calculate-k-0-n-1-k-k-with-root-of-x-2-2x-5-0-

Question Number 76514 by mathmax by abdo last updated on 28/Dec/19

c a l c u l a t e \prod_{k = 0}^{n - 1} (α^{k} + \overset{-^{k}}{α}) w i t h α r o o t o f x^{2} + 2 x + 5 = 0

Commented by mathmax by abdo last updated on 28/Dec/19

l e t A_{n} = \prod_{k = 0}^{n - 1} (α^{k} + \overset{-^{k}}{α}) l e t s o l v e x^{2} + 2 x + 5 = 0

Δ^{^{'}} = 1 - 5 = - 4 = {(2 i)}^{2} / \Rightarrow α = - 1 + 2 i a n d \bar{α} = - 1 - 2 i

α = \sqrt{5} e^{- i a r c t a n (2)} a n d \bar{α} = \sqrt{5} e^{i a r c t a n (2)} \Rightarrow

α^{k} + \overset{-^{k}}{α} = {(\sqrt{5})}^{k} e^{- i k a r c t a n (2)} + {(\sqrt{5})}^{k} e^{i k a r c t a n (2)}

= {(\sqrt{5})}^{k} (2 c o s (k a r c t a n 2) \Rightarrow A_{n} = \prod_{k = 0}^{n - 1} 2 {(\sqrt{5})}^{k} c o s (k a r c t a n (2))

= 2^{n} \prod_{k = 0}^{n - 1} {(\sqrt{5})}^{k} \prod_{k = 0}^{n - 1} c o s (k a r c t a n (2))

= 2^{n} {(\sqrt{5})}^{\sum_{k = 0}^{n - 1} k} \prod_{k = 0}^{n - 1} c o s (k a r c t a n (2))

= 2^{n} {(\sqrt{5})}^{\frac{(n - 1) n}{2}} \prod_{k = 0}^{n - 1} c o s (k a r c t a n (2))

Answered by john santu last updated on 28/Dec/19

{(x + 1)}^{2} + 4 = 0 \to α = 2 i - 1

α + α^{- 1} = 2 i - 1 + \frac{1}{2 i - 1} = 2 i - 1 + \frac{- 2 i - 1}{(2 i - 1) (- 2 i - 1)}

= \frac{10 i - 5 - 2 i - 1}{5} = \frac{8 i - 6}{5}

α^{2} + α^{- 2} = {(\frac{8 i - 6}{5})}^{2} - 2