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Question Number 214888 by Emmanuel07 last updated on 22/Dec/24

Commented by mr W last updated on 23/Dec/24
$i got a_n =cot {[(1/( (√5)))(tan^(−1) 2−((3π)/8))+(π/8)](((1−(√5))/2))^n −[(1/( (√5)))(tan^(−1) 2−((3π)/8))−(π/8)](((1+(√5))/2))^n } which delivers: a_0 =1 a_1 =2 a_2 =(1/3) a_3 =−(1/7) a_4 =−((11)/2) a_5 =(3/(79)) ......$
$i g o t$ $a_{n} = \cot {[\frac{1}{\sqrt{5}} (\tan^{- 1} 2 - \frac{3 π}{8}) + \frac{π}{8}] {(\frac{1 - \sqrt{5}}{2})}^{n} - [\frac{1}{\sqrt{5}} (\tan^{- 1} 2 - \frac{3 π}{8}) - \frac{π}{8}] {(\frac{1 + \sqrt{5}}{2})}^{n}}$ $w h i c h d e l i v e r s :$ $a_{0} = 1$ $a_{1} = 2$ $a_{2} = \frac{1}{3}$ $a_{3} = - \frac{1}{7}$ $a_{4} = - \frac{11}{2}$ $a_{5} = \frac{3}{79}$ $. . . . . .$
Commented by mr W last updated on 23/Dec/24

$h a s s o m e b o d y g o t a m o r e s i m p l e$ $f o r m u l a f o r a_{n} i n t e r m s o f n ?$
	Answered by maths2 last updated on 23/Dec/24

	$a_{n + 1} = \frac{a_{n} a_{n - 1} - 1}{a_{n} + a_{n - 1}}$ $a_{n} = c o t (b_{n}) \Rightarrow c o t (b_{n + 1}) = \frac{c o t (b_{n}) c o t (b_{n - 1}) - 1}{c o t (b_{n}) + \cot (b_{n - 1})} =$ $\cot (b_{n} + b_{n - 1}) \Rightarrow b_{n + 1} = b_{n} + b_{n - 1}$ $b_{n + 1} - b_{n} - b_{n - 1} = 0$ $b_{0} = \frac{π}{4}; b_{1} = \cot^{- 1} (2)$
	Answered by mr W last updated on 24/Dec/24
	$a_(n−1) a_n −a_(n−1) a_(n+1) =1+a_n a_(n+1) a_(n−1) (a_n −a_(n+1) )=1+a_n a_(n+1) ((1+a_n a_(n+1) )/(a_n −a_(n+1) ))=a_(n−1) let a_n =cot θ_n ((1+cot θ_n cot θ_(n+1) )/(cot θ_n −cot θ_(n+1) ))=cot θ_(n−1) cot (θ_(n+1) −θ_n )=cot θ_(n−1) ⇒θ_(n+1) −θ_n =θ_(n−1) ⇒θ_(n+1) −θ_n −θ_(n−1) =0 ← Fibonacci r^2 −r−1=0 ⇒r=((1±(√5))/2) ⇒θ_n =A(((1+(√5))/2))^n +B(((1−(√5))/2))^n θ_0 =A+B=cot^(−1) a_0 =cot^(−1) 1=(π/4) ⇒((A+B)/2)=(π/8) θ_1 =((A+B)/2)+(( (√5)(A−B))/2)=cot^(−1) a_1 =cot^(−1) 2 ⇒(( A−B)/2)=(1/( (√5)))(cot^(−1) 2−(π/8)) ⇒A=(π/8)+(1/( (√5)))(cot^(−1) 2−(π/8)) ⇒B=(π/8)−(1/( (√5)))(cot^(−1) 2−(π/8)) ⇒θ_n =[(π/8)+(1/( (√5)))(cot^(−1) 2−(π/8))](((1+(√5))/2))^n +[(π/8)−(1/( (√5)))(cot^(−1) 2−(π/8))](((1−(√5))/2))^n ⇒a_n =cot {[(π/8)+(1/( (√5)))(cot^(−1) 2−(π/8))](((1+(√5))/2))^n +[(π/8)−(1/( (√5)))(cot^(−1) 2−(π/8))](((1−(√5))/2))^n }$
	$a_{n - 1} a_{n} - a_{n - 1} a_{n + 1} = 1 + a_{n} a_{n + 1}$ $a_{n - 1} (a_{n} - a_{n + 1}) = 1 + a_{n} a_{n + 1}$ $\frac{1 + a_{n} a_{n + 1}}{a_{n} - a_{n + 1}} = a_{n - 1}$ $l e t a_{n} = \cot θ_{n}$ $\frac{1 + \cot θ_{n} \cot θ_{n + 1}}{\cot θ_{n} - \cot θ_{n + 1}} = \cot θ_{n - 1}$ $\cot (θ_{n + 1} - θ_{n}) = \cot θ_{n - 1}$ $\Rightarrow θ_{n + 1} - θ_{n} = θ_{n - 1}$ $\Rightarrow θ_{n + 1} - θ_{n} - θ_{n - 1} = 0 \leftarrow F i b o n a c c i$ $r^{2} - r - 1 = 0$ $\Rightarrow r = \frac{1 \pm \sqrt{5}}{2}$ $\Rightarrow θ_{n} = A {(\frac{1 + \sqrt{5}}{2})}^{n} + B {(\frac{1 - \sqrt{5}}{2})}^{n}$ $θ_{0} = A + B = \cot^{- 1} a_{0} = \cot^{- 1} 1 = \frac{π}{4}$ $\Rightarrow \frac{A + B}{2} = \frac{π}{8}$ $θ_{1} = \frac{A + B}{2} + \frac{\sqrt{5} (A - B)}{2} = \cot^{- 1} a_{1} = \cot^{- 1} 2$ $\Rightarrow \frac{A - B}{2} = \frac{1}{\sqrt{5}} (\cot^{- 1} 2 - \frac{π}{8})$ $\Rightarrow A = \frac{π}{8} + \frac{1}{\sqrt{5}} (\cot^{- 1} 2 - \frac{π}{8})$ $\Rightarrow B = \frac{π}{8} - \frac{1}{\sqrt{5}} (\cot^{- 1} 2 - \frac{π}{8})$ $\Rightarrow θ_{n} = [\frac{π}{8} + \frac{1}{\sqrt{5}} (\cot^{- 1} 2 - \frac{π}{8})] {(\frac{1 + \sqrt{5}}{2})}^{n} + [\frac{π}{8} - \frac{1}{\sqrt{5}} (\cot^{- 1} 2 - \frac{π}{8})] {(\frac{1 - \sqrt{5}}{2})}^{n}$ $\Rightarrow a_{n} = \cot {[\frac{π}{8} + \frac{1}{\sqrt{5}} (\cot^{- 1} 2 - \frac{π}{8})] {(\frac{1 + \sqrt{5}}{2})}^{n} + [\frac{π}{8} - \frac{1}{\sqrt{5}} (\cot^{- 1} 2 - \frac{π}{8})] {(\frac{1 - \sqrt{5}}{2})}^{n}}$
	Commented by ajfour last updated on 24/Dec/24

