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Question Number 125202 by bemath last updated on 09/Dec/20

Solve the reccurence relation a_n = 2(a_(n−1) −a_(n−2) ) ; given a_0 =1 and a_1 = 0.

$$\:{Solve}\:{the}\:{reccurence}\:{relation} \\ $$$${a}_{{n}} \:=\:\mathrm{2}\left({a}_{{n}−\mathrm{1}} −{a}_{{n}−\mathrm{2}} \right)\:;\:{given}\:{a}_{\mathrm{0}} =\mathrm{1}\: \\ $$$${and}\:{a}_{\mathrm{1}} =\:\mathrm{0}. \\ $$

Answered by liberty last updated on 09/Dec/20

The characteristic equation :≡ x^2 −2x+2=0 and have the roots are { ((α=1+i)),((α^− = 1−i)) :} Expressing α and α^− in trigonometric form we have { ((α=(√2) (cos (π/4)+ i sin (π/4)))),((α^− = (√2) (cos (π/4)−i sin (π/4)))) :} The general solution is a_n = P (α)^n +Q(α^− )^n by the Moivre′s Theorem give a_n = ((√2))^n [ P(cos ((nπ)/4)+i sin ((nπ)/4))+Q(cos ((nπ)/4)−i sin ((nπ)/4)) ] a_n = ((√2))^n [ (P+Q)cos ((nπ)/4)+(P−Q)i sin ((nπ)/4) ] The initial condition imply that { ((P+Q=1)),(((√2) (((√2)/2) (P+Q)+((√2)/2) (P−Q)i)=0)) :} we get (P−Q)i = −1 thus the required solution given by a_n = ((√2))^n (cos ((nπ)/4)−sin ((nπ)/4)); for n ≥ 0

$${The}\:{characteristic}\:{equation}\::\equiv\:{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}=\mathrm{0} \\ $$$${and}\:{have}\:{the}\:{roots}\:{are}\:\begin{cases}{\alpha=\mathrm{1}+{i}}\\{\overset{−} {\alpha}\:=\:\mathrm{1}−{i}}\end{cases} \\ $$$${Expressing}\:\alpha\:{and}\:\overset{−} {\alpha}\:{in}\:{trigonometric}\:{form} \\ $$$${we}\:{have}\:\begin{cases}{\alpha=\sqrt{\mathrm{2}}\:\left(\mathrm{cos}\:\frac{\pi}{\mathrm{4}}+\:{i}\:\mathrm{sin}\:\frac{\pi}{\mathrm{4}}\right)}\\{\overset{−} {\alpha}\:=\:\sqrt{\mathrm{2}}\:\left(\mathrm{cos}\:\frac{\pi}{\mathrm{4}}−{i}\:\mathrm{sin}\:\frac{\pi}{\mathrm{4}}\right)}\end{cases} \\ $$$${The}\:{general}\:{solution}\:{is}\:{a}_{{n}} \:=\:{P}\:\left(\alpha\right)^{{n}} +{Q}\left(\overset{−} {\alpha}\right)^{{n}} \\ $$$${by}\:{the}\:{Moivre}'{s}\:{Theorem}\:{give} \\ $$$$\:{a}_{{n}} \:=\:\left(\sqrt{\mathrm{2}}\right)^{{n}} \:\left[\:{P}\left(\mathrm{cos}\:\frac{{n}\pi}{\mathrm{4}}+{i}\:\mathrm{sin}\:\frac{{n}\pi}{\mathrm{4}}\right)+{Q}\left(\mathrm{cos}\:\frac{{n}\pi}{\mathrm{4}}−{i}\:\mathrm{sin}\:\frac{{n}\pi}{\mathrm{4}}\right)\:\right] \\ $$$$\:{a}_{{n}} \:=\:\left(\sqrt{\mathrm{2}}\right)^{{n}} \:\left[\:\left({P}+{Q}\right)\mathrm{cos}\:\frac{{n}\pi}{\mathrm{4}}+\left({P}−{Q}\right){i}\:\mathrm{sin}\:\frac{{n}\pi}{\mathrm{4}}\:\right] \\ $$$${The}\:{initial}\:{condition}\:{imply}\:{that} \\ $$$$\begin{cases}{{P}+{Q}=\mathrm{1}}\\{\sqrt{\mathrm{2}}\:\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\left({P}+{Q}\right)+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\left({P}−{Q}\right){i}\right)=\mathrm{0}}\end{cases} \\ $$$${we}\:{get}\:\left({P}−{Q}\right){i}\:=\:−\mathrm{1} \\ $$$${thus}\:{the}\:{required}\:{solution}\:{given}\:{by}\: \\ $$$$\:{a}_{{n}} \:=\:\left(\sqrt{\mathrm{2}}\right)^{{n}} \:\left(\mathrm{cos}\:\frac{{n}\pi}{\mathrm{4}}−\mathrm{sin}\:\frac{{n}\pi}{\mathrm{4}}\right);\:{for}\:{n}\:\geqslant\:\mathrm{0} \\ $$$$ \\ $$

