if-x-1-x-a-find-x-n-1-x-n-

Question Number 78931 by mr W last updated on 21/Jan/20

i f x + \frac{1}{x} = a

f i n d x^{n} + \frac{1}{x^{n}} = ?

Commented by mathmax by abdo last updated on 22/Jan/20

x + \frac{1}{x} = a \Rightarrow x^{2} + 1 = a x \Rightarrow x^{2} - a x + 1 = 0

Δ = a^{2} - 4 \Rightarrow x_{1} = \frac{a + \sqrt{a^{2} - 4}}{2} a n d x_{2} = \frac{a - \sqrt{a^{2} - 4}}{2}

c a s e 1 x = x_{1} \Rightarrow x^{n} + \frac{1}{x^{n}} = {(\frac{a + \sqrt{a^{2} - 4}}{2})}^{n} + {(\frac{2}{a + \sqrt{a^{2} - 4}})}^{n}

= \frac{1}{2^{n}} {(a + \sqrt{a^{2} - 4})}^{n} + 2^{n} {(\frac{a - \sqrt{a^{2} - 4}}{4})}^{n}

= \frac{1}{2^{n}} {(a + \sqrt{a^{2} - 4})}^{n} + \frac{1}{2^{n}} {(a - \sqrt{a^{2} - 4})}^{n}

c a s e 2 x = x_{2} w e g e t t h e s a m e v a l u e d u e t o x_{2} = c o n j (x_{1})

Commented by mr W last updated on 22/Jan/20

t h a n k y o u s i r!

U_{n} = \frac{1}{2^{n}} {{(a + \sqrt{a^{2} - 4})}^{n} + {(a - \sqrt{a^{2} - 4})}^{n}}

c a n t h e t e r m {(a + \sqrt{a^{2} - 4})}^{n} + {(a - \sqrt{a^{2} - 4})}^{n}

b e s i m p l i f i e d m o r e ?

i f a \in Q, i s s l s o U_{n} \in Q ?

Commented by mind is power last updated on 22/Jan/20

y e a h a \in Q, U n \in Q

b e c a u s e

U_{n + 1} = {a U}_{n} - U_{n - 1} \dots .

U_{0} = 2, U_{1} = a \in Q B y i n d i c t i o n

\forall n \in N, U_{n} \in Q i f a \in Z U_{n} \in Z a l s o

Commented by mr W last updated on 22/Jan/20

t h a n k s s i r!

i t h i n k s o t o o . t h a t m e a n s i n f i n a l

U_{n} = \frac{1}{2^{n}} {{(a + \sqrt{a^{2} - 4})}^{n} + {(a - \sqrt{a^{2} - 4})}^{n}}

t h e t e r m \sqrt{a^{2} - 4} s h o u l d d i s a p p e a r .

h o w c a n t h e t e r m {(a + \sqrt{a^{2} - 4})}^{n} + {(a - \sqrt{a^{2} - 4})}^{n}

b e s i m p l i f i e d s u c h t h a t \sqrt{a^{2} - 4} d i s a p p e a r s ?

Commented by mind is power last updated on 22/Jan/20

y e a h

{(a - b)}^{n} + {(a + b)}^{n}

= \underset{k = 0}{\sum^{n}} C_{n}^{k} {a^{n - k} {(- b)}^{k} + a^{n - k} b^{k}}

= \underset{k = 0}{\sum^{E (\frac{n - 1}{2})}} C_{n}^{2 k + 1} {- a^{n - (2 k + 1)} b^{2 k + 1} + a^{n - (2 k + 1)} b^{2 k + 1}} + \underset{k = 0}{\sum^{E (\frac{n}{2})}} C_{n}^{2 k} {a^{n - 2 k} b^{2 k} + a^{n - 2 k} b^{2 k}}

= \underset{k = 0}{\sum^{E (\frac{n}{2})}} 2 C_{n}^{2 k} a^{n - 2 k} b^{2 k}

{(a + \sqrt{a^{2} - 4})}^{n} + {(a - \sqrt{a^{2} - 4})}^{n}, b = \sqrt{a^{2} - 4}

w e G e t

2 \underset{k = 0}{\sum^{E (\frac{n}{2})}} C_{n}^{2 k} a^{n - 2 k} {(\sqrt{a^{2} - 4})}^{2 k} i f a ⩾ 2

= 2 \underset{k = 0}{\sum^{E (\frac{n}{2})}} C_{n}^{2 k} a^{n - 2 k} {(a^{2} - 4)}^{k}

Commented by mr W last updated on 22/Jan/20

i f a^{2} - 4 < 0, i . e . - 2 < a < 2 :

a \pm \sqrt{a^{2} - 4} = a \pm \sqrt{4 - a^{2}} i = 2 (\frac{a}{2} \pm \frac{\sqrt{4 - a^{2}}}{2} i)

= 2 [\cos (\pm φ) + i \sin (\pm φ)] w i t h φ = \cos^{- 1} \frac{a}{2}

U_{n} = \frac{1}{2^{n}} {{(a + \sqrt{a^{2} - 4})}^{n} + {(a - \sqrt{a^{2} - 4})}^{n}}

= 2 \cos (n φ)

= 2 \cos (n \cos^{- 1} \frac{a}{2})

Commented by mind is power last updated on 22/Jan/20

n i c e s i r

Commented by mr W last updated on 22/Jan/20

i^{'} m w o n d e r i n g t h a t U_{n} i s a l w a y s < 2 i f

a < 2 a n d \cos (n \cos^{- 1} \frac{a}{2}) \in Q i f a \in Q .

Answered by mind is power last updated on 21/Jan/20

l e t U_{n} = x^{n} + \frac{1}{x^{n}}

U_{0} = 2, U_{1} = a

U_{n + 1} = (x + \frac{1}{x}) (x^{n} + \frac{1}{x^{n}}) - x^{n - 1} + \frac{1}{x^{n - 1}}

U_{n + 1} = {a U}_{n} + U_{n - 1}

\Rightarrow U_{n + 1} - {a U}_{n} - U_{n - 1} = 0

X^{2} - a X - 1 = 0

\Rightarrow X_{1} = \frac{a - \sqrt{a^{2} + 4}}{2}, X_{2}^{} = \frac{a + \sqrt{a^{2} + 4}}{2}

U_{n} = c {(X_{1})}^{n} + b {(X_{2})}^{n}

c + b = 2

{c X}_{1} + {b X}_{2} = a

\frac{a}{2} (c + b) + \frac{\sqrt{a^{2} + 4}}{2} (b - c) = a

\Rightarrow b = c

\Rightarrow U_{n} = {(\frac{a - \sqrt{a^{2} + 4}}{2})}^{n} + {(\frac{a + \sqrt{a^{2} + 4}}{2})}^{n}

Commented by mr W last updated on 21/Jan/20

t h a n k s a l o t s i r!

p l e a s e c h e c k t y p o s :

U_{n + 1} = {a U}_{n} - U_{n - 1}

\Rightarrow U_{n} = {(\frac{a - \sqrt{a^{2} - 4}}{2})}^{n} + {(\frac{a + \sqrt{a^{2} - 4}}{2})}^{n}

Commented by mind is power last updated on 21/Jan/20

y e a h S o r r y S i r i m a c k t h i s T y p o f e r r o r