If-and-are-the-roots-of-x-2-x-1-0-Find-23-23-without-demoivre-s-theorem-

Question Number 74742 by TawaTawa last updated on 30/Nov/19

If α and β are the roots of x^{2} - x + 1 = 0,

Find α^{23} + β^{23} without {demoivre}^{'} s theorem .

Commented by abdomathmax last updated on 02/Dec/19

x^{2} - x + 1 = 0 \to Δ = 1 - 4 = - 3 \Rightarrow α = \frac{1 + i \sqrt{3}}{2}

a n d β = \frac{1 - i \sqrt{3}}{2} = \bar{α} w e h a v e x^{2} - x + 1 = 0 \Rightarrow

x^{2} = x - 1 \Rightarrow x^{2 n} = {(x - 1)}^{n} \Rightarrow x^{2 n + 1} = x {(x - 1)}^{n} \Rightarrow

α^{23} = α^{2 \times 11 + 1} = α {(α - 1)}^{11} a l s o

β^{23} = {(\bar{α})}^{2 \times 11 + 1} = \bar{α} {(\bar{α} - 1)}^{11} \Rightarrow

α^{23} + β^{23} = α {(α - 1)}^{11} + \bar{α} {(\bar{α} - 1)}^{11}

= 2 R e (α {(α - 1)}^{11}) b u t α {(α - 1)}^{11}

= \frac{1 + i \sqrt{3}}{2} {(\frac{i \sqrt{3} - 1}{2})}^{11} = \frac{1}{2^{12}} (1 + i \sqrt{3}) \sum_{k = 0}^{11} C_{11}^{k} {(i \sqrt{3})}^{k} {(- 1)}^{11 - k}

= \frac{1}{2^{12}} (1 + i \sqrt{3}) {\sum_{p = 0}^{5} C_{11}^{2 p} {(- 1)}^{p} 3^{p} {(- 1)}^{11 - 2 p}

+ \sum_{p = 1}^{5} C_{11}^{2 p + 1} i {(- 1)}^{p} \sqrt{3} \times 3^{p} {(- 1)}^{10 - 2 p}}

r e s t t o e x t r a c t R e s l o f t h i s q u a n t i t y \dots .

Answered by Smail last updated on 30/Nov/19

α = - j a n d β = - j^{2}

{(- j)}^{23} + {(- j^{2})}^{23} = - (j^{23} + (j^{46})

= - (j^{21} j^{2} + {j j}^{45}) = - ({(j^{3})}^{7} j^{2} + j {(j^{3})}^{15})

= - (j^{2} + j) = 1

T h u s, α^{23} + β^{23} = 1 w i t h j = e^{2 i π / 3}

Commented by TawaTawa last updated on 30/Nov/19

God bless you sir

Commented by TawaTawa last updated on 30/Nov/19

But is it possible without complex number and

cube root of unity

Answered by mind is power last updated on 30/Nov/19

(x + 1) (x^{2} - x + 1) = x^{3} + 1 = 0

\Rightarrow α^{3} = β^{3} = - 1

\Rightarrow \forall k \in N α^{3} = β^{3} = {(- 1)}^{k} = 1 if k \equiv 0 (2), - 1 else

23 = 3.7 + 2

\Rightarrow α^{3.7 + 2} + β^{3.7 + 2}

= (- 1) . α^{2} - {(β)}^{2} = - α^{2} - β^{2}

α^{2} = - 1 + α, β = - 1 + β

= - (α + β) + 2

α + β = - \frac{- 1}{1} = 1, som of root

= - 1 + 2 = 1

Commented by TawaTawa last updated on 30/Nov/19

God bless you sir .

Commented by TawaTawa last updated on 30/Nov/19

Can we use it for :

x^{2} - 2 x + 4 = 0

Find α^{23} + β^{23}

Commented by mind is power last updated on 30/Nov/19

\Leftrightarrow \frac{x^{2}}{4} - \frac{x}{2} + 1 = 0

let Y = \frac{x}{2} \Leftrightarrow Y^{2} - Y + 1 = 0 \dots E

root of E are \frac{α}{2}, \frac{β}{2}

by previous Quation

we have {(\frac{α}{2})}^{23} + {(\frac{β}{2})}^{23} = 1

\Rightarrow α^{23} + β^{23} = 2^{23}

Commented by TawaTawa last updated on 30/Nov/19

God bless you sir

Commented by TawaTawa last updated on 30/Nov/19

I appreciate sir .

Commented by TawaTawa last updated on 30/Nov/19

Sir, is there any way this method cannot be used appart

from complex number ?

Or the method can work for any question ? ?

Commented by TawaTawa last updated on 30/Nov/19

Something like : {3 x}^{2} + 5 x + 7 = 0

Find α^{23} + β^{23}

Commented by mind is power last updated on 30/Nov/19

in this you have too solve Equation and

use complex anslysis

previous one our equation \Rightarrow X^{3} = - 1 witch redious calclus

ther {3 x}^{2} + 5 x + 7 = 0

\Rightarrow x = \frac{- 5 - i \sqrt{59}}{6}, x_{2} = \frac{- 5 + i \sqrt{59}}{6}, isee no other methode i will try later

Commented by TawaTawa last updated on 30/Nov/19

I can use complex number

Help me find other method sir

Commented by TawaTawa last updated on 30/Nov/19

God bless you sir