factorise-p-x-1-x-x-2-x-3-x-5-inside-C-x-and-R-x-calculate-p-e-i-pi-5-and-p-cos-pi-5-

Question Number 67234 by prof Abdo imad last updated on 24/Aug/19

f a c t o r i s e p (x) = 1 + x + x^{2} + x^{3} + x^{5}

i n s i d e C [x] a n d R [x]

c a l c u l a t e p (e^{i \frac{π}{5}}) a n d p (c o s (\frac{π}{5}))

Commented by mathmax by abdo last updated on 24/Aug/19

s o r r y p (x) = 1 + x + x^{2} + x^{3} + x^{4}

Commented by mathmax by abdo last updated on 25/Aug/19

1) p (x) = 0 \Leftrightarrow 1 + x + x^{2} + x^{3} + x^{4} = 0 \Leftrightarrow 1 - x^{5} = 0 a n d x \neq 1

x^{5} = 1 l e t x = {r e}^{i θ} (e) \Rightarrow r^{5} e^{i 5 θ} = e^{i (2 k π)} \Rightarrow r = 1 a n d 5 θ = 2 k π \Rightarrow

θ_{k} = \frac{2 k π}{5} s o t h e r o o t s a r e z_{k} = e^{i (\frac{2 k π}{5})} w i t h k \in [[1, 4]]

p (x) = \prod_{k = 1}^{4} (x - z_{k}) = \prod_{k = 1}^{4} (x - e^{i \frac{2 k π}{5}}) = (x - e^{i \frac{2 π}{5}}) (x - e^{i \frac{4 π}{5}}) (x - e^{i \frac{6 π}{5}}) (x - e^{i \frac{8 π}{5}})

b u t e^{i \frac{8 π}{5}} = e^{i (2 π - \frac{2 π}{5})} = e^{- \frac{i 2 π}{5}} a n d e^{i \frac{6 π}{5}} = e^{- i \frac{4 π}{5}} \Rightarrow

p (x) = (x - e^{i \frac{2 π}{5}}) (x - e^{- i \frac{2 π}{5}}) (x - e^{i \frac{4 π}{5}}) (x - e^{- i \frac{4 π}{5}})

= (x^{2} - 2 R e (e^{i \frac{2 π}{5}}) x + 1) (x^{2} - 2 R e (e^{i \frac{4 π}{5}}) x + 1) i s t h e d e c o m p o s i t i o n

i n s i d e C [x]

w e h a v e c o s (\frac{π}{5}) = \frac{1 + \sqrt{5}}{4} \Rightarrow c o s (\frac{2 π}{5}) = 2 {c o s}^{2} (\frac{π}{5}) - 1

= 2 {(\frac{1 + \sqrt{5}}{4})}^{2} - 1 = \frac{1}{8} (6 + 2 \sqrt{5}) - 1 = \frac{6 + 2 \sqrt{5} - 8}{8} = \frac{- 2 + 2 \sqrt{5}}{8} = \frac{\sqrt{5} - 1}{4}

c o s (\frac{4 π}{5}) = c o s (π - \frac{π}{5}) = - c o s (\frac{π}{5}) \Rightarrow

p (x) = (x^{2} - \frac{\sqrt{5} - 1}{2} x + 1) (x^{2} + \frac{\sqrt{5} - 1}{2} x + 1)

Commented by mathmax by abdo last updated on 25/Aug/19

w e h a v e f o r x \neq 1 p (x) = \frac{1 - x^{5}}{1 - x} \Rightarrow p (e^{i \frac{π}{5}}) = \frac{1 - {(e^{\frac{i π}{5}})}^{5}}{1 - e^{i \frac{π}{5}}}

= \frac{2}{1 - e^{\frac{i π}{5}}} a l s o w e h a v e p (c o s (\frac{π}{5})) = \frac{1 - {c o s}^{5} (\frac{π}{5})}{1 - c o s (\frac{π}{5})}

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

Answered by mind is power last updated on 24/Aug/19

p (x) = 1 + x + x^{2} + x^{3} + x^{4} ?

Answered by MJS last updated on 25/Aug/19

(x^{4} + x^{3} + x^{2} + x + 1) (x - 1) = x^{5} - 1

x^{5} = 1

x_{0} = 1

x_{1} = - \frac{1 - \sqrt{5}}{4} + \frac{\sqrt{10 + 2 \sqrt{5}}}{4} i

x_{2} = - \frac{1 + \sqrt{5}}{4} + \frac{\sqrt{10 - 2 \sqrt{5}}}{4} i

x_{3} = - \frac{1 + \sqrt{5}}{4} - \frac{\sqrt{10 - 2 \sqrt{5}}}{4} i

x_{4} = - \frac{1 - \sqrt{5}}{4} - \frac{\sqrt{10 + 2 \sqrt{5}}}{4} i

\Rightarrow

x^{4} + x^{3} + x^{2} + x + 1 =

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

= (x + \frac{1 - \sqrt{5}}{4} - \frac{\sqrt{10 + 2 \sqrt{5}}}{4} i) (x + \frac{1 - \sqrt{5}}{4} + \frac{\sqrt{10 + 2 \sqrt{5}}}{4} i) (x + \frac{1 + \sqrt{5}}{4} - \frac{\sqrt{10 - 2 \sqrt{5}}}{4} i) (x + \frac{1 + \sqrt{5}}{4} + \frac{\sqrt{10 - 2 \sqrt{5}}}{4} i)

p (e^{i \frac{π}{5}}) = 1 + \sqrt{5 + 2 \sqrt{5}} i

p (\cos \frac{π}{5}) = \frac{67}{32} + \frac{19 \sqrt{5}}{32}

Commented by mathmax by abdo last updated on 25/Aug/19

t h a n k s s i r .