	$\frac{a_{n - 1}}{a_{n + 1}} - \frac{a_{n - 1}}{a_{n}} = \frac{1}{a_{n} a_{n + 1} + 1}$ $\frac{1}{r^{2}} - \frac{1}{r} = \frac{1}{r^{3} + 1}$ $(r^{3} + 1) (1 - r) = r^{2}$ $r^{4} - r^{3} + r^{2} + r - 1 = 0$ $(r^{2} - \frac{1}{r^{2}}) - (r - \frac{1}{r}) = - 1$ $s a y t b e t w o r e a l r o o t s a r e r_{1} & r_{2} .$ $a_{n} = {A r}_{1}^{n} + {B r}_{2}^{n}$ $c a n w e d o t h i s ? i^{'} v e f o r g o t t e n t h i s$ $s e q u e n c e a n d s e r i e s t o p i c . b e e n 20 y .$
	Commented by mr W last updated on 24/Dec/24

	$n o s i r .$ $i f w e a s s u m e a_{n} = {A r}^{n}$ $\frac{a_{n - 1}}{a_{n + 1}} - \frac{a_{n - 1}}{a_{n}} = \frac{1}{a_{n} a_{n + 1} + 1}$ $\Rightarrow \frac{{A r}^{n - 1}}{{A r}^{n + 1}} - \frac{{A r}^{n - 1}}{{A r}^{n}} = \frac{1}{A^{2} r^{2 n + 1} + 1}$ $\Rightarrow \frac{1}{r^{2}} - \frac{1}{r} = \frac{1}{A^{2} r^{2 n + 1} + 1}!!!$ $w e {c a n}^{'} t f i n d r w h i c h i s i n d e p e n d e n t$ $f r o m n a n d A .$ $t h a t m e a n s a_{n} c a n n o t b e e x p r e s s e d$ $a s t y p e {A r}^{n} .$ $g e n e r a l l y c h a r a c t e r i s t i c e q u a t i o n$ $e x i s t s o n l y f o r l i n e a r r e c u r r e n c e$ $r e l a t i o n s$ $a_{n} = c_{1} a_{n - 1} + c_{2} a_{n - 2} + . . . + c_{k} a_{n - k}$ $w i t h c_{i} = c o n s t a n t s .$
	Commented by ajfour last updated on 24/Dec/24

	$T h a n k s, i l l r e v i s e .$
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