x + (1/x) =ϕ (Golden ratio) then x^( 2000) +(( 1)/x^( 2000) ) = ?


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Question Number 176554 by mnjuly1970 last updated on 21/Sep/22

$x + \frac{1}{x} = φ (G o l d e n r a t i o)$ $t h e n$ $x^{2000} + \frac{1}{x^{2000}} = ?$
	Answered by Peace last updated on 21/Sep/22
	$U_n =x^n +(1/x^n );ϕU_n =U_(n+1) +U_(n−1) U_(n+1) −ϕU_n +U_(n−1) =0∴ X^2 −ϕX+1=0 X=((ϕ−i(√(4−ϕ^2 )))/2),((ϕ+i(√(4−ϕ^2 )))/2),U_0 =2,U_1 =ϕ U_n =a(((ϕ−i(√(4−ϕ^2 )))/2))^n +b(((ϕ+i(√(4−ϕ^2 )))/2))^n a+b=2,((a+b)/2)ϕ+((i(√(4−ϕ^2 )))/2)(b−a)=ϕ b−a=0⇒a=b=1 ((ϕ+i(√(4−ϕ^2 )))/2)=((ϕ/2)+i(√(1−((ϕ/2))^2 ))) =e^(itg^− ((√(((2/ϕ))^2 −1)))) U_n =2Re{e^(intg^− ((√(((2/ϕ))^2 −1)))) } U_n =2cos(ntg^− (√((4−ϕ^2 )/ϕ^2 )))$
	$U_{n} = x^{n} + \frac{1}{x^{n}}; φ U_{n} = U_{n + 1} + U_{n - 1}$ $U_{n + 1} - φ U_{n} + U_{n - 1} = 0 ∴ X^{2} - φ X + 1 = 0$ $X = \frac{φ - i \sqrt{4 - φ^{2}}}{2}, \frac{φ + i \sqrt{4 - φ^{2}}}{2}, U_{0} = 2, U_{1} = φ$ $U_{n} = a {(\frac{φ - i \sqrt{4 - φ^{2}}}{2})}^{n} + b {(\frac{φ + i \sqrt{4 - φ^{2}}}{2})}^{n}$ $a + b = 2, \frac{a + b}{2} φ + \frac{i \sqrt{4 - φ^{2}}}{2} (b - a) = φ$ $b - a = 0 \Rightarrow a = b = 1$ $\frac{φ + i \sqrt{4 - φ^{2}}}{2} = (\frac{φ}{2} + i \sqrt{1 - {(\frac{φ}{2})}^{2}})$ $= e^{{i t g}^{-} (\sqrt{{(\frac{2}{φ})}^{2} - 1})}$ $U_{n} = 2 R e {e^{{i n t g}^{-} (\sqrt{{(\frac{2}{φ})}^{2} - 1})}}$ $U_{n} = 2 c o s ({n t g}^{-} \sqrt{\frac{4 - φ^{2}}{φ^{2}}})$
	Commented by Tawa11 last updated on 23/Sep/22

	$Great sir$
	Commented by Peace last updated on 25/Sep/22

	$W i t h e p l e a s u r$
	Answered by Frix last updated on 22/Sep/22

	$x + \frac{1}{x} = φ$ $x^{2} - φ x + 1 = 0$ $[x^{5} - 1 = 0$ $(x + 1) (x^{4} - x^{3} + x^{2} - x + 1) = 0$ $(x + 1) (x^{2} - \frac{1 - \sqrt{5}}{2} x + 1) (x^{2} - \frac{1 + \sqrt{5}}{2} + 1) = 0]$ $\Rightarrow$ $x = \cos \frac{π}{5} \pm i \sin \frac{π}{5} = e^{\pm i \frac{π}{5}}$ ${(e^{\pm i \frac{π}{5}})}^{2000} = e^{\pm 400 i π} = e^{\pm i π} = 1$ $\Rightarrow x^{2000} + \frac{1}{x^{2000}} = e^{\pm i π} + e^{\mp i π} = 1 + 1 = 2$
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