Given-a-0-1-a-n-1-1-2-3a-n-5a-n-2-4-n-0-n-I-find-a-n-

Question Number 184280 by mr W last updated on 04/Jan/23

G i v e n {\begin{cases} a_{0} = 1 \\ a_{n + 1} = \frac{1}{2} (3 a_{n} + \sqrt{5 a_{n}^{2} - 4}) \end{cases}

\forall n ⩾ 0, n \in I

f i n d a_{n} .

Commented by mr W last updated on 04/Jan/23

u n s o l v e d o l d q u e s t i o n

Commented by mr W last updated on 04/Jan/23

i f o u n d f i n a l l y a s o l u t i o n :

a_{n} = \frac{2}{\sqrt{5}} \cosh [(2 n + 1) \cosh^{- 1} \frac{\sqrt{5}}{2}]

w h i c h d e l i v e r s :

a_{0} = 1

a_{1} = 2

a_{2} = 5

a_{3} = 13

a_{4} = 34

a_{5} = 89

\dots

Commented by mr W last updated on 04/Jan/23

m o r e o t h e r s o l u t i o n s ?

Commented by SEKRET last updated on 04/Jan/23

a_{n} = \frac{a_{n - 1} + a_{n + 1}}{3}

Commented by SEKRET last updated on 04/Jan/23

a_{0} = 1

a_{1} = 2 \cdot 1 + 0

a_{2} = 2 + (2 + 1)

a_{3} = 5 + (5 + 2 + 1)

a_{4} = 13 + (13 + 5 + 2 + 1)

a_{5} = 34 + (34 + 13 + 5 + 2 + 1)

\dots \dots \dots . .

a_{n} = a_{n - 1} + \underset{k = 0}{\sum^{n - 1}} a_{k}

Commented by SEKRET last updated on 04/Jan/23

a_{n} = \frac{2}{\sqrt{5}} \cosh ((2 n + 1) \cosh^{- 1} \frac{\sqrt{5}}{2})

a_{0} = 1 a_{1} = 2 a_{3} = 13 a_{4} = 34 a_{5} = 89

a_{6} = 233 a_{7} = 610 a_{8} = 1597

Commented by SEKRET last updated on 04/Jan/23

how did you find it

Commented by mr W last updated on 04/Jan/23

t h a n k s!

b u t t h e q u e s t i o n a s k s f o r a_{n} i n

e x p l i c i t f o r m . r e c u r s i v e f o r m i s

a l r e a d y g i v e n i n t h e q u e s t i o n .

Commented by SEKRET last updated on 04/Jan/23

right

Commented by Frix last updated on 04/Jan/23

{It}^{'} s each 2 nd Fibonacci - Number \Rightarrow

a_{n} = \frac{1}{\sqrt{5}} ({(\frac{1 + \sqrt{5}}{2})}^{2 n + 1} - {(\frac{1 - \sqrt{5}}{2})}^{2 n + 1})

Commented by mr W last updated on 04/Jan/23

h o w d i d y o u g e t t h i s ?

Commented by manxsol last updated on 05/Jan/23

t r u \bar{e}, g o o d o b s e r v a t i o n S r . F r i x

\begin{array}{ccc} n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ f_{n} & 1 & 2 & 3 & 5 & 8 & 13 & 21 & 34 & 55 & 89 \\ a_{n} & 2 & 5 & 13 & 34 & 89 \end{array}

a_{n} = f_{2 n - 2} a_{1} = 1

t h i s s o l u t i o n i s v a l i d ?

a_{n} = f_{2 n - 2}

Commented by mr W last updated on 05/Jan/23

s o f a r a s i t i s n o t d e v i a t e d f r o m t h e

g i v e n e q u a t i o n a_{n + 1} = \frac{1}{2} (3 a_{n} + \sqrt{5 a_{n}^{2} - 4}),

i {d o n}^{'} t t h i n k i t i s m a t h e m a t i c a l l y

s t r i c t, e v e n t h o u g h i t i s c o r r e c t .

t h e q u e s t i o n i s n o t :

g i v e n : 1, 2, 5, 13, 34, \dots f i n d a_{n} = ? .

a n d e v e n s o, t h e r e a r e i n f i n i t e

s o l u t i o n s f o r a_{n} a s w e k n o w .

Answered by mr W last updated on 05/Jan/23

a_{n + 1} = \frac{1}{2} (3 a_{n} + \sqrt{5 a_{n}^{2} - 4})

2 a_{n + 1} - 3 a_{n} = \sqrt{5 a_{n}^{2} - 4}

a_{n + 1}^{2} - 3 a_{n + 1} a_{n} + a_{n}^{2} + 1 = 0

{(a_{n + 1} - a_{n})}^{2} + 1 = a_{n + 1} a_{n}

l e t

a_{n + 1} + a_{n} = u

a_{n + 1} - a_{n} = v

\Rightarrow u^{2} - 5 v^{2} = 4

{(\frac{u}{2})}^{2} - {(\frac{\sqrt{5} v}{2})}^{2} = 1

\cosh^{2} x_{n} - \sinh^{2} x_{n} = 1

\frac{u}{2} = \cosh x_{n} \Rightarrow u = 2 \cosh x_{n}

\frac{\sqrt{5} v}{2} = \sinh x_{n} \Rightarrow v = \frac{2 \sinh x_{n}}{\sqrt{5}}

a_{n} = \frac{1}{2} (u - v) = \cosh x_{n} - \frac{\sinh x_{n}}{\sqrt{5}}

a_{n} = \frac{2}{\sqrt{5}} (\frac{\sqrt{5}}{2} \cosh x_{n} - \frac{1}{2} \sinh x_{n})

