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


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

$S o l v e t h e r e c c u r e n c e r e l a t i o n$ $a_{n} = 2 (a_{n - 1} - a_{n - 2}); g i v e n a_{0} = 1$ $a n d a_{1} = 0 .$
	Answered by liberty last updated on 09/Dec/20
	$The characteristic equation :≡ x^2 −2x+2=0 and have the roots are { ((α=1+i)),((α^− = 1−i)) :} Expressing α and α^− in trigonometric form we have { ((α=(√2) (cos (π/4)+ i sin (π/4)))),((α^− = (√2) (cos (π/4)−i sin (π/4)))) :} The general solution is a_n = P (α)^n +Q(α^− )^n by the Moivre′s Theorem give a_n = ((√2))^n [ P(cos ((nπ)/4)+i sin ((nπ)/4))+Q(cos ((nπ)/4)−i sin ((nπ)/4)) ] a_n = ((√2))^n [ (P+Q)cos ((nπ)/4)+(P−Q)i sin ((nπ)/4) ] The initial condition imply that { ((P+Q=1)),(((√2) (((√2)/2) (P+Q)+((√2)/2) (P−Q)i)=0)) :} we get (P−Q)i = −1 thus the required solution given by a_n = ((√2))^n (cos ((nπ)/4)−sin ((nπ)/4)); for n ≥ 0$
	$T h e c h a r a c t e r i s t i c e q u a t i o n :\equiv x^{2} - 2 x + 2 = 0$ $a n d h a v e t h e r o o t s a r e {\begin{cases} α = 1 + i \\ \bar{α} = 1 - i \end{cases}$ $E x p r e s s i n g α a n d \bar{α} i n t r i g o n o m e t r i c f o r m$ $w e h a v e {\begin{cases} α = \sqrt{2} (\cos \frac{π}{4} + i \sin \frac{π}{4}) \\ \bar{α} = \sqrt{2} (\cos \frac{π}{4} - i \sin \frac{π}{4}) \end{cases}$ $T h e g e n e r a l s o l u t i o n i s a_{n} = P {(α)}^{n} + Q {(\bar{α})}^{n}$ $b y t h e {M o i v r e}^{'} s T h e o r e m g i v e$ $a_{n} = {(\sqrt{2})}^{n} [P (\cos \frac{n π}{4} + i \sin \frac{n π}{4}) + Q (\cos \frac{n π}{4} - i \sin \frac{n π}{4})]$ $a_{n} = {(\sqrt{2})}^{n} [(P + Q) \cos \frac{n π}{4} + (P - Q) i \sin \frac{n π}{4}]$ $T h e i n i t i a l c o n d i t i o n i m p l y t h a t$ ${\begin{cases} P + Q = 1 \\ \sqrt{2} (\frac{\sqrt{2}}{2} (P + Q) + \frac{\sqrt{2}}{2} (P - Q) i) = 0 \end{cases}$ $w e g e t (P - Q) i = - 1$ $t h u s t h e r e q u i r e d s o l u t i o n g i v e n b y$ $a_{n} = {(\sqrt{2})}^{n} (\cos \frac{n π}{4} - \sin \frac{n π}{4}); f o r n ⩾ 0$
	Answered by mathmax by abdo last updated on 09/Dec/20
