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