Prove that ∀n∈IN^∗ Σ_(k=1) ^(2^n −1) (1/(sin^2 (((kπ)/2^(n+1) ))))= ((2^(2n+1) −2)/3) Give in terms of n Σ_(k=1) ^(2^n −1) (1/(sin^4 (((kπ)/2^(n+1) ))))


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Question Number 194638 by Erico last updated on 12/Jul/23

$Prove that \forall n \in {IN}^{*}$ $\underset{k = 1}{\sum^{2^{n} - 1}} \frac{1}{{s i n}^{2} (\frac{k π}{2^{n + 1}})} = \frac{2^{2 n + 1} - 2}{3}$ $Give in terms of n \underset{k = 1}{\sum^{2^{n} - 1}} \frac{1}{{s i n}^{4} (\frac{k π}{2^{n + 1}})}$
Commented by witcher3 last updated on 12/Jul/23

$\frac{1}{\sin^{4} (x)} = \frac{{(\sin^{2} (x) + \cos^{2} (x))}^{2}}{\sin^{4} (x)} = (1 + \frac{2}{{tg}^{2} (x)} + \frac{1}{{tg}^{4} (x)})$ $Σ \frac{1}{{tg}^{2} (x)} . . known By 1 st Quation$ $Σ {tg}^{4} (x) . . . withe same methode Polynomial roots$ $p_{4} = Σ x_{i}^{4} . . . use newtoon identity$
	Answered by witcher3 last updated on 12/Jul/23
	$Σ_(k=1) ^(2^n −1) (1/(sin^2 (((kπ)/2^(n+1) ))))=Σ_(k=1) ^(2^n −1) (1+(1/(tg^2 (((kπ)/2^(n+1) ))))),k→2^n −k =Σ_(k=1) ^(2^n −1) (1+tg^2 (((kπ)/2^(n+1) ))),⇔Σ_(k=1) ^2^(n−1) tg^2 (((kπ)/2^(n+1) ))=Σ_(k=1) ^(2^n −1) (1/(tg^2 (((kπ)/2^(n+1) ))))...I Let S=Σ_(k=1,k≠2^n ) ^(2^(n+1) −1) tg^2 (((kπ)/2^(n+1) ))=Σ_(k=1) ^(2^n −1) tg^2 (((kπ)/2^(n+1) ))+Σ_(2^n +1) ^(2^n −1) tg^2 (((kπ)/2^(n+1) )) for the 2nd Sum k→2^n +k ⇒S=Σ_(k=1) ^(2^n −1) tg^2 (((kπ)/2^(n+1) ))+tg^2 (((2^n +k)/2^(n+1) ))=Σ_(k=1) ^(2^n −1) tg^2 (((kπ)/2^(n+1) ))+(1/(tg^2 (((kπ)/2^(n+1) )))) ⇒S=2Σ_(k=1) ^(2^n −1) tg^2 (((kπ)/2^(n+1) ))...A=[1,2^(n+1) −1]−{2^n } let′s find polynonial withe Roots tg(((kπ)/2^(n+1) )),k∈A tg(2^(n+1) x)=0⇔2^(n+1) x=kπ⇔x=((kπ)/2^(n+1) ),k∈A∪{2^n ,0} if we tack,((tg(2^(n+1) x))/(tg(x)))=0⇒x=((kπ)/2^(n+1) ),k∈A we Know tg(mx)=((Σ_(k=0) ^([((m−1)/2)]) (−1)^k ((m),((2k+1)) )tg^(2k+1) (x))/(Σ_(k=0) ^([(m/2)]) (−1)^k ((m),((2k)) ).tg^(2k) (x)_ )) ⇒((tg(2^(n+1) x))/(tg(x)))=0⇔Σ_(k=0) ^(2^n −1) (−1)^k ((2^(n+1) ),((2k+1)) )tg^(2k) (x)=p(tg(x)) Σ_(k=0) ^(2^n −1) (−1)^k ((2^(n+1) ),((2k+1)) )t^(2k) =p(t),deg(p)=2^(n+1) −2 and root of P.are tan(((kπ)/2^(n+1) )),k∈A