a_{n} = \frac{2}{\sqrt{5}} (\cosh x_{n} \cosh α - \sinh x_{n} \sinh α) = k \cosh (x_{n} - α)

a_{n} = \frac{2}{\sqrt{5}} \cosh (x_{n} - α) w i t h α = \cosh^{- 1} \frac{\sqrt{5}}{2}

a_{n + 1} = \frac{1}{2} (u + v) = \cosh x_{n} + \frac{\sinh x_{n}}{\sqrt{5}}

a_{n + 1} = \frac{2}{\sqrt{5}} (\frac{\sqrt{5}}{2} \cosh x_{n} + \frac{1}{2} \sinh x_{n})

a_{n + 1} = \frac{2}{\sqrt{5}} (\cosh x_{n} \cosh α + \sinh x_{n} \sinh α)

a_{n + 1} = \frac{2}{\sqrt{5}} \cosh (x_{n} + α)

o n t h e o t h e r s i d e :

a_{n + 1} = \frac{2}{\sqrt{5}} \cosh (x_{n + 1} - α)

\Rightarrow \cosh (x_{n + 1} - α) = \cosh (x_{n} + α)

\Rightarrow x_{n + 1} - α = x_{n} + α

\Rightarrow x_{n + 1} = x_{n} + 2 α \leftarrow A . P . w i t h c . d . = 2 α

\Rightarrow x_{n} = x_{0} + 2 n α

a_{n} = \frac{2}{\sqrt{5}} \cosh (x_{0} + 2 n α - α)

\Rightarrow a_{n} = \frac{2}{\sqrt{5}} \cosh [x_{0} + (2 n - 1) \cosh^{- 1} \frac{\sqrt{5}}{2}]

a_{0} = \frac{2}{\sqrt{5}} \cosh (x_{0} - \cosh^{- 1} \frac{\sqrt{5}}{2}) = 1

\Rightarrow x_{0} = 2 \cosh^{- 1} \frac{\sqrt{5}}{2}

\Rightarrow a_{n} = \frac{2}{\sqrt{5}} \cosh [(2 n + 1) \cosh^{- 1} \frac{\sqrt{5}}{2}]

a d d i t i o n a l l y :

\cosh^{- 1} \frac{\sqrt{5}}{2} = \ln (\frac{\sqrt{5}}{2} + \sqrt{{(\frac{\sqrt{5}}{2})}^{2} - 1}) = \ln \frac{\sqrt{5} + 1}{2}

(2 n + 1) \cosh^{- 1} \frac{\sqrt{5}}{2} = \ln {(\frac{\sqrt{5} + 1}{2})}^{(2 n + 1)}

e^{- (2 n + 1) \cosh^{- 1} \frac{\sqrt{5}}{2}} = {(\frac{\sqrt{5} + 1}{2})}^{- (2 n + 1)} = {(\frac{\sqrt{5} - 1}{2})}^{(2 n + 1)}

a_{n} = \frac{2}{\sqrt{5}} \cosh [(2 n + 1) \cosh^{- 1} \frac{\sqrt{5}}{2}]

a_{n} = \frac{1}{\sqrt{5}} [e^{(2 n + 1) \cosh^{- 1} \frac{\sqrt{5}}{2}} + e^{- (2 n + 1) \cosh^{- 1} \frac{\sqrt{5}}{2}}]

\Rightarrow a_{n} = \frac{1}{\sqrt{5}} [{(\frac{\sqrt{5} + 1}{2})}^{(2 n + 1)} + {(\frac{\sqrt{5} - 1}{2})}^{(2 n + 1)}]

Commented by SEKRET last updated on 05/Jan/23

beatiful solution

Commented by Frix last updated on 05/Jan/23

Remember that \cosh α = \frac{e^{α} + e^{- α}}{2}

a_{n} = \frac{1}{\sqrt{5}} ({(\frac{1 + \sqrt{5}}{2})}^{2 n + 1} - {(\frac{1 - \sqrt{5}}{2})}^{2 n + 1}) =

= \frac{1}{\sqrt{5}} ({(\frac{1 + \sqrt{5}}{2})}^{2 b + 1} + {(\frac{2}{1 + \sqrt{5}})}^{2 n + 1}) =

= \frac{1}{\sqrt{5}} (t^{x} + t^{- x}) =

= \frac{1}{\sqrt{5}} (e^{x \ln t} + e^{- x \ln t}) =

= \frac{2}{\sqrt{5}} \cosh (x \ln t) =

= \frac{2}{\sqrt{5}} \cosh ((2 n + 1) \ln \frac{1 + \sqrt{5}}{2}) =

\ln (x + \sqrt{x^{2} - 1}) = \cosh^{- 1} x

\ln \frac{1 + \sqrt{5}}{2} = \ln (\frac{\sqrt{5}}{2} + \sqrt{\frac{5}{4} - 1}) = \cosh^{- 1} \frac{\sqrt{5}}{2}

= \frac{2}{\sqrt{5}} \cosh ((2 n + 1) \cosh^{- 1} \frac{\sqrt{5}}{2})

\dots . which is your result

Commented by mr W last updated on 05/Jan/23

t h a n k s s i r!

i^{'} v e a l s o a d d e d t h e d e d u c t i o n t o

a_{n} = \frac{1}{\sqrt{5}} [{(\frac{\sqrt{5} + 1}{2})}^{(2 n + 1)} + {(\frac{\sqrt{5} - 1}{2})}^{(2 n + 1)}] .

s e e a b o v e .

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