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Question Number 73487 by abdomathmax last updated on 13/Nov/19

let α and β roots of the equation x^2 −x+2=0 simplify A_p = α^p +β^p and calculate Σ_(p=0) ^(n−1) A_p and Σ_(p=0) ^(n−1) A_p ^2

$${let}\:\:\:\:\alpha\:{and}\:\beta\:{roots}\:{of}\:\:{the}\:{equation}\:\:{x}^{\mathrm{2}} −{x}+\mathrm{2}=\mathrm{0} \\ $$$${simplify}\:\:\:{A}_{{p}} =\:\alpha^{{p}} \:+\beta^{{p}} \:{and}\:{calculate} \\ $$$$\sum_{{p}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{A}_{{p}} \:\:{and}\:\sum_{{p}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:{A}_{{p}} ^{\mathrm{2}} \\ $$

Commented by mathmax by abdo last updated on 15/Nov/19

letsolve x^2 −x +2 =0 →Δ=1−8=−7 ⇒ α=((1+i(√7))/2) and β=((1−i(√7))/2) ∣α∣=(1/2)(√(1+7))=((2(√2))/2) =(√2) ⇒α=(√2)e^(iarctan((√7))) and β =(√2)e^(−iarctan((√7))) A_p =α^p +β^p =((√2))^p e^(ip arctan((√7))) +((√2))^p e^(−ip arctan((√7))) =2^(p/2) ×2 Re(e^(ip arctan((√7))) ) =2^(1+(p/2)) cos(p arctan((√7))) Σ_(p=0) ^(n−1) A_p =Σ_(p=0) ^(n−1) 2^(1+(p/2)) cos(parctan((√7))) =Re(2Σ_(p=0) ^(n−1) ((√2))^p e^(ip arctan((√7))) ) we have Σ_(p=0) ^(n−1) ((√2))^p e^(ip arctan((√7))) ) =Σ_(p=0) ^(n−1) ((√2)e^(i arctan((√7))) )^p =((1−((√2)e^(i arctan((√7))) )^n )/(1−(√2)e^(i arctan((√7))) )) =((1−((√2))^n e^(in arctan((√7))) )/(1−(√2)e^(isrctan((√7))) )) =(((1−((√2))^n ( cos(narctan((√7)) +isin(narctan((√7))))/(1−(√2)(cos(arctan((√7))+isin(arctan((√7))))) rest to extract Re (of this quantity...)

$${letsolve}\:{x}^{\mathrm{2}} −{x}\:+\mathrm{2}\:=\mathrm{0}\:\rightarrow\Delta=\mathrm{1}−\mathrm{8}=−\mathrm{7}\:\:\Rightarrow \\ $$$$\alpha=\frac{\mathrm{1}+{i}\sqrt{\mathrm{7}}}{\mathrm{2}}\:\:{and}\:\beta=\frac{\mathrm{1}−{i}\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$$$\mid\alpha\mid=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}+\mathrm{7}}=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}\:=\sqrt{\mathrm{2}}\:\Rightarrow\alpha=\sqrt{\mathrm{2}}{e}^{{iarctan}\left(\sqrt{\mathrm{7}}\right)} \:\:{and}\:\beta\:=\sqrt{\mathrm{2}}{e}^{−{iarctan}\left(\sqrt{\mathrm{7}}\right)} \\ $$$${A}_{{p}} =\alpha^{{p}} \:+\beta^{{p}} \:=\left(\sqrt{\mathrm{2}}\right)^{{p}} \:{e}^{{ip}\:{arctan}\left(\sqrt{\mathrm{7}}\right)} \:+\left(\sqrt{\mathrm{2}}\right)^{{p}} \:{e}^{−{ip}\:{arctan}\left(\sqrt{\mathrm{7}}\right)} \\ $$$$=\mathrm{2}^{\frac{{p}}{\mathrm{2}}} ×\mathrm{2}\:{Re}\left({e}^{{ip}\:{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)\:=\mathrm{2}^{\mathrm{1}+\frac{{p}}{\mathrm{2}}} \:\:{cos}\left({p}\:{arctan}\left(\sqrt{\mathrm{7}}\right)\right) \\ $$$$\sum_{{p}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{A}_{{p}} =\sum_{{p}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\mathrm{2}^{\mathrm{1}+\frac{{p}}{\mathrm{2}}} {cos}\left({parctan}\left(\sqrt{\mathrm{7}}\right)\right) \\ $$$$={Re}\left(\mathrm{2}\sum_{{p}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left(\sqrt{\mathrm{2}}\right)^{{p}} \:{e}^{{ip}\:{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)\:\:{we}\:{have} \\ $$$$\left.\sum_{{p}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\sqrt{\mathrm{2}}\right)^{{p}} \:{e}^{{ip}\:{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)\:=\sum_{{p}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\sqrt{\mathrm{2}}{e}^{{i}\:{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)^{{p}} \\ $$$$=\frac{\mathrm{1}−\left(\sqrt{\mathrm{2}}{e}^{{i}\:{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)^{{n}} }{\mathrm{1}−\sqrt{\mathrm{2}}{e}^{{i}\:{arctan}\left(\sqrt{\mathrm{7}}\right)} }\:=\frac{\mathrm{1}−\left(\sqrt{\mathrm{2}}\right)^{{n}} \:{e}^{{in}\:{arctan}\left(\sqrt{\mathrm{7}}\right)} }{\mathrm{1}−\sqrt{\mathrm{2}}{e}^{{isrctan}\left(\sqrt{\mathrm{7}}\right)} } \\ $$$$=\frac{\left(\mathrm{1}−\left(\sqrt{\mathrm{2}}\right)^{{n}} \left(\:{cos}\left({narctan}\left(\sqrt{\mathrm{7}}\right)\:+{isin}\left({narctan}\left(\sqrt{\mathrm{7}}\right)\right)\right.\right.\right.}{\mathrm{1}−\sqrt{\mathrm{2}}\left({cos}\left({arctan}\left(\sqrt{\mathrm{7}}\right)+{isin}\left({arctan}\left(\sqrt{\mathrm{7}}\right)\right)\right.\right.} \\ $$$${rest}\:{to}\:{extract}\:\:{Re}\:\left({of}\:{this}\:{quantity}...\right) \\ $$

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