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Question Number 194637 by manxsol last updated on 12/Jul/23
x+y=1 x^2 +y^2 =2 x^(11) +y^(11) =?
$$ \\ $$$${x}+{y}=\mathrm{1} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2} \\ $$$${x}^{\mathrm{11}} +{y}^{\mathrm{11}} =? \\ $$$$ \\ $$$$ \\ $$
Commented by Rasheed.Sindhi last updated on 14/Jul/23
See Q#191527
$${See}\:{Q}#\mathrm{191527} \\ $$
Answered by Rasheed.Sindhi last updated on 12/Jul/23
y=1−x⇒y^2 =(1−x)^2 y^2 =2−x^2 1−2x+x^2 =2−x^2 2x^2 −2x−1=0 x=((2±(√(4−8)))/4)=((2±2i)/4)=((1±i)/2) y=1−x=1−((1±i)/2)=((2−1∓i)/2)=((1∓i)/2) x+y=((1±i)/2)+((1∓i)/2)=(2/2)=1 satisfy x+y=1 x^2 +y^2 =(((1±i)/2))^2 +(((1∓i)/2))^2 =((2i)/4)+((−2i)/4)=0 Do not satisfy x^2 +y^2 =2 x^(11) +y^(11) =(((1±i)/2))^(11) +(((1∓i)/2))^(11) =((−1)/(32)) ???
$${y}=\mathrm{1}−{x}\Rightarrow{y}^{\mathrm{2}} =\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} =\mathrm{2}−{x}^{\mathrm{2}} \\ $$$$\mathrm{1}−\mathrm{2}{x}+{x}^{\mathrm{2}} =\mathrm{2}−{x}^{\mathrm{2}} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{2}\pm\sqrt{\mathrm{4}−\mathrm{8}}}{\mathrm{4}}=\frac{\mathrm{2}\pm\mathrm{2}{i}}{\mathrm{4}}=\frac{\mathrm{1}\pm{i}}{\mathrm{2}} \\ $$$${y}=\mathrm{1}−{x}=\mathrm{1}−\frac{\mathrm{1}\pm{i}}{\mathrm{2}}=\frac{\mathrm{2}−\mathrm{1}\mp{i}}{\mathrm{2}}=\frac{\mathrm{1}\mp{i}}{\mathrm{2}} \\ $$$${x}+{y}=\frac{\mathrm{1}\pm{i}}{\mathrm{2}}+\frac{\mathrm{1}\mp{i}}{\mathrm{2}}=\frac{\mathrm{2}}{\mathrm{2}}=\mathrm{1} \\ $$$$\:{satisfy}\:{x}+{y}=\mathrm{1} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\left(\frac{\mathrm{1}\pm{i}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}\mp{i}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}{i}}{\mathrm{4}}+\frac{−\mathrm{2}{i}}{\mathrm{4}}=\mathrm{0} \\ $$$${Do}\:{not}\:{satisfy}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2} \\ $$$${x}^{\mathrm{11}} +{y}^{\mathrm{11}} =\left(\frac{\mathrm{1}\pm{i}}{\mathrm{2}}\right)^{\mathrm{11}} +\left(\frac{\mathrm{1}\mp{i}}{\mathrm{2}}\right)^{\mathrm{11}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{−\mathrm{1}}{\mathrm{32}}\:\:\:\:\:\:\:??? \\ $$
Commented by Tinku Tara last updated on 13/Jul/23
2x^2 −2x−1=0 ⇒x=((2±(√(4+8)))/4)=((1±(√3))/2) y=1−x=((1∓(√3))/2) Mistake in your solution −8 should be +8 (((1+(√3))/2))^(11) +(((1−(√3))/2))^(11)
$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{8}}}{\mathrm{4}}=\frac{\mathrm{1}\pm\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${y}=\mathrm{1}−{x}=\frac{\mathrm{1}\mp\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{Mistake}\:\mathrm{in}\:\mathrm{your}\:\mathrm{solution}\:−\mathrm{8}\:\mathrm{should} \\ $$$$\mathrm{be}\:+\mathrm{8} \\ $$$$\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{11}} +\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{11}} \\ $$
Commented by Rasheed.Sindhi last updated on 13/Jul/23
Thanks sir for pointing out my mistake!
