decompose-inside-R-x-p-x-x-2n-1-1-then-find-k-1-n-sin-kpi-2n-1-

Question Number 30743 by abdo imad last updated on 25/Feb/18

d e c o m p o s e i n s i d e R [x] p (x) = x^{2 n + 1} - 1 t h e n f i n d

\prod_{k = 1}^{n} s i n (\frac{k π}{2 n + 1}) .

Commented by abdo imad last updated on 02/Mar/18

t h e r o o t s o f p (x) ? z^{2 n + 1} - 1 = 0 \Leftrightarrow z^{2 n + 1} = e^{i (2 k π)}

s o t h e r o o t s a r e z_{k} = e^{i \frac{2 k π}{2 n + 1}} w i t h k \in [[0, 2 n]] \Rightarrow

p (x) = \prod_{k = 0}^{2 n} (x - z_{k}) = (x - 1) \prod_{k = 1}^{2 n} (x - e^{i \frac{2 k π}{2 n + 1}}) \Rightarrow

\frac{p (x)}{x - 1} = \prod_{k = 1}^{2 n} (x - e^{i \frac{2 k π}{2 n + 1}}) \Rightarrow {l i m}_{x \to 1} \frac{p (x)}{x - 1} = \prod_{k = 1}^{2 n} (1 - e^{i \frac{2 k π}{2 n + 1}})

\Rightarrow (2 n + 1) = \prod_{k = 1}^{2 n} (1 - c o s (\frac{2 k π}{2 n + 1}) - i s i n (\frac{2 k π}{2 n + 1}))

= \prod_{k = 1}^{2 n} (2 {s i n}^{2} (\frac{k π}{2 n + 1}) - 2 i s i n (\frac{k π}{2 n + 1}) c o s (\frac{k π}{2 n + 1}))

= 2^{2 n} {(- 1)}^{n} \prod_{k = 1}^{2 n} s i n (\frac{k π}{2 n + 1}) e^{i (\frac{k π}{2 n + 1})}

= 2^{2 n} {(- i)}^{n} (\prod_{k = 1}^{2 n} s i n (\frac{k π}{2 n + 1})) e^{i \frac{π}{2 n + 1} \frac{2 n (2 n + 1)}{2}}

= 2^{2 n} \prod_{k = 1}^{2 n} s i n (\frac{k π}{2 n + 1}) b u t

\prod_{k = 1}^{2 n} s i n (\frac{k π}{2 n + 1}) = \prod_{k = 1}^{n} s i n (\frac{k π}{2 n + 1}) \prod_{k = n + 1}^{2 n} s i n (\frac{k π}{2 n + 1}) t h e

c h . k - n = j g i v e

\prod_{k = n + 1}^{2 n} s i n (\frac{k π}{2 n + 1}) = \prod_{j = 1}^{n} s i n (\frac{(n + j) π}{2 n + 1})

= \prod_{j = 1}^{n} s i n (π - \frac{n π + j π}{2 n + 1}) = \prod_{j = 1}^{n} s i n (\frac{2 n π + π - n π - j π}{2 n + 1})

= \prod_{j = 1}^{n} s i n (\frac{(n + 1 - j) π}{2 n + 1}) = \prod_{k = 1}^{n} s i n (\frac{k π}{2 n + 1}) b y c h . n + 1 - j = k \Rightarrow

2 n + 1 = {(2^{n})}^{2} {(\prod_{k = 1}^{n} s i n (\frac{k π}{2 n + 1}))}^{2} \Rightarrow

\prod_{k = 1}^{n} s i n (\frac{k π}{2 n + 1}) = \frac{\sqrt{2 n + 1}}{2^{n}} .