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Question Number 142987 by mathmax by abdo last updated on 08/Jun/21
find the sequence u_n wich verify u_(n+1) =u_n −λu_(n−1) λ real
$$\mathrm{find}\:\mathrm{the}\:\mathrm{sequence}\:\mathrm{u}_{\mathrm{n}} \mathrm{wich}\:\mathrm{verify}\:\mathrm{u}_{\mathrm{n}+\mathrm{1}} =\mathrm{u}_{\mathrm{n}} −\lambda\mathrm{u}_{\mathrm{n}−\mathrm{1}} \\ $$$$\lambda\:\mathrm{real} \\ $$
Answered by Dwaipayan Shikari last updated on 08/Jun/21
u_n =mr^n mr^(n+1) =mr^n −λmr^(n−1) ⇒r^2 −r+λ=0⇒r=((1±(√(1−4λ)))/2) λ≤(1/4) r∈R if r∈C then −∞<λ<∞ u_n =Λ(((1+(√(1−4λ)))/2))^n −Φ(((1−(√(1−4λ)))/2))^n u_0 =Λ−Φ u_1 =((Λ−Φ)/2)+((Λ+Φ)/2)(√(1−4λ)) ⇒(1/( (√(1−4λ))))(2u_1 −u_0 )=Λ+Φ Λ=(1/2).(((2u_1 −u_0 ))/( (√(1−4λ))))+(u_0 /2) Φ=(1/2).(((2u_1 −u_0 ))/( (√(1−4λ))))−(u_0 /2) u_n =(((2u_1 −u_0 ))/(2(√(1−4λ)))){(((1+(√(1−4λ)))/2))^n −(((1−(√(1−4λ)))/2))^n }+(u_0 /2){(((1+(√(1−4λ)))/2))^n −(((1−(√(1−4λ)))/2))^n }
$${u}_{{n}} ={mr}^{{n}} \\ $$$${mr}^{{n}+\mathrm{1}} ={mr}^{{n}} −\lambda{mr}^{{n}−\mathrm{1}} \\ $$$$\Rightarrow{r}^{\mathrm{2}} −{r}+\lambda=\mathrm{0}\Rightarrow{r}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{4}\lambda}}{\mathrm{2}}\:\:\:\:\:\lambda\leqslant\frac{\mathrm{1}}{\mathrm{4}}\:\:{r}\in\mathbb{R} \\ $$$${if}\:{r}\in\mathbb{C}\:\:{then}\:−\infty<\lambda<\infty \\ $$$${u}_{{n}} =\Lambda\left(\frac{\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{4}\lambda}}{\mathrm{2}}\right)^{{n}} −\Phi\left(\frac{\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{4}\lambda}}{\mathrm{2}}\right)^{{n}} \\ $$$${u}_{\mathrm{0}} =\Lambda−\Phi \\ $$$${u}_{\mathrm{1}} =\frac{\Lambda−\Phi}{\mathrm{2}}+\frac{\Lambda+\Phi}{\mathrm{2}}\sqrt{\mathrm{1}−\mathrm{4}\lambda}\:\Rightarrow\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{4}\lambda}}\left(\mathrm{2}{u}_{\mathrm{1}} −{u}_{\mathrm{0}} \right)=\Lambda+\Phi \\ $$$$\Lambda=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\left(\mathrm{2}{u}_{\mathrm{1}} −{u}_{\mathrm{0}} \right)}{\:\sqrt{\mathrm{1}−\mathrm{4}\lambda}}+\frac{{u}_{\mathrm{0}} }{\mathrm{2}} \\ $$$$\Phi=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\left(\mathrm{2}{u}_{\mathrm{1}} −{u}_{\mathrm{0}} \right)}{\:\sqrt{\mathrm{1}−\mathrm{4}\lambda}}−\frac{{u}_{\mathrm{0}} }{\mathrm{2}} \\ $$$${u}_{{n}} =\frac{\left(\mathrm{2}{u}_{\mathrm{1}} −{u}_{\mathrm{0}} \right)}{\mathrm{2}\sqrt{\mathrm{1}−\mathrm{4}\lambda}}\left\{\left(\frac{\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{4}\lambda}}{\mathrm{2}}\right)^{{n}} −\left(\frac{\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{4}\lambda}}{\mathrm{2}}\right)^{{n}} \right\}+\frac{{u}_{\mathrm{0}} }{\mathrm{2}}\left\{\left(\frac{\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{4}\lambda}}{\mathrm{2}}\right)^{{n}} −\left(\frac{\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{4}\lambda}}{\mathrm{2}}\right)^{{n}} \right\} \\ $$

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