Answered by mathmax by abdo last updated on 09/Dec/20

a_n =2(a_(n−1) −a_(n−2) ) ⇒a_n −2a_(n−1) +2a_(n−2) =0 ⇒ a_(n+2) −2a_(n+1) +2a_n =0 →r^2 −2r+2=0 Δ^′ =1−2=−1 ⇒r_1 =1+i and r_2 =1−i ⇒ a_n =α r_1 ^n +β r_2 ^n we have r_1 =(√2)e^((iπ)/4) and r_2 =(√2)e^(−((iπ)/4)) ⇒ a_n =α((√2))^n e^((inπ)/4) +β ((√2))^n e^(−((inπ)/4)) a_0 =1 =α+β a_1 =0 =α(√2)e^((iπ)/4) +β(√2)e^(−((iπ)/4)) we get the system { ((α+β=1 → Δ_s = determinant (((1 1)),(((√2)e^((iπ)/4) (√2)e^(−((iπ)/4)) ))))),(((√2)e^((iπ)/4) α +(√2)e^(−((iπ)/4)) β =0 )) :} =(√2)(e^(−((iπ)/4)) −e^((iπ)/4) )=(√2)(−2i×((√2)/2))=−2i α =( determinant (((1 1)),((0 (√2)e^(−((iπ)/4)) )))/(−2i))=(i/2)(√2)e^(−((iπ)/4)) β =( determinant (((1 1)),(((√2)e^((iπ)/4) 0)))/(−2i))=((−(√2)e^((iπ)/4) )/(−2i))=−((√2)/2)i e^((iπ)/4) ⇒ a_n =(i/2)(√2)e^((−iπ)/4) ((√2))^n e^((inπ)/4) −((√2)/2)i e^((iπ)/4) ((√2))^n e^(−((inπ)/4)) =(i/2)((√2))^(n+1) e^((i(n−1)π)/4) −(i/2)((√2))^(n+1) e^(−((i(n−1)π)/4)) =i((√2))^(n−1) e^((i(n−1)π)/4) −i((√2))^(n−1) e^(−((i(n−1)π)/4)) =i((√2))^(n−1) (2i sin((((n−1)π)/4)))=−((√2))^(n+1) sin((((n−1)π)/4))

