Question Number 76431 by john santu last updated on 27/Dec/19
$${how}\:{to}\:{find}\:{x}^{\mathrm{2048}} \:+\:{x}^{−\mathrm{2048}} \\ $$$${if}\:\:{x}+{x}^{−\mathrm{1}} =\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}} \\ $$
Commented by mathmax by abdo last updated on 28/Dec/19
$${we}\:{have}\:{x}+{x}^{−\mathrm{1}} \:=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\Rightarrow\left({x}+{x}^{−\mathrm{1}} \right)^{{n}} \:=\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{x}^{{k}} \:\left({x}^{−\mathrm{1}} \right)^{{n}−{k}} \:=\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{x}^{{k}} \:{x}^{{k}−{n}} \:=\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \:\:\Rightarrow\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{x}^{\mathrm{2}{k}} \:=\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \:{x}^{{n}} \\ $$$${change}\:{x}\:{by}\:\frac{\mathrm{1}}{{x}}\:\Rightarrow\sum_{{k}=\mathrm{0}} ^{{n}} \:{x}^{−\mathrm{2}{k}} \:=\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}\:} {x}^{−{n}} \:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} {x}^{\mathrm{2}{k}} \:+\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{x}^{−\mathrm{2}{k}} \:=\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \left({x}^{{n}} \:+{x}^{−{n}} \right)\:\Rightarrow \\ $$$${x}^{{n}} \:+{x}^{−{n}\:} =\frac{\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{x}^{\mathrm{2}{k}} \:+\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{x}^{−\mathrm{2}{k}} }{\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} }\:=\frac{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} \:+\left({x}^{−\mathrm{2}} \:+\mathrm{1}\right)^{{n}} }{\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} }\:\Rightarrow \\ $$$${x}^{\mathrm{2048}} \:+{x}^{−\mathrm{2048}} \:\:=\frac{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2048}} \:+\left(\mathrm{1}+{x}^{−\mathrm{2}} \right)^{\mathrm{2048}} }{\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2048}} } \\ $$
Answered by Kunal12588 last updated on 27/Dec/19
$${x}+{x}^{−\mathrm{1}} =\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}+{x}^{−\mathrm{2}} =\frac{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}=\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +{x}^{−\mathrm{2}} =\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} +{x}^{−\mathrm{4}} =\frac{\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}−\mathrm{2}=−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${x}^{\mathrm{8}} +{x}^{−\mathrm{8}} =\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$$${x}^{\mathrm{16}} +{x}^{−\mathrm{16}} =−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\vdots \\ $$$${x}^{\mathrm{2048}} +{x}^{−\mathrm{2048}} =\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$
Commented by john santu last updated on 27/Dec/19
$${thanks}\:{sir} \\ $$
Commented by mr W last updated on 27/Dec/19
$${what}\:{if}\:{it}\:{is}\:{to}\:{find}\:{x}^{\mathrm{2020}} +{x}^{−\mathrm{2020}} \:? \\ $$
Commented by Kunal12588 last updated on 27/Dec/19
$${I}\:{can}\:{find}\:{x} \\ $$$${x}=\frac{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)\pm{i}\sqrt{\mathrm{2}\sqrt{\mathrm{5}}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}}{\mathrm{4}} \\ $$$${but}\:{how}\:{to}\:{get}\:{x}^{\mathrm{2020}} \:{and}\:\:\frac{\mathrm{1}}{{x}^{\mathrm{2020}} }? \\ $$
Commented by Kunal12588 last updated on 27/Dec/19
$${f}_{{n}} ={x}^{{n}} +{x}^{−{n}} \\ $$$${given}\::\:{f}_{\mathrm{1}} =\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}} \\ $$$${find}:\:\:{f}_{\mathrm{2048}} ,{f}_{\mathrm{2020}} \:{and}\:{f}_{{n}} \\ $$