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Question Number 125760 by snipers237 last updated on 13/Dec/20

L e t n ⩾ 1 a n d i n t e g e r, P_{n} (X) = {(1 + X)}^{n} - {(1 - X)}^{n}

1) F a c t o r i z e P_{n}

2) D e d u c e S_{n} = \underset{k = 1}{\prod^{n}} [4 + {c o t a n}^{2} (\frac{k π}{2 n + 1})]

Answered by mathmax by abdo last updated on 13/Dec/20

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

1 - x = e^{\frac{i 2 k π}{n}} + e^{\frac{i 2 k π}{n}} x \Rightarrow (1 + e^{\frac{i 2 k π}{n}}) x = 1 - e^{\frac{i 2 k π}{n}} \Rightarrow so the roots are

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

= \frac{{2 \sin}^{2} (\frac{k π}{n}) - 2 isin (\frac{k π}{n}) \cos (\frac{k π}{n})}{{2 \cos}^{2} (\frac{k π}{n}) + 2 isin (\frac{k π}{n}) \cos (\frac{k π}{n})} = \frac{- isin (\frac{k π}{n}) e^{\frac{ik π}{n}}}{\cos (\frac{k π}{n}) e^{\frac{ik π}{n}}} = - itan (\frac{k π}{n})

k \in [[o, n - 1]] \Rightarrow P_{n} (x) = λ \prod_{k = 0}^{n - 1} (x + itan (\frac{k π}{n})) let find λ

we have P_{n} (x) = {(x + 1)}^{n} - {(- 1)}^{n} {(x - 1)}^{n}

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

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

= \sum_{k = 0}^{n} C_{n}^{k} (1 - {(- 1)}^{k}) x^{k} = 2 \sum_{p = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 p + 1} x^{2 p + 1}

\Rightarrow λ = 2 C_{n}^{2 [\frac{n - 1}{2}] + 1} \Rightarrow P_{n} (x) = 2 C_{n}^{2 [\frac{n - 1}{2}] + 1} \prod_{k = 0}^{n - 1} (x + itan (\frac{k π}{n}))

= 2 x C_{n}^{2 [\frac{n - 1}{2}] + 1} \prod_{k = 1}^{n - 1} (x + itan (\frac{k π}{n}))

P_{n} (2) . P_{n} (- 2) = - 16 λ_{n}^{2} \prod_{k = 1}^{n} (2 + itan (\frac{k π}{n})) \prod_{k = 1}^{n} (- 2 + itan (\frac{k π}{n}))

= - 16 λ_{n}^{2} {(- 1)}^{n} \prod_{k = 1}^{n} (2 + itan (\frac{k π}{n})) (2 - itan (\frac{k π}{n}))

= 16 λ_{n}^{2} {(- 1)}^{n + 1} \prod_{k = 1}^{n} (4 + \tan^{2} (\frac{k π}{n})) \dots be continued \dots

Answered by mindispower last updated on 13/Dec/20

P_{n} (X) = 0

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

\frac{1 + x}{1 - x} = e^{2 i k π / n}

k \in [0, n [

x = \frac{1 - e^{\frac{2 i k π}{n}}}{1 + e^{\frac{2 i k π}{n}}} = - \frac{e^{\frac{i k π}{n}} - e^{- \frac{i k π}{n}}}{e^{\frac{i k π}{n}} + e^{- i \frac{k π}{n}}} = - i t g (\frac{k π}{n})

k \in [0, n - 1]

i f n = 2 m, k \neq m

P_{n} (x) = a \prod_{k \in [0, n - 1]} (X + i t g (\frac{k π}{n})) i f n = 2 m + 1

= \prod_{k \in [0, n - 1] \neq \frac{n}{2}} (X + i t g (\frac{k π}{n})), n = 2 m

o n e e a s y w a y t o o s e e t h i s

i f n = 2 m, {d e g p}_{2 m} ⩽ 2 m - 1 n u m b e r o f r o o t s ⩽ 2 m - 1

2)

P_{2 n + 1} = 2 \prod_{0 ⩽ k ⩽ 2 n} (X + i t g (\frac{k π}{2 n + 1}))

P_{2 n + 1} = 2 x . \underset{k = 1}{\prod^{n}} (X + i t g (\frac{k π}{2 n + 1}) . \underset{k = n + 1}{\prod^{2 n}} (X + i t g (\frac{k π}{2 n + 1}))

= 2 x \underset{k = 1}{\prod^{n}} (X + i t g (\frac{k π}{2 n + 1})) \underset{k = 1}{\prod^{n}} (X + i t g (\frac{(2 n + 1 - k) π}{2 n + 1}))

= 2 x \underset{k = 1}{\prod^{n}} (X + i t g (\frac{k π}{2 n + 1})) \underset{k = 1}{\prod^{n}} (X + i t g (π - \frac{k π}{2 n + 1}))

= 2 x \underset{k = 1}{\prod^{n}} (X + i t g (\frac{k π}{2 n + 1})) \underset{k = 1}{\prod^{n}} (X - i t g (\frac{k π}{2 n + 1}))

= 2 x \underset{k = 1}{\prod^{n}} (X^{2} + {t g}^{2} (\frac{k π}{2 n + 1})) = p_{2 n + 1} (x)

x = \frac{1}{2}

\Rightarrow p_{2 n + 1} (\frac{1}{2}) = \underset{k = 1}{\prod^{n}} (\frac{1}{4} + \frac{1}{{c o t a n}^{2} (\frac{k π}{2 n + 1})}) . . E

Missing \left or extra \right

T_{n} = \frac{1}{4^{n}} \underset{k = 1}{\prod^{n}} {t g}^{2} (\frac{k π}{2 n + 1})

2 x \underset{k = 1}{\prod^{n}} (X^{2} + {t g}^{2} (\frac{k π}{2 n + 1})) = p_{2 n + 1} (x)

\Rightarrow \underset{k = 1}{\prod^{n}} (X^{2} + {t g}^{2} (\frac{k π}{2 n + 1})) = \frac{P_{2 n + 1} (x)}{2 x}

\Rightarrow \lim_{x \to 0} Π (X^{2} + {t g}^{2} (\frac{k π}{2 n + 1}) = \frac{1}{2} . \lim_{x \to 0} \frac{P_{2 n + 1} (x)}{x}

= \frac{1}{2} P_{2 n + 1}^{^{'}} (0) = \frac{1}{2} ((2 n + 1) + (2 n + 1)) = 2 n + 1

\Rightarrow \underset{k = 1}{\prod^{n}} {t g}^{2} (\frac{k π}{2 n + 1}) = 2 n + 1

T_{n} = \frac{1}{4^{n} (2 n + 1)}

E \Leftrightarrow P_{2 n + 1} (\frac{1}{2}) = \frac{1}{4^{n} (2 n + 1)} . S_{n} = {(\frac{3}{2})}^{n} - \frac{1}{2^{n}}

S_{n} = (2 n + 1) (6^{n} - 2^{n})