$$\mathrm{a}_{\mathrm{n}} =\mathrm{2}\left(\mathrm{a}_{\mathrm{n}−\mathrm{1}} −\mathrm{a}_{\mathrm{n}−\mathrm{2}} \right)\:\Rightarrow\mathrm{a}_{\mathrm{n}} −\mathrm{2a}_{\mathrm{n}−\mathrm{1}} +\mathrm{2a}_{\mathrm{n}−\mathrm{2}} =\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{a}_{\mathrm{n}+\mathrm{2}} −\mathrm{2a}_{\mathrm{n}+\mathrm{1}} +\mathrm{2a}_{\mathrm{n}} =\mathrm{0}\:\rightarrow\mathrm{r}^{\mathrm{2}} −\mathrm{2r}+\mathrm{2}=\mathrm{0} \\ $$$$\Delta^{'} \:=\mathrm{1}−\mathrm{2}=−\mathrm{1}\:\Rightarrow\mathrm{r}_{\mathrm{1}} =\mathrm{1}+\mathrm{i}\:\:\mathrm{and}\:\mathrm{r}_{\mathrm{2}} =\mathrm{1}−\mathrm{i}\:\Rightarrow \\ $$$$\mathrm{a}_{\mathrm{n}} =\alpha\:\mathrm{r}_{\mathrm{1}} ^{\mathrm{n}} \:+\beta\:\mathrm{r}_{\mathrm{2}} ^{\mathrm{n}} \:\:\:\:\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{r}_{\mathrm{1}} =\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \:\mathrm{and}\:\mathrm{r}_{\mathrm{2}} =\sqrt{\mathrm{2}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \:\Rightarrow \\ $$$$\mathrm{a}_{\mathrm{n}} =\alpha\left(\sqrt{\mathrm{2}}\right)^{\mathrm{n}} \mathrm{e}^{\frac{\mathrm{in}\pi}{\mathrm{4}}} +\beta\:\left(\sqrt{\mathrm{2}}\right)^{\mathrm{n}} \:\mathrm{e}^{−\frac{\mathrm{in}\pi}{\mathrm{4}}} \\ $$$$\mathrm{a}_{\mathrm{0}} =\mathrm{1}\:=\alpha+\beta \\ $$$$\mathrm{a}_{\mathrm{1}} =\mathrm{0}\:=\alpha\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \:+\beta\sqrt{\mathrm{2}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \:\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{system} \\ $$$$\begin{cases}{\alpha+\beta=\mathrm{1}\:\:\:\:\:\:\:\:\:\:\rightarrow\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Delta_{\mathrm{s}} =\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \:\:\:\:\:\:\:\:\sqrt{\mathrm{2}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} }\end{vmatrix}}\\{\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \alpha\:+\sqrt{\mathrm{2}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \beta\:=\mathrm{0}\:\:}\end{cases} \\ $$$$=\sqrt{\mathrm{2}}\left(\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} −\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)=\sqrt{\mathrm{2}}\left(−\mathrm{2i}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)=−\mathrm{2i} \\ $$$$\alpha\:=\frac{\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\sqrt{\mathrm{2}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} }\end{vmatrix}}{−\mathrm{2i}}=\frac{\mathrm{i}}{\mathrm{2}}\sqrt{\mathrm{2}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \\ $$$$\beta\:=\frac{\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \:\:\:\mathrm{0}}\end{vmatrix}}{−\mathrm{2i}}=\frac{−\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} }{−\mathrm{2i}}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i}\:\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \:\Rightarrow \\ $$$$\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{i}}{\mathrm{2}}\sqrt{\mathrm{2}}\mathrm{e}^{\frac{−\mathrm{i}\pi}{\mathrm{4}}} \left(\sqrt{\mathrm{2}}\right)^{\mathrm{n}} \:\mathrm{e}^{\frac{\mathrm{in}\pi}{\mathrm{4}}} \:\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i}\:\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \:\left(\sqrt{\mathrm{2}}\right)^{\mathrm{n}} \:\mathrm{e}^{−\frac{\mathrm{in}\pi}{\mathrm{4}}} \\ $$$$=\frac{\mathrm{i}}{\mathrm{2}}\left(\sqrt{\mathrm{2}}\right)^{\mathrm{n}+\mathrm{1}} \mathrm{e}^{\frac{\mathrm{i}\left(\mathrm{n}−\mathrm{1}\right)\pi}{\mathrm{4}}} −\frac{\mathrm{i}}{\mathrm{2}}\left(\sqrt{\mathrm{2}}\right)^{\mathrm{n}+\mathrm{1}} \:\mathrm{e}^{−\frac{\mathrm{i}\left(\mathrm{n}−\mathrm{1}\right)\pi}{\mathrm{4}}} \\ $$$$=\mathrm{i}\left(\sqrt{\mathrm{2}}\right)^{\mathrm{n}−\mathrm{1}} \:\mathrm{e}^{\frac{\mathrm{i}\left(\mathrm{n}−\mathrm{1}\right)\pi}{\mathrm{4}}} −\mathrm{i}\left(\sqrt{\mathrm{2}}\right)^{\mathrm{n}−\mathrm{1}} \:\mathrm{e}^{−\frac{\mathrm{i}\left(\mathrm{n}−\mathrm{1}\right)\pi}{\mathrm{4}}} \\ $$$$=\mathrm{i}\left(\sqrt{\mathrm{2}}\right)^{\mathrm{n}−\mathrm{1}} \left(\mathrm{2i}\:\mathrm{sin}\left(\frac{\left(\mathrm{n}−\mathrm{1}\right)\pi}{\mathrm{4}}\right)\right)=−\left(\sqrt{\mathrm{2}}\right)^{\mathrm{n}+\mathrm{1}} \:\mathrm{sin}\left(\frac{\left(\mathrm{n}−\mathrm{1}\right)\pi}{\mathrm{4}}\right) \\ $$

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