x-7-x-6-x-5-x-4-x-3-x-2-x-1-0-k-1-7-x-k-2-x-k-k-th-root-of-the-equation-x-k-real-part-of-the-root-

Question Number 30849 by Penguin last updated on 27/Feb/18

x^{7} + x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1 = 0

\underset{k = 1}{\sum^{7}} {[ℜ (x_{k})]}^{2} = ?

x_{k} = k^{th} root of the equation

ℜ (x_{k}) = real part of the root

Commented by prof Abdo imad last updated on 27/Feb/18

z r o o t f o r t h i s e q u a t i o n \Leftrightarrow z^{8} = 1 a n d z \neq o

t h e r o o t s o f t h i s e q u a t i o n a r e z_{k} = e^{i \frac{k π}{4}} a n d

k \in [[1, 7]] \Rightarrow R e (z_{k}) = c o s (\frac{k π}{4}) \Rightarrow

{(R e (z_{k}))}^{2} = {c o s}^{2} (\frac{k π}{4}) a n d

\sum_{k = 1}^{7} {(R e (z_{k}))}^{2} = \sum_{k = 1}^{7} \frac{1 + c o s (\frac{k π}{2})}{2}

= \frac{7}{2} + \frac{1}{2} \sum_{k = 1}^{7} c o s (\frac{k π}{2}) b u t

\sum_{k = 1}^{7} c o s (\frac{k π}{2}) = \sum_{k = 0}^{7} c o s (\frac{k π}{2}) - 1

= R e (\sum_{k = 0}^{7} e^{i \frac{k π}{2}}) - 1

R e (\frac{1 - {(e^{i \frac{π}{2}})}^{8}}{1 - e^{i \frac{π}{2}}}) - 1 = 0 - 1 = - 1 \Rightarrow

\sum_{k = 1}^{7} {(R e (z_{k}))}^{2} = \frac{7}{2} - \frac{1}{2} = 3 .

Commented by Penguin last updated on 27/Feb/18

\sum_{k = 1}^{7} {(R e (z_{k}))}^{2} = \sum_{k = 1}^{7} \frac{1 + c o s (\frac{k π}{2})}{2}

how did you get this result ?

Commented by MJS last updated on 27/Feb/18

I tried to solve it, must admit that

this might not always be possible

(but I always love to try \dots)

x_{1} = - 1

x_{2} = - i

x_{3} = i

x_{4} = - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i

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

x_{6} = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i

x_{7} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i

\underset{k = 1}{\sum^{7}} {[ℜ (x_{k})]}^{2} = 3

Commented by rahul 19 last updated on 27/Feb/18

y a h h, s h o r t m e t h o d s!

Commented by abdo imad last updated on 28/Feb/18

l o o k m y p r o o f i h a v e u s e d t h e f o r m u l a {c o s}^{2} α = \frac{1 + c o s (2 α)}{2} .