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Question Number 78931 by mr W last updated on 21/Jan/20
if x+(1/x)=a find x^n +(1/x^n )=?
$${if}\:{x}+\frac{\mathrm{1}}{{x}}={a} \\ $$$${find}\:{x}^{{n}} +\frac{\mathrm{1}}{{x}^{{n}} }=? \\ $$
Commented by mathmax by abdo last updated on 22/Jan/20
x+(1/x)=a ⇒x^2 +1=ax ⇒x^2 −ax +1 =0 Δ=a^2 −4 ⇒x_1 =((a+(√(a^2 −4)))/2) and x_2 =((a−(√(a^2 −4)))/2) case 1 x=x_1 ⇒x^n +(1/x^n ) =(((a+(√(a^2 −4)))/2))^n +((2/(a+(√(a^2 −4)))))^n =(1/2^n )(a+(√(a^2 −4)))^n +2^n (((a−(√(a^2 −4)))/4))^n =(1/2^n )(a+(√(a^2 −4)))^n +(1/2^n )(a−(√(a^2 −4)))^n case 2 x =x_2 we get the same value due to x_2 =conj(x_1 )
$${x}+\frac{\mathrm{1}}{{x}}={a}\:\Rightarrow{x}^{\mathrm{2}} +\mathrm{1}={ax}\:\:\Rightarrow{x}^{\mathrm{2}} −{ax}\:+\mathrm{1}\:=\mathrm{0} \\ $$$$\Delta={a}^{\mathrm{2}} −\mathrm{4}\:\Rightarrow{x}_{\mathrm{1}} =\frac{{a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}\:{and}\:{x}_{\mathrm{2}} =\frac{{a}−\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}} \\ $$$${case}\:\mathrm{1}\:{x}={x}_{\mathrm{1}} \:\Rightarrow{x}^{{n}} +\frac{\mathrm{1}}{{x}^{{n}} }\:=\left(\frac{{a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}\right)^{{n}} \:+\left(\frac{\mathrm{2}}{{a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}\right)^{{n}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\left({a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}\right)^{{n}} \:+\mathrm{2}^{{n}} \left(\frac{{a}−\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{4}}\right)^{{n}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\left({a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}\right)^{{n}} \:+\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\left({a}−\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}\right)^{{n}} \\ $$$${case}\:\mathrm{2}\:\:{x}\:={x}_{\mathrm{2}} \:\:\:{we}\:{get}\:{the}\:{same}\:{value}\:{due}\:{to}\:{x}_{\mathrm{2}} ={conj}\left({x}_{\mathrm{1}} \right) \\ $$
Commented by mr W last updated on 22/Jan/20
thank you sir! U_n =(1/2^n ){(a+(√(a^2 −4)))^n +(a−(√(a^2 −4)))^n } can the term (a+(√(a^2 −4)))^n +(a−(√(a^2 −4)))^n be simplified more? if a∈Q, is slso U_n ∈Q?
$${thank}\:{you}\:{sir}! \\ $$$${U}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\left\{\left({a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}\right)^{{n}} +\left({a}−\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}\right)^{{n}} \right\} \\ $$$${can}\:{the}\:{term}\:\left({a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}\right)^{{n}} +\left({a}−\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}\right)^{{n}} \\ $$$${be}\:{simplified}\:{more}? \\ $$$${if}\:{a}\in\mathbb{Q},\:{is}\:{slso}\:{U}_{{n}} \in\mathbb{Q}? \\ $$
Commented by mind is power last updated on 22/Jan/20
yeah a∈Q ,Un∈Q because U_(n+1) =aU_n −U_(n−1) .... U_0 =2,U_1 =a∈Q By indiction ∀n∈N ,U_n ∈Q if a∈Z U_n ∈Z also
$${yeah}\:{a}\in\mathbb{Q}\:,{Un}\in\mathbb{Q} \\ $$$${because}\: \\ $$$${U}_{{n}+\mathrm{1}} ={aU}_{{n}} −{U}_{{n}−\mathrm{1}} …. \\ $$$${U}_{\mathrm{0}} =\mathrm{2},{U}_{\mathrm{1}} \:={a}\in{Q}\:{By}\:{indiction} \\ $$$$\forall{n}\in\mathbb{N}\:\:,{U}_{{n}} \in\mathbb{Q}\:{if}\:{a}\in\mathbb{Z}\:\:{U}_{{n}} \in\mathbb{Z}\:{also} \\ $$
Commented by mr W last updated on 22/Jan/20
thanks sir! i think so too. that means in final U_n =(1/2^n ){(a+(√(a^2 −4)))^n +(a−(√(a^2 −4)))^n } the term (√(a^2 −4)) should disappear. how can the term (a+(√(a^2 −4)))^n +(a−(√(a^2 −4)))^n be simplified such that (√(a^2 −4)) disappears?
