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Question Number 180785 by Vynho last updated on 17/Nov/22

solve (1−x−x^2 ...)(2−x−x^2 ...)

$${solve}\:\left(\mathrm{1}−{x}−{x}^{\mathrm{2}} ...\right)\left(\mathrm{2}−{x}−{x}^{\mathrm{2}} ...\right) \\ $$

Answered by Rasheed.Sindhi last updated on 17/Nov/22

( 1−(x+x^2 +...) )( 2−(x+x^2 +...) ) If ∣x∣<1: (1−(x/(1−x)))(2−(x/(1−x))) ((1−x−x)/(1−x))∙((2−2x−x)/(1−x)) =(((1−2x)(2−3x))/((1−x)^2 ))=((6x^2 −7x+2)/(x^2 −2x+1))

$$\left(\:\mathrm{1}−\left({x}+{x}^{\mathrm{2}} +...\right)\:\right)\left(\:\mathrm{2}−\left({x}+{x}^{\mathrm{2}} +...\right)\:\right) \\ $$$${If}\:\mid{x}\mid<\mathrm{1}: \\ $$$$\left(\mathrm{1}−\frac{{x}}{\mathrm{1}−{x}}\right)\left(\mathrm{2}−\frac{{x}}{\mathrm{1}−{x}}\right) \\ $$$$\frac{\mathrm{1}−{x}−{x}}{\mathrm{1}−{x}}\centerdot\frac{\mathrm{2}−\mathrm{2}{x}−{x}}{\mathrm{1}−{x}} \\ $$$$=\frac{\left(\mathrm{1}−\mathrm{2}{x}\right)\left(\mathrm{2}−\mathrm{3}{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }=\frac{\mathrm{6}{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}} \\ $$

Answered by mr W last updated on 17/Nov/22

let t=x+x^2 +...=x(1+x+x^2 +...) ⇒(1−t)(2−t)=0 ⇒t=1 or 2 ⇒x(1+x+x^2 +...)=1 or 2 x<1 (x/(1−x))=1 or 2 ⇒x=(1/2) or (2/3)

$${let}\:{t}={x}+{x}^{\mathrm{2}} +...={x}\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +...\right) \\ $$$$\Rightarrow\left(\mathrm{1}−{t}\right)\left(\mathrm{2}−{t}\right)=\mathrm{0} \\ $$$$\Rightarrow{t}=\mathrm{1}\:{or}\:\mathrm{2} \\ $$$$\Rightarrow{x}\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +...\right)=\mathrm{1}\:{or}\:\mathrm{2} \\ $$$$\:{x}<\mathrm{1} \\ $$$$\frac{{x}}{\mathrm{1}−{x}}=\mathrm{1}\:{or}\:\mathrm{2} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}}\:{or}\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$

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