let-P-n-x-x-1-n-x-1-n-1-fartorize-inside-C-x-P-n-x-2-calculate-k-1-p-cotan-kpi-2p-1-

Question Number 73059 by mathmax by abdo last updated on 05/Nov/19

l e t P_{n} (x) = {(x + 1)}^{n} - {(x - 1)}^{n}

1) f a r t o r i z e i n s i d e C (x) P_{n} (x)

2) c a l c u l a t e \prod_{k = 1}^{p} c o t a n (\frac{k π}{2 p + 1})

Commented by mathmax by abdo last updated on 06/Nov/19

1) P_{n} (x) = 0 \Leftrightarrow {(x + 1)}^{n} = {(x - 1)}^{n} \Leftrightarrow {(\frac{x - 1}{x + 1})}^{n} = 1 l e t z = \frac{x - 1}{x + 1}

(e) \Rightarrow z^{n} = 1 \Rightarrow z^{n} = e^{i 2 k π} \Rightarrow z_{k} = e^{\frac{i 2 k π}{n}} a n d k \in [[0, n - 1]]

z = \frac{x - 1}{x + 1} \Rightarrow z x + z = x - 1 \Rightarrow (z - 1) x = - z - 1 \Rightarrow x = \frac{1 + z}{1 - z} \Rightarrow

t h e r o o t s o f P_{n} (x) = 0 a r e x_{k} = \frac{1 + z_{k}}{1 - z_{k}} = \frac{1 + e^{\frac{i 2 k π}{n}}}{1 - e^{\frac{i 2 k π}{n}}}

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

= \frac{c o s (\frac{k π}{n}) e^{i \frac{k π}{n}}}{- i s i n (\frac{k π}{n}) e^{\frac{i k π}{n}}} = i c o t a n (\frac{k π}{n}) \Rightarrow x_{k} = i c o t a n (\frac{k π}{n})

w i t h k \in [[1, n - 1]] a n d P_{n} (x) = a \prod_{k = 1}^{n - 1} (x - i c o t a n (\frac{k π}{n}))

l e t f i n d a w e h a v e P_{n} (x) = \sum_{k = 0}^{n} C_{n}^{k} x^{k} - \sum_{k = 0}^{n} C_{n}^{k} x^{k} {(- 1)}^{n - k}

= \sum_{k = 0}^{n} {1 - {(- 1)}^{n - k}} C_{n}^{k} x^{k} \Rightarrow a = 2 C_{n}^{n - 1} = 2 n \Rightarrow

P_{n} (x) = 2 n \prod_{k = 1}^{n - 1} (x - i c o t a n (\frac{k π}{n}))

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

P_{2 n + 1} (0) = 2 (2 n + 1) \prod_{k = 1}^{2 n} (- i c o t a n (\frac{k π}{2 n + 1}))

= 2 (2 n + 1) {(- i)}^{2 n} \prod_{k = 1}^{2 n} c o t a n (\frac{k π}{2 n + 1}) a n d P_{n} (0) = 1 - {(- 1)}^{n}

\Rightarrow P_{2 n + 1} (0) = 2 \Rightarrow (2 n + 1) {(- 1)}^{n} \prod_{k = 1}^{2 n} c o t a n (\frac{k π}{2 n + 1}) = 1 \Rightarrow

{(- 1)}^{n} \prod_{k = 1}^{2 n} c o t a n (\frac{k π}{2 n + 1}) = \frac{1}{2 n + 1} w e h a v e

\prod_{k = 1}^{2 n} c o t a n (\frac{k π}{2 n + 1}) = \prod_{k = 1}^{n} c o t a n (\frac{k π}{2 n + 1}) \prod_{k = n + 1}^{2 n} c o t a n (\frac{k π}{2 n + 1})

=_{k - n = p} \prod_{k = 1}^{n} c o t a n (\frac{k π}{2 n + 1}) \times \prod_{p = 1}^{n} c o t a n (\frac{(n + p) π}{2 n + 1})

r e s t t o p r o v e t h s t \prod_{p = 1}^{n} c o t a n (\frac{(n + p) π}{2 n + 1}) = {(- 1)}^{n} \prod_{p = 1}^{n} c o t a n (\frac{p π}{2 n + 1})

\Rightarrow {(\prod_{k = 1}^{n} c o t a n (\frac{k π}{2 n + 1}))}^{2} = \frac{1}{2 n + 1} \Rightarrow \prod_{k = 1}^{n} c o t a n (\frac{k π}{2 n + 1}) = \frac{1}{\sqrt{2 n + 1}}

Commented by mind is power last updated on 06/Nov/19

nice

Commented by mathmax by abdo last updated on 06/Nov/19

t h a n k s s i r .

Answered by mind is power last updated on 06/Nov/19

p_{n} (x) = 0 \Rightarrow \frac{x + 1}{x - 1} = e^{\frac{2 ik π}{n}}, k ⩽ n

\Rightarrow x (1 - e^{\frac{2 ik π}{n}}) = - e^{\frac{2 ik π}{n}} - 1

\Rightarrow k \neq 0

x = \frac{e^{\frac{2 ik π}{n}} + 1}{e^{\frac{2 ik π}{n}} - 1} = \frac{e^{i \frac{k π}{n}} (2 \cos (\frac{k π}{n}))}{e^{i \frac{k π}{n}} (2 isin (\frac{k π}{n}))} = - icot (\frac{k π}{n})

p_{n} (x) = a \underset{k = 1}{\prod^{n - 1}} (x + icot (\frac{k π}{n})

let n = 2 p + 1

p_{n} (x) = \underset{k = 1}{\prod^{2 p}} (x + icot (\frac{k π}{2 p + 1}))

= \underset{k = 1}{\prod^{p}} (x + icot (\frac{k π}{2 p + 1})) . \underset{k = p + 1}{\prod^{2 p}} (x + icot (\frac{k π}{2 p + 1})

= \underset{k = 1}{\prod^{p}} (x + icot (\frac{k π}{2 p + 1})) . \underset{k = 0}{\prod^{p - 1}} (x + icot (\frac{2 p - k) π}{2 p + 1}))

= \underset{k = 1}{\prod^{p}} (x + icot (\frac{k π}{2 p + 1})) \underset{k = 0}{\prod^{p - 1}} (x - icot (\frac{k + 1}{2 p + 1}) π) = {(1 + x)}^{n} - {(x - 1)}^{n}

pour x = 0

ona \underset{k = 1}{\prod^{p}} (icot (\frac{k π}{2 p + 1})) (\underset{k = 1}{\prod^{p}} - icot (\frac{k π}{2 p + 1})) =

\Rightarrow {\underset{k = 1}{\prod^{p}} \cot (\frac{k π}{2 p + 1})}^{2} = \frac{2}{a}

\Rightarrow \underset{k = 1}{\prod^{p}} \cot (\frac{k π}{2 p + 1}) = \sqrt{\frac{2}{a}} = \frac{1}{\sqrt{2 p + 1}}

a = {2 C}_{n}^{1} = 2 n = 4 p + 2

cause \forall k \in [1, n] 0 < \frac{k π}{2 p + 1} < \frac{π}{2} \Rightarrow \cot (\frac{k π}{2 p + 1}) > 0