$${thanks}\:{sir}! \\ $$$${i}\:{think}\:{so}\:{too}.\:{that}\:{means}\:{in}\:{final} \\ $$$${U}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\left\{\left({a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}\right)^{{n}} +\left({a}−\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}\right)^{{n}} \right\} \\ $$$${the}\:{term}\:\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}\:{should}\:{disappear}. \\ $$$${how}\:{can}\:{the}\:{term}\:\left({a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}\right)^{{n}} +\left({a}−\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}\right)^{{n}} \\ $$$${be}\:{simplified}\:{such}\:{that}\:\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}\:{disappears}? \\ $$
Commented by mind is power last updated on 22/Jan/20
yeah (a−b)^n +(a+b)^n =Σ_(k=0) ^n C_n ^k {a^(n−k) (−b)^k +a^(n−k) b^k } =Σ_(k=0) ^(E(((n−1)/2))) C_n ^(2k+1) {−a^(n−(2k+1)) b^(2k+1) +a^(n−(2k+1)) b^(2k+1) }+Σ_(k=0) ^(E((n/2))) C_n ^(2k) {a^(n−2k) b^(2k) +a^(n−2k) b^(2k) } =Σ_(k=0) ^(E((n/2))) 2C_n ^(2k) a^(n−2k) b^(2k) (a+(√(a^2 −4)))^n +(a−(√(a^2 −4)))^n ,b=(√(a^2 −4)) we Get 2Σ_(k=0) ^(E((n/2))) C_n ^(2k) a^(n−2k) ((√(a^2 −4)))^(2k) ifa≥2 =2Σ_(k=0) ^(E((n/2))) C_n ^(2k) a^(n−2k) (a^2 −4)^k
$${yeah} \\ $$$$\left({a}−{b}\right)^{{n}} +\left({a}+{b}\right)^{{n}} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} \left\{{a}^{{n}−{k}} \left(−{b}\right)^{{k}} +{a}^{{n}−{k}} {b}^{{k}} \right\} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{E}\left(\frac{{n}−\mathrm{1}}{\mathrm{2}}\right)} {\sum}}{C}_{{n}} ^{\mathrm{2}{k}+\mathrm{1}} \left\{−{a}^{{n}−\left(\mathrm{2}{k}+\mathrm{1}\right)} {b}^{\mathrm{2}{k}+\mathrm{1}} +{a}^{{n}−\left(\mathrm{2}{k}+\mathrm{1}\right)} {b}^{\mathrm{2}{k}+\mathrm{1}} \right\}+\underset{{k}=\mathrm{0}} {\overset{{E}\left(\frac{{n}}{\mathrm{2}}\right)} {\sum}}{C}_{{n}} ^{\mathrm{2}{k}} \left\{{a}^{{n}−\mathrm{2}{k}} {b}^{\mathrm{2}{k}} +{a}^{{n}−\mathrm{2}{k}} {b}^{\mathrm{2}{k}} \right\} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{E}\left(\frac{{n}}{\mathrm{2}}\right)} {\sum}}\mathrm{2}{C}_{{n}} ^{\mathrm{2}{k}} {a}^{{n}−\mathrm{2}{k}} {b}^{\mathrm{2}{k}} \\ $$$$\left({a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}\right)^{{n}} +\left({a}−\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}\right)^{{n}} ,{b}=\sqrt{{a}^{\mathrm{2}} −\mathrm{4}} \\ $$$${we}\:{Get} \\ $$$$\mathrm{2}\underset{{k}=\mathrm{0}} {\overset{{E}\left(\frac{{n}}{\mathrm{2}}\right)} {\sum}}{C}_{{n}} ^{\mathrm{2}{k}} {a}^{{n}−\mathrm{2}{k}} \left(\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}\right)^{\mathrm{2}{k}} \:\:\:{ifa}\geqslant\mathrm{2} \\ $$$$=\mathrm{2}\underset{{k}=\mathrm{0}} {\overset{{E}\left(\frac{{n}}{\mathrm{2}}\right)} {\sum}}{C}_{{n}} ^{\mathrm{2}{k}} {a}^{{n}−\mathrm{2}{k}} \left({a}^{\mathrm{2}} −\mathrm{4}\right)^{{k}} \\ $$
Commented by mr W last updated on 22/Jan/20
if a^2 −4<0, i.e. −2<a<2: a±(√(a^2 −4))=a±(√(4−a^2 ))i=2((a/2)±((√(4−a^2 ))/2)i) =2[cos (±ϕ)+i sin (±ϕ)] with ϕ=cos^(−1) (a/2) U_n =(1/2^n ){(a+(√(a^2 −4)))^n +(a−(√(a^2 −4)))^n } =2 cos (nϕ) =2 cos (n cos^(−1) (a/2))
$${if}\:{a}^{\mathrm{2}} −\mathrm{4}<\mathrm{0},\:{i}.{e}.\:−\mathrm{2}<{a}<\mathrm{2}: \\ $$$${a}\pm\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}={a}\pm\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }{i}=\mathrm{2}\left(\frac{{a}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}{\mathrm{2}}{i}\right) \\ $$$$=\mathrm{2}\left[\mathrm{cos}\:\left(\pm\varphi\right)+{i}\:\mathrm{sin}\:\left(\pm\varphi\right)\right]\:{with}\:\varphi=\mathrm{cos}^{−\mathrm{1}} \frac{{a}}{\mathrm{2}} \\ $$$${U}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\left\{\left({a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}\right)^{{n}} +\left({a}−\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}\right)^{{n}} \right\} \\ $$$$=\mathrm{2}\:\mathrm{cos}\:\left({n}\varphi\right) \\ $$$$=\mathrm{2}\:\mathrm{cos}\:\left({n}\:\mathrm{cos}^{−\mathrm{1}} \frac{{a}}{\mathrm{2}}\right) \\ $$
Commented by mind is power last updated on 22/Jan/20
nice sir
$${nice}\:{sir} \\ $$
Commented by mr W last updated on 22/Jan/20
i′m wondering that U_n is always <2 if a<2 and cos (n cos^(−1) (a/2)) ∈ Q if a∈Q.