$$\mathbb{T}\boldsymbol{\mathrm{han}}\Bbbk\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{sir}}\:\mathrm{for}\:\mathrm{pointing}\:\mathrm{out}\:\mathrm{my} \\ $$$$\mathrm{mistake}! \\ $$
Answered by Rasheed.Sindhi last updated on 12/Jul/23
x+y=1, x^2 +y^2 =2 , x^(11) +y^(11) =? (x+y)^2 =(1)^2 x^2 +y^2 +2xy=1 2+2xy=1 xy=−(1/2) (x+y)(x^2 +y^2 )=(1)(2) x^3 +y^3 +xy(x+y)=2 x^3 +y^3 +(−(1/2))(1)=2 x^3 +y^3 =2+(1/2)=(5/2) (x^3 +y^3 )(x+y)=((5/2))(1) x^4 +y^4 +xy(x^2 +y^2 )=(5/2) x^4 +y^4 +(−(1/2))(2)=(5/2) x^4 +y^4 =(7/2) (x^3 +y^3 )(x^2 +y^2 )=((5/2))(2) x^5 +y^5 +x^2 y^2 (x+y)=5 x^5 +y^5 +(−(1/2))^2 (1)=5 x^5 +y^5 =5−(1/4)=((19)/4) (x^4 +y^4 )^2 =((7/2))^2 x^8 +y^8 +2x^4 y^4 =((49)/4) x^8 +y^8 +2(−(1/2))^4 =((49)/4) x^8 +y^8 =((49)/4)−(1/8)=((97)/8) (x^8 +y^8 )(x^3 +y^3 )=(((97)/8))((5/2)) x^(11) +y^(11) +x^3 y^3 (x^5 +y^5 )=((485)/(16)) x^(11) +y^(11) +(−(1/2))^3 (((19)/4))=((485)/(16)) x^(11) +y^(11) =((485)/(16))+((19)/(32))=((970+19)/(32))=((989)/(32))
$${x}+{y}=\mathrm{1},\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2}\:,\:{x}^{\mathrm{11}} +{y}^{\mathrm{11}} =? \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} =\left(\mathrm{1}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{xy}=\mathrm{1} \\ $$$$\mathrm{2}+\mathrm{2}{xy}=\mathrm{1} \\ $$$${xy}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left({x}+{y}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)=\left(\mathrm{1}\right)\left(\mathrm{2}\right) \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{xy}\left({x}+{y}\right)=\mathrm{2} \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} +\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{1}\right)=\mathrm{2} \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} \right)\left({x}+{y}\right)=\left(\frac{\mathrm{5}}{\mathrm{2}}\right)\left(\mathrm{1}\right) \\ $$$${x}^{\mathrm{4}} +{y}^{\mathrm{4}} +{xy}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} +{y}^{\mathrm{4}} +\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{2}\right)=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} +{y}^{\mathrm{4}} =\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} \right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)=\left(\frac{\mathrm{5}}{\mathrm{2}}\right)\left(\mathrm{2}\right) \\ $$$${x}^{\mathrm{5}} +{y}^{\mathrm{5}} +{x}^{\mathrm{2}} {y}^{\mathrm{2}} \left({x}+{y}\right)=\mathrm{5} \\ $$$${x}^{\mathrm{5}} +{y}^{\mathrm{5}} +\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \left(\mathrm{1}\right)=\mathrm{5} \\ $$$${x}^{\mathrm{5}} +{y}^{\mathrm{5}} =\mathrm{5}−\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{19}}{\mathrm{4}} \\ $$$$\left({x}^{\mathrm{4}} +{y}^{\mathrm{4}} \right)^{\mathrm{2}} =\left(\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{8}} +{y}^{\mathrm{8}} +\mathrm{2}{x}^{\mathrm{4}} {y}^{\mathrm{4}} =\frac{\mathrm{49}}{\mathrm{4}} \\ $$$${x}^{\mathrm{8}} +{y}^{\mathrm{8}} +\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} =\frac{\mathrm{49}}{\mathrm{4}} \\ $$$${x}^{\mathrm{8}} +{y}^{\mathrm{8}} =\frac{\mathrm{49}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{8}}=\frac{\mathrm{97}}{\mathrm{8}} \\ $$$$\left({x}^{\mathrm{8}} +{y}^{\mathrm{8}} \right)\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} \right)=\left(\frac{\mathrm{97}}{\mathrm{8}}\right)\left(\frac{\mathrm{5}}{\mathrm{2}}\right) \\ $$$${x}^{\mathrm{11}} +{y}^{\mathrm{11}} +{x}^{\mathrm{3}} {y}^{\mathrm{3}} \left({x}^{\mathrm{5}} +{y}^{\mathrm{5}} \right)=\frac{\mathrm{485}}{\mathrm{16}} \\ $$$${x}^{\mathrm{11}} +{y}^{\mathrm{11}} +\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} \left(\frac{\mathrm{19}}{\mathrm{4}}\right)=\frac{\mathrm{485}}{\mathrm{16}} \\ $$$${x}^{\mathrm{11}} +{y}^{\mathrm{11}} =\frac{\mathrm{485}}{\mathrm{16}}+\frac{\mathrm{19}}{\mathrm{32}}=\frac{\mathrm{970}+\mathrm{19}}{\mathrm{32}}=\frac{\mathrm{989}}{\mathrm{32}} \\ $$
