if x+(1/x)=a find x^n +(1/x^n )=?


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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
$thank you sir! U_n =(1/2^n ){(a+(√(a^2 −4)))^n +(a−(√(a^2 −4)))^n } can the term (a+(√(a^2 −4)))^n +(a−(√(a^2 −4)))^n be simplified more? if a∈Q, is slso U_n ∈Q?$
$t h a n k y o u s i r!$ $U_{n} = \frac{1}{2^{n}} {{(a + \sqrt{a^{2} - 4})}^{n} + {(a - \sqrt{a^{2} - 4})}^{n}}$ $c a n t h e t e r m {(a + \sqrt{a^{2} - 4})}^{n} + {(a - \sqrt{a^{2} - 4})}^{n}$ $b e s i m p l i f i e d m o r e ?$ $i f a \in Q, i s s l s o U_{n} \in Q ?$
Commented by mind is power last updated on 22/Jan/20

$y e a h a \in Q, U n \in Q$ $b e c a u s e$ $U_{n + 1} = {a U}_{n} - U_{n - 1} . . . .$ $U_{0} = 2, U_{1} = a \in Q B y i n d i c t i o n$ $\forall n \in N, U_{n} \in Q i f a \in Z U_{n} \in Z a l s o$
Commented by mr W last updated on 22/Jan/20
$thanks sir! i think so too. that means in final U_n =(1/2^n ){(a+(√(a^2 −4)))^n +(a−(√(a^2 −4)))^n } the term (√(a^2 −4)) should disappear. how can the term (a+(√(a^2 −4)))^n +(a−(√(a^2 −4)))^n be simplified such that (√(a^2 −4)) disappears?$
$t h a n k s s i r!$ $i t h i n k s o t o o . t h a t m e a n s i n f i n a l$ $U_{n} = \frac{1}{2^{n}} {{(a + \sqrt{a^{2} - 4})}^{n} + {(a - \sqrt{a^{2} - 4})}^{n}}$ $t h e t e r m \sqrt{a^{2} - 4} s h o u l d d i s a p p e a r .$ $h o w c a n t h e t e r m {(a + \sqrt{a^{2} - 4})}^{n} + {(a - \sqrt{a^{2} - 4})}^{n}$ $b e s i m p l i f i e d s u c h t h a t \sqrt{a^{2} - 4} d i s a p p e a r s ?$
Commented by mind is power last updated on 22/Jan/20
$yeah (a−b)^n +(a+b)^n =Σ_(k=0) ^n C_n ^k {a^(n−k) (−b)^k +a^(n−k) b^k } =Σ_(k=0) ^(E(((n−1)/2))) C_n ^(2k+1) {−a^(n−(2k+1)) b^(2k+1) +a^(n−(2k+1)) b^(2k+1) }+Σ_(k=0) ^(E((n/2))) C_n ^(2k) {a^(n−2k) b^(2k) +a^(n−2k) b^(2k) } =Σ_(k=0) ^(E((n/2))) 2C_n ^(2k) a^(n−2k) b^(2k) (a+(√(a^2 −4)))^n +(a−(√(a^2 −4)))^n ,b=(√(a^2 −4)) we Get 2Σ_(k=0) ^(E((n/2))) C_n ^(2k) a^(n−2k) ((√(a^2 −4)))^(2k) ifa≥2 =2Σ_(k=0) ^(E((n/2))) C_n ^(2k) a^(n−2k) (a^2 −4)^k$
$y e a h$ ${(a - b)}^{n} + {(a + b)}^{n}$ $= \underset{k = 0}{\sum^{n}} C_{n}^{k} {a^{n - k} {(- b)}^{k} + a^{n - k} b^{k}}$ $= \underset{k = 0}{\sum^{E (\frac{n - 1}{2})}} C_{n}^{2 k + 1} {- a^{n - (2 k + 1)} b^{2 k + 1} + a^{n - (2 k + 1)} b^{2 k + 1}} + \underset{k = 0}{\sum^{E (\frac{n}{2})}} C_{n}^{2 k} {a^{n - 2 k} b^{2 k} + a^{n - 2 k} b^{2 k}}$ $= \underset{k = 0}{\sum^{E (\frac{n}{2})}} 2 C_{n}^{2 k} a^{n - 2 k} b^{2 k}$ ${(a + \sqrt{a^{2} - 4})}^{n} + {(a - \sqrt{a^{2} - 4})}^{n}, b = \sqrt{a^{2} - 4}$ $w e G e t$ $2 \underset{k = 0}{\sum^{E (\frac{n}{2})}} C_{n}^{2 k} a^{n - 2 k} {(\sqrt{a^{2} - 4})}^{2 k} i f a ⩾ 2$ $= 2 \underset{k = 0}{\sum^{E (\frac{n}{2})}} C_{n}^{2 k} a^{n - 2 k} {(a^{2} - 4)}^{k}$
Commented by mr W last updated on 22/Jan/20
$if a^2 −4<0, i.e. −2<a<2: a±(√(a^2 −4))=a±(√(4−a^2 ))i=2((a/2)±((√(4−a^2 ))/2)i) =2[cos (±ϕ)+i sin (±ϕ)] with ϕ=cos^(−1) (a/2) U_n =(1/2^n ){(a+(√(a^2 −4)))^n +(a−(√(a^2 −4)))^n } =2 cos (nϕ) =2 cos (n cos^(−1) (a/2))$
$i f a^{2} - 4 < 0, i . e . - 2 < a < 2 :$ $a \pm \sqrt{a^{2} - 4} = a \pm \sqrt{4 - a^{2}} i = 2 (\frac{a}{2} \pm \frac{\sqrt{4 - a^{2}}}{2} i)$ $= 2 [\cos (\pm φ) + i \sin (\pm φ)] w i t h φ = \cos^{- 1} \frac{a}{2}$ $U_{n} = \frac{1}{2^{n}} {{(a + \sqrt{a^{2} - 4})}^{n} + {(a - \sqrt{a^{2} - 4})}^{n}}$ $= 2 \cos (n φ)$ $= 2 \cos (n \cos^{- 1} \frac{a}{2})$
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

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