$${i}'{m}\:{wondering}\:{that}\:{U}_{{n}} {is}\:{always}\:<\mathrm{2}\:{if} \\ $$$${a}<\mathrm{2}\:{and}\:\mathrm{cos}\:\left({n}\:\mathrm{cos}^{−\mathrm{1}} \frac{{a}}{\mathrm{2}}\right)\:\in\:{Q}\:{if}\:{a}\in{Q}. \\ $$
Answered by mind is power last updated on 21/Jan/20
let U_n =x^n +(1/x^n ) U_0 =2,U_1 =a U_(n+1) =(x+(1/x))(x^n +(1/x^n ))−x^(n−1) +(1/x^(n−1) ) U_(n+1) =aU_n +U_(n−1) ⇒U_(n+1) −aU_n −U_(n−1) =0 X^2 −aX−1=0 ⇒X_1 =((a−(√(a^2 +4)))/2),X_2 ^ =((a+(√(a^2 +4)))/2) U_n =c(X_1 )^n +b(X_2 )^n c+b=2 cX_1 +bX_2 =a (a/2)(c+b)+((√(a^2 +4))/2)(b−c)=a ⇒b=c ⇒U_n =(((a−(√(a^2 +4)))/2))^n +(((a+(√(a^2 +4)))/2))^n
$${let}\:{U}_{{n}} ={x}^{{n}} +\frac{\mathrm{1}}{{x}^{{n}} } \\ $$$${U}_{\mathrm{0}} =\mathrm{2},{U}_{\mathrm{1}} ={a} \\ $$$${U}_{{n}+\mathrm{1}} =\left({x}+\frac{\mathrm{1}}{{x}}\right)\left({x}^{{n}} +\frac{\mathrm{1}}{{x}^{{n}} }\right)−{x}^{{n}−\mathrm{1}} +\frac{\mathrm{1}}{{x}^{{n}−\mathrm{1}} } \\ $$$${U}_{{n}+\mathrm{1}} ={aU}_{{n}} +{U}_{{n}−\mathrm{1}} \\ $$$$\Rightarrow{U}_{{n}+\mathrm{1}} −{aU}_{{n}} −{U}_{{n}−\mathrm{1}} =\mathrm{0} \\ $$$${X}^{\mathrm{2}} −{aX}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{X}_{\mathrm{1}} =\frac{{a}−\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}},{X}_{\mathrm{2}} ^{} =\frac{{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}} \\ $$$${U}_{{n}} ={c}\left({X}_{\mathrm{1}} \right)^{{n}} +{b}\left({X}_{\mathrm{2}} \right)^{{n}} \\ $$$${c}+{b}=\mathrm{2} \\ $$$${cX}_{\mathrm{1}} +{bX}_{\mathrm{2}} ={a} \\ $$$$\frac{{a}}{\mathrm{2}}\left({c}+{b}\right)+\frac{\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}\left({b}−{c}\right)={a} \\ $$$$\Rightarrow{b}={c} \\ $$$$\Rightarrow{U}_{{n}} =\left(\frac{{a}−\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}\right)^{{n}} +\left(\frac{{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}\right)^{{n}} \\ $$
Commented by mr W last updated on 21/Jan/20
thanks alot sir! please check typos: U_(n+1) =aU_n −U_(n−1) ⇒U_n =(((a−(√(a^2 −4)))/2))^n +(((a+(√(a^2 −4)))/2))^n
$${thanks}\:{alot}\:{sir}! \\ $$$${please}\:{check}\:{typos}: \\ $$$${U}_{{n}+\mathrm{1}} ={aU}_{{n}} −{U}_{{n}−\mathrm{1}} \\ $$$$\Rightarrow{U}_{{n}} =\left(\frac{{a}−\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}\right)^{{n}} +\left(\frac{{a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}\right)^{{n}} \\ $$
Commented by mind is power last updated on 21/Jan/20
yeah Sorry Sir i mack this Typ of error
$${yeah}\:{Sorry}\:{Sir}\:{i}\:{mack}\:{this}\:{Typ}\:{of}\:{error}\: \\ $$

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