrm{1} \\ $$$${Subtracting}\:\:{x}^{\mathrm{2}} \:{from}\:{both}\:{sides},\:{dividing}\:{by}\: \\ $$$$\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\:{and}\:{iterating} \\ $$$$\Rightarrow{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)=\mathrm{1}−{x}^{\mathrm{2}} \Rightarrow{x}^{\mathrm{2}} =\mathrm{1}\Rightarrow{x}=\underset{−} {+}\mathrm{1} \\ $$$${But}\:{by}\:{assumption}\:{that}\:{x}\neq\underset{−} {+}\mathrm{1}\:{initially} \\ $$$$\Rightarrow{x}=\underset{−} {+}\mathrm{1} \\ $$

Commented by Fridunatjan08 last updated on 07/Jun/24

thanks sir.

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

Answered by Berbere last updated on 07/Jun/24

x^2 (Σ_(k=0) ^(100) (1−x^2 )^k )=1⇒x≠0 x^2 .((1−(1−x^2 )^(101) )/x^2 )=1 ⇔1−(1−x^2 )^(101) =1⇒x^2 =1⇒x∈{−1,1}

$${x}^{\mathrm{2}} \left(\underset{{k}=\mathrm{0}} {\overset{\mathrm{100}} {\sum}}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{{k}} \right)=\mathrm{1}\Rightarrow{x}\neq\mathrm{0} \\ $$$${x}^{\mathrm{2}} .\frac{\mathrm{1}−\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{101}} }{{x}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Leftrightarrow\mathrm{1}−\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{101}} =\mathrm{1}\Rightarrow{x}^{\mathrm{2}} =\mathrm{1}\Rightarrow{x}\in\left\{−\mathrm{1},\mathrm{1}\right\} \\ $$

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