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Question Number 151699 by iloveisrael last updated on 22/Aug/21
Σ_(x^3 +2022x−2021=0) (((1+x)/(1−x))) =?
∑x3+2022x−2021=0(1+x1−x)=?
Answered by mr W last updated on 22/Aug/21
x^3 +2022x−2021=0 (x−1+1)^3 +2022(x−1+1)−2021=0 let x−1=t (t+1)^3 +2022(t+1)−2021=0 t^3 +3t^2 +3t+1+2022t+2022−2021=0 t^3 +3t^2 +2025t+2=0 (2/t^3 )+((2025)/t^2 )+(3/t)+1=0 ⇒Σ(1/t)=−((2025)/2) ((1+x)/(1−x))=−((x−1+2)/(x−1))=−1−(2/(x−1))=−1−(2/t) Σ((1+x)/(1−x))=Σ(−1−(2/t))=−3−2Σ(1/t) =−3−2×(−((2025)/2)) =2022
x3+2022x−2021=0(x−1+1)3+2022(x−1+1)−2021=0letx−1=t(t+1)3+2022(t+1)−2021=0t3+3t2+3t+1+2022t+2022−2021=0t3+3t2+2025t+2=02t3+2025t2+3t+1=0⇒Σ1t=−202521+x1−x=−x−1+2x−1=−1−2x−1=−1−2tΣ1+x1−x=Σ(−1−2t)=−3−2Σ1t=−3−2×(−20252)=2022
Commented by iloveisrael last updated on 22/Aug/21
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Answered by aleks041103 last updated on 22/Aug/21
((1+x)/(1−x))=((1+1−(1−x))/(1−x))=(2/(1−x))−1 x^3 +2022x−2021=0 has 3 roots by the fundamental theorem of algebra. ⇒Σ_(i=1) ^3 ((2/(1−x_i ))−1)=2(Σ_(i=1) ^3 (1/(1−x_i )))−3 Σ_(i=1) ^3 (1/(1−x_i ))=(((1−x_2 )(1−x_3 )+(1−x_1 )(1−x_3 )+(1−x_1 )(1−x_2 ))/((1−x_1 )(1−x_2 )(1−x_3 )))= =((3−2(x_1 +x_2 +x_3 )+(x_1 x_2 +x_2 x_3 +x_1 x_3 ))/((1−x_1 )(1−x_2 )(1−x_3 ))) Since x_(1,2,3) are roots of the cubic polynomial then x^3 +2022x−2021=(x−x_1 )(x−x_2 )(x−x_3 ) Then (1−x_1 )(1−x_2 )(1−x_3 )=1^3 +2022.1−2021=2 Also by Vieta′s formulas x_1 +x_2 +x_3 =0 x_1 x_2 +x_2 x_3 +x_1 x_3 =2022 Therefore Σ_(i=1) ^3 (1/(1−x_i ))=((3−2(0)+2022)/2)=((2025)/2) Then: Σ_(x^3 +2022x−2021=0) (((1+x)/(1−x))) =2022
1+x1−x=1+1−(1−x)1−x=21−x−1x3+2022x−2021=0has3rootsbythefundamentaltheoremofalgebra.⇒∑3i=1(21−xi−1)=2(∑3i=111−xi)−3∑3i=111−xi=(1−x2)(1−x3)+(1−x1)(1−x3)+(1−x1)(1−x2)(1−x1)(1−x2)(1−x3)==3−2(x1+x2+x3)+(x1x2+x2x3+x1x3)(1−x1)(1−x2)(1−x3)Sincex1,2,3arerootsofthecubicpolynomialthenx3+2022x−2021=(x−x1)(x−x2)(x−x3)Then(1−x1)(1−x2)(1−x3)=13+2022.1−2021=2AlsobyVieta′sformulasx1+x2+x3=0x1x2+x2x3+x1x3=2022Therefore∑3i=111−xi=3−2(0)+20222=20252Then:∑x3+2022x−2021=0(1+x1−x)=2022
Commented by iloveisrael last updated on 22/Aug/21
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