Solve-the-reccurence-relation-a-n-2-a-n-1-a-n-2-given-a-0-1-and-a-1-0-

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

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}) \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})