Answered by AST last updated on 13/Jul/23
(x+y)^2 −2xy=2⇒xy=((−1)/2) x^(11) +y^(11) =(x^6 +y^6 )(x^5 +y^5 )−(xy)^5 (x+y) (x^6 +y^6 )=[x^2 +y^2 ][(x^2 +y^2 )^2 −3(xy)^2 ]=2(4−(3/4))=((26)/4) x^5 +y^5 =(x^2 +y^2 )[(x+y){(x+y)^2 −3xy}]−(xy)^2 (x+y) =2[1+(3/2)]−(1/4)=((19)/4) ⇒x^(11) +y^(11) =((26)/4)×((19)/4)+(1/(32))=((989)/(32))
$$\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{xy}=\mathrm{2}\Rightarrow{xy}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$${x}^{\mathrm{11}} +{y}^{\mathrm{11}} =\left({x}^{\mathrm{6}} +{y}^{\mathrm{6}} \right)\left({x}^{\mathrm{5}} +{y}^{\mathrm{5}} \right)−\left({xy}\right)^{\mathrm{5}} \left({x}+{y}\right) \\ $$$$\left({x}^{\mathrm{6}} +{y}^{\mathrm{6}} \right)=\left[{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right]\left[\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{3}\left({xy}\right)^{\mathrm{2}} \right]=\mathrm{2}\left(\mathrm{4}−\frac{\mathrm{3}}{\mathrm{4}}\right)=\frac{\mathrm{26}}{\mathrm{4}} \\ $$$${x}^{\mathrm{5}} +{y}^{\mathrm{5}} =\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\left[\left({x}+{y}\right)\left\{\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{3}{xy}\right\}\right]−\left({xy}\right)^{\mathrm{2}} \left({x}+{y}\right) \\ $$$$=\mathrm{2}\left[\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}\right]−\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{19}}{\mathrm{4}} \\ $$$$\Rightarrow{x}^{\mathrm{11}} +{y}^{\mathrm{11}} =\frac{\mathrm{26}}{\mathrm{4}}×\frac{\mathrm{19}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{32}}=\frac{\mathrm{989}}{\mathrm{32}} \\ $$
Answered by manxsol last updated on 15/Jul/23
thanks Rasheed,AST,Tinku tara my solution r(cosθ+sinθ)=1 r=(√2) sin(θ+45)=sin(150) θ=105 (cos105^(11) +sin105^(11) ) (√2)^(11) [((((√6)+(√2))/4))^(11) −((((√6)−(√2))/4))^(11) ] (√2)^(11) .(((√2)^(11) )/2^(22) )[((√3)+1)^(11) −((√3) −1)^(11) ] (1/(2048))[2Σ_1 ^6 C_(2k−1) ^(11) 3^(6−k) ] (3^6 /(1024))Σ_1 ^6 C_(2k−1) ^(11) 3^(−k) 30.90625=(((1+(√3))/2))^(11) +(((1−(√3))/2))^(11)
$${thanks}\:{Rasheed},{AST},{Tinku}\:{tara} \\ $$$${my}\:{solution} \\ $$$${r}\left({cos}\theta+{sin}\theta\right)=\mathrm{1} \\ $$$${r}=\sqrt{\mathrm{2}} \\ $$$${sin}\left(\theta+\mathrm{45}\right)={sin}\left(\mathrm{150}\right) \\ $$$$\theta=\mathrm{105} \\ $$$$\left({cos}\mathrm{105}^{\mathrm{11}} +{sin}\mathrm{105}^{\mathrm{11}} \right) \\ $$$$\sqrt{\mathrm{2}}\:^{\mathrm{11}} \left[\left(\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{4}}\right)^{\mathrm{11}} −\left(\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{4}}\right)^{\mathrm{11}} \right] \\ $$$$\sqrt{\mathrm{2}}\:^{\mathrm{11}} .\frac{\sqrt{\mathrm{2}}\:^{\mathrm{11}} }{\mathrm{2}^{\mathrm{22}} }\left[\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)^{\mathrm{11}} −\left(\sqrt{\mathrm{3}}\:−\mathrm{1}\right)^{\mathrm{11}} \right] \\ $$$$\frac{\mathrm{1}}{\mathrm{2048}}\left[\mathrm{2}\underset{\mathrm{1}} {\overset{\mathrm{6}} {\sum}}{C}_{\mathrm{2}{k}−\mathrm{1}} ^{\mathrm{11}} \mathrm{3}^{\mathrm{6}−{k}} \right] \\ $$$$\frac{\mathrm{3}^{\mathrm{6}} }{\mathrm{1024}}\underset{\mathrm{1}} {\overset{\mathrm{6}} {\sum}}{C}_{\mathrm{2}{k}−\mathrm{1}} ^{\mathrm{11}} \:\mathrm{3}^{−{k}} \\ $$$$\mathrm{30}.\mathrm{90625}=\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{11}} +\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{11}} \\ $$$$ \\ $$$$ \\ $$

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