If a_(n + 3) = (a_(n − 1) /a_(n + 1) ) , and a_0 = 1, a_2 = 2 find a


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Question Number 93173 by I want to learn more last updated on 11/May/20

$If a_{n + 3} = \frac{a_{n - 1}}{a_{n + 1}}, and a_{0} = 1, a_{2} = 2$ $find a_{n}$
	Answered by prakash jain last updated on 12/May/20
	$a_(n+3) =(a_(n−1) /a_(n+1) ) b_n =ln a_n ln a_(n+3) =ln a_(n−1) −ln a_(n+1) b_(n+3) =b_(n−1) −b_(n+1) b_(n+3) +b_(n+1) −b_(n−1) =0 Characteristic equation x^4 +x^2 −1=0 x^2 =((−1±(√5))/2) x=±(√((−1±(√5))/2)) x=±(√(((√5)−1)/2)) and ±i(√(((√5)+1)/2)) b_n =c_1 ((√((√(5−1))/2)))^n +c_2 (−(√(((√5)−1)/2)))^n +c_3 (i(√(((√5)+1)/2)))^n +c_4 (−i(√(((√5)+1)/2)))^n b_n =(1/2^(n/2) ){(c_1 +(−1)^n c_2 )((√5)−1)^(n/2) +i(c_3 +(−1)^n c_4 )((√5)+1)^(n/2) a_n =exp((1/2^(n/2) ){(c_1 +(−1)^n c_2 )((√5)−1)^(n/2) +i(c_3 +(−1)^n c_4 )((√5)+1)^(n/2) ) where c_(1,) c_2 ,c_3 and c_4 are arbitarary constants. You have only given two initial conditions. You need 4 to determine c_1 to c_4$
	$a_{n + 3} = \frac{a_{n - 1}}{a_{n + 1}}$ $b_{n} = \ln a_{n}$ $\ln a_{n + 3} = \ln a_{n - 1} - \ln a_{n + 1}$ $b_{n + 3} = b_{n - 1} - b_{n + 1}$ $b_{n + 3} + b_{n + 1} - b_{n - 1} = 0$ $Characteristic equation$ $x^{4} + x^{2} - 1 = 0$ $x^{2} = \frac{- 1 \pm \sqrt{5}}{2}$ $x = \pm \sqrt{\frac{- 1 \pm \sqrt{5}}{2}}$ $x = \pm \sqrt{\frac{\sqrt{5} - 1}{2}} and \pm i \sqrt{\frac{\sqrt{5} + 1}{2}}$ $b_{n} = c_{1} {(\sqrt{\frac{\sqrt{5 - 1}}{2}})}^{n} + c_{2} {(- \sqrt{\frac{\sqrt{5} - 1}{2}})}^{n}$ $+ c_{3} {(i \sqrt{\frac{\sqrt{5} + 1}{2}})}^{n} + c_{4} {(- i \sqrt{\frac{\sqrt{5} + 1}{2}})}^{n}$ $b_{n} = \frac{1}{2^{n / 2}} {(c_{1} + {(- 1)}^{n} c_{2}) {(\sqrt{5} - 1)}^{n / 2} + i (c_{3} + {(- 1)}^{n} c_{4}) {(\sqrt{5} + 1)}^{n / 2}$ $a_{n} = \exp (\frac{1}{2^{n / 2}} {(c_{1} + {(- 1)}^{n} c_{2}) {(\sqrt{5} - 1)}^{n / 2} + i (c_{3} + {(- 1)}^{n} c_{4}) {(\sqrt{5} + 1)}^{n / 2})$ $where c_{1,} c_{2}, c_{3} and c_{4} are arbitarary$ $constants .$ $You have only given two initial$ $conditions . You need 4 to$ $determine c_{1} to c_{4}$
	Commented by prakash jain last updated on 12/May/20

	$e x p (x) = e^{x}$
	Commented by I want to learn more last updated on 12/May/20

	$Thanks sir, i appreciate$
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