card (A)=2^(n+1) −2 so we have all root of P if f(x)=Σ_(k=0) ^n a_k x^k ;x_i root of P Σx_i =−(a_(n−1) /a_n ),Σ_(1≤i<j≤n) x_i x_j =(a_(n−2) /a_n ) Σ_(k∈A) tg(((kπ)/2^(n+1) ))=0 Σ_(k_1 <k_2 ,(k_1 ,k_2 )∈A^2 ) tg(((k_1 π)/2^(n+1) ))tg(((k_2 π)/2^(n+1) ))=−( ((2^(n+1) ),((2^(n+1) −3)) )/ ((2^(n+1) ),((2^(n+1) −1)) )) =−(1/6)(2^(n+1) −1)(2^(n+1) −2) Σx_i ^2 =(Σx_i )^2 −2Σx_i x_j =(1/3)(2^(n+1) −1)(2^(n+1) −2) S=2Σ_(k=1) ^(2^n −1) tg^2 (((kπ)/2^(n+1) )) Σ_(k=1) ^(2^n −1) tg^2 (((kπ)/2^(n+1) ))=(1/6)(2^(2n+2) −3.2^(n+1) +2)=((2^(2n+1) −3.2^n +1)/3) S_1 =Σ_(k=1) ^(2^n −1) (1/(sin^2 (((kπ)/2^(n+1) ))))=Σ_(k=1) ^(2^n −1) (1+(1/(tg(((kπ)/2^(n+1) ))))) =2^n −1+Σtg^2 (((kπ)/2^(m+1) ))=2^n −1+((2^(2n+1) −3.2^n +1)/3) ⇔Σ_(k=1) ^(2^n −1) (1/(sin^2 (((kπ)/2^(n+1) ))))=((2^(2n+1) −2)/3)$
	$\underset{k = 1}{\sum^{2^{n} - 1}} \frac{1}{\sin^{2} (\frac{k π}{2^{n + 1}})} = \underset{k = 1}{\sum^{2^{n} - 1}} (1 + \frac{1}{{tg}^{2} (\frac{k π}{2^{n + 1}})}), k \to 2^{n} - k$ $= \underset{k = 1}{\sum^{2^{n} - 1}} (1 + {tg}^{2} (\frac{k π}{2^{n + 1}})), \Leftrightarrow \underset{k = 1}{\sum^{2^{n - 1}}} {tg}^{2} (\frac{k π}{2^{n + 1}}) = \underset{k = 1}{\sum^{2^{n} - 1}} \frac{1}{{tg}^{2} (\frac{k π}{2^{n + 1}})} . . . I$ $Let$ $S = \underset{k = 1, k \neq 2^{n}}{\sum^{2^{n + 1} - 1}} {tg}^{2} (\frac{k π}{2^{n + 1}}) = \underset{k = 1}{\sum^{2^{n} - 1}} {tg}^{2} (\frac{k π}{2^{n + 1}}) + \underset{2^{n} + 1}{\sum^{2^{n} - 1}} {tg}^{2} (\frac{k π}{2^{n + 1}})$ $for the 2 nd Sum k \to 2^{n} + k$ $\Rightarrow S = \underset{k = 1}{\sum^{2^{n} - 1}} {tg}^{2} (\frac{k π}{2^{n + 1}}) + {tg}^{2} (\frac{2^{n} + k}{2^{n + 1}}) = \underset{k = 1}{\sum^{2^{n} - 1}} {tg}^{2} (\frac{k π}{2^{n + 1}}) + \frac{1}{{tg}^{2} (\frac{k π}{2^{n + 1}})}$ $\Rightarrow S = 2 \underset{k = 1}{\sum^{2^{n} - 1}} {tg}^{2} (\frac{k π}{2^{n + 1}}) . . . A = [1, 2^{n + 1} - 1] - {2^{n}}$ ${let}^{'} s find polynonial withe Roots tg (\frac{k π}{2^{n + 1}}), k \in A$ $tg (2^{n + 1} x) = 0 \Leftrightarrow 2^{n + 1} x = k π \Leftrightarrow x = \frac{k π}{2^{n + 1}}, k \in A \cup {2^{n}, 0}$ $if we