	$a_n =2(a_(n−1) −a_(n−2) ) ⇒a_n −2a_(n−1) +2a_(n−2) =0 ⇒ a_(n+2) −2a_(n+1) +2a_n =0 →r^2 −2r+2=0 Δ^′ =1−2=−1 ⇒r_1 =1+i and r_2 =1−i ⇒ a_n =α r_1 ^n +β r_2 ^n we have r_1 =(√2)e^((iπ)/4) and r_2 =(√2)e^(−((iπ)/4)) ⇒ a_n =α((√2))^n e^((inπ)/4) +β ((√2))^n e^(−((inπ)/4)) a_0 =1 =α+β a_1 =0 =α(√2)e^((iπ)/4) +β(√2)e^(−((iπ)/4)) we get the system { ((α+β=1 → Δ_s = determinant (((1 1)),(((√2)e^((iπ)/4) (√2)e^(−((iπ)/4)) ))))),(((√2)e^((iπ)/4) α +(√2)e^(−((iπ)/4)) β =0 )) :} =(√2)(e^(−((iπ)/4)) −e^((iπ)/4) )=(√2)(−2i×((√2)/2))=−2i α =( determinant (((1 1)),((0 (√2)e^(−((iπ)/4)) )))/(−2i))=(i/2)(√2)e^(−((iπ)/4)) β =( determinant (((1 1)),(((√2)e^((iπ)/4) 0)))/(−2i))=((−(√2)e^((iπ)/4) )/(−2i))=−((√2)/2)i e^((iπ)/4) ⇒ a_n =(i/2)(√2)e^((−iπ)/4) ((√2))^n e^((inπ)/4) −((√2)/2)i e^((iπ)/4) ((√2))^n e^(−((inπ)/4)) =(i/2)((√2))^(n+1) e^((i(n−1)π)/4) −(i/2)((√2))^(n+1) e^(−((i(n−1)π)/4)) =i((√2))^(n−1) e^((i(n−1)π)/4) −i((√2))^(n−1) e^(−((i(n−1)π)/4)) =i((√2))^(n−1) (2i sin((((n−1)π)/4)))=−((√2))^(n+1) sin((((n−1)π)/4))$
	$a_{n} = 2 (a_{n - 1} - a_{n - 2}) \Rightarrow a_{n} - {2 a}_{n - 1} + {2 a}_{n - 2} = 0 \Rightarrow$ $a_{n + 2} - {2 a}_{n + 1} + {2 a}_{n} = 0 \to r^{2} - 2 r + 2 = 0$ $Δ^{^{'}} = 1 - 2 = - 1 \Rightarrow r_{1} = 1 + i and r_{2} = 1 - i \Rightarrow$ $a_{n} = α r_{1}^{n} + β r_{2}^{n} we have r_{1} = \sqrt{2} e^{\frac{i π}{4}} and r_{2} = \sqrt{2} e^{- \frac{i π}{4}} \Rightarrow$ $a_{n} = α {(\sqrt{2})}^{n} e^{\frac{in π}{4}} + β {(\sqrt{2})}^{n} e^{- \frac{in π}{4}}$ $a_{0} = 1 = α + β$ $a_{1} = 0 = α \sqrt{2} e^{\frac{i π}{4}} + β \sqrt{2} e^{- \frac{i π}{4}} we get the system$ ${\begin{cases} α + β = 1 \to Δ_{s} = \| \begin{array}{c} 1 1 \\ \sqrt{2} e^{\frac{i π}{4}} \sqrt{2} e^{- \frac{i π}{4}} \end{array} \| \\ \sqrt{2} e^{\frac{i π}{4}} α + \sqrt{2} e^{- \frac{i π}{4}} β = 0 \end{cases}$ $= \sqrt{2} (e^{- \frac{i π}{4}} - e^{\frac{i π}{4}}) = \sqrt{2} (- 2 i \times \frac{\sqrt{2}}{2}) = - 2 i$ $α = \frac{\| \begin{matrix} 1 1 \\ 0 \sqrt{2} e^{- \frac{i π}{4}} \end{matrix} \|}{- 2 i} = \frac{i}{2} \sqrt{2} e^{- \frac{i π}{4}}$ $β = \frac{\| \begin{matrix} 1 1 \\ \sqrt{2} e^{\frac{i π}{4}} 0 \end{matrix} \|}{- 2 i} = \frac{- \sqrt{2} e^{\frac{i π}{4}}}{- 2 i} = - \frac{\sqrt{2}}{2} i e^{\frac{i π}{4}} \Rightarrow$ $a_{n} = \frac{i}{2} \sqrt{2} e^{\frac{- i π}{4}} {(\sqrt{2})}^{n} e^{\frac{in π}{4}} - \frac{\sqrt{2}}{2} i e^{\frac{i π}{4}} {(\sqrt{2})}^{n} e^{- \frac{in π}{4}}$ $= \frac{i}{2} {(\sqrt{2})}^{n + 1} e^{\frac{i (n - 1) π}{4}} - \frac{i}{2} {(\sqrt{2})}^{n + 1} e^{- \frac{i (n - 1) π}{4}}$ $= i {(\sqrt{2})}^{n - 1} e^{\frac{i (n - 1) π}{4}} - i {(\sqrt{2})}^{n - 1} e^{- \frac{i (n - 1) π}{4}}$ $= i {(\sqrt{2})}^{n - 1} (2 i \sin (\frac{(n - 1) π}{4})) = - {(\sqrt{2})}^{n + 1} \sin (\frac{(n - 1) π}{4})$
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