tack, \frac{tg (2^{n + 1} x)}{tg (x)} = 0 \Rightarrow x = \frac{k π}{2^{n + 1}}, k \in A$ $we Know$ $tg (mx) = \frac{\underset{k = 0}{\sum^{[\frac{m - 1}{2}]}} {(- 1)}^{k} (\begin{matrix} m \\ 2 k + 1 \end{matrix}) {tg}^{2 k + 1} (x)}{\underset{k = 0}{\sum^{[\frac{m}{2}]}} {(- 1)}^{k} (\begin{matrix} m \\ 2 k \end{matrix}) . {tg}^{2 k} {(x)}_{}}$ $\Rightarrow \frac{tg (2^{n + 1} x)}{tg (x)} = 0 \Leftrightarrow \underset{k = 0}{\sum^{2^{n} - 1}} {(- 1)}^{k} (\begin{matrix} 2^{n + 1} \\ 2 k + 1 \end{matrix}) {tg}^{2 k} (x) = p (tg (x))$ $\underset{k = 0}{\sum^{2^{n} - 1}} {(- 1)}^{k} (\begin{matrix} 2^{n + 1} \\ 2 k + 1 \end{matrix}) t^{2 k} = p (t), \deg (p) = 2^{n + 1} - 2$ $and root of P . are \tan (\frac{k π}{2^{n + 1}}), k \in A card (A) = 2^{n + 1} - 2$ $so we have all root of P$ $if f (x) = \underset{k = 0}{\sum^{n}} a_{k} x^{k}; x_{i} root of P$ $Σ x_{i} = - \frac{a_{n - 1}}{a_{n}}, \sum_{1 ⩽ i < j ⩽ n} x_{i} x_{j} = \frac{a_{n - 2}}{a_{n}}$ $\sum_{k \in A} tg (\frac{k π}{2^{n + 1}}) = 0$ $\sum_{k_{1} < k_{2}, (k_{1}, k_{2}) \in A^{2}} tg (\frac{k_{1} π}{2^{n + 1}}) tg (\frac{k_{2} π}{2^{n + 1}}) = - \frac{(\begin{matrix} 2^{n + 1} \\ 2^{n + 1} - 3 \end{matrix})}{(\begin{matrix} 2^{n + 1} \\ 2^{n + 1} - 1 \end{matrix})}$ $= - \frac{1}{6} (2^{n + 1} - 1) (2^{n + 1} - 2)$ $Σ x_{i}^{2} = {(Σ x_{i})}^{2} - 2 Σ x_{i} x_{j} = \frac{1}{3} (2^{n + 1} - 1) (2^{n + 1} - 2)$ $S = 2 \underset{k = 1}{\sum^{2^{n} - 1}} {tg}^{2} (\frac{k π}{2^{n + 1}})$ $\underset{k = 1}{\sum^{2^{n} - 1}} {tg}^{2} (\frac{k π}{2^{n + 1}}) = \frac{1}{6} (2^{2 n + 2} - 3 . 2^{n + 1} + 2) = \frac{2^{2 n + 1} - 3 . 2^{n} + 1}{3}$ $S_{1} = \underset{k = 1}{\sum^{2^{n} - 1}} \frac{1}{\sin^{2} (\frac{k π}{2^{n + 1}})} = \underset{k = 1}{\sum^{2^{n} - 1}} (1 + \frac{1}{tg (\frac{k π}{2^{n + 1}})})$ $= 2^{n} - 1 + Σ {tg}^{2} (\frac{k π}{2^{m + 1}}) = 2^{n} - 1 + \frac{2^{2 n + 1} - 3 . 2^{n} + 1}{3}$ $\Leftrightarrow \underset{k = 1}{\sum^{2^{n} - 1}} \frac{1}{\sin^{2} (\frac{k π}{2^{n + 1}})} = \frac{2^{2 n + 1} - 2}{3}$
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