calculate interms of n U_n =Σ_(1≤i<j≤n) sin(((iπ)/n))sin(((jπ)/n))


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Question Number 72021 by mathmax by abdo last updated on 23/Oct/19

$c a l c u l a t e i n t e r m s o f n U_{n} = \sum_{1 ⩽ i < j ⩽ n} s i n (\frac{i π}{n}) s i n (\frac{j π}{n})$
Commented bymathmax by abdo last updated on 26/Oct/19
$we have (Σ_(i=1) ^n sin(x_i ))^2 =Σ_(i=1) ^n sin^2 (x_i ) +2Σ_(1≤i<j≤n) sin( x_i )sin(x_j ) let x_i =((iπ)/n) ⇒(Σ_(i=1) ^n sin(((iπ)/n))^2 )=Σ_(i=1) ^n sin^2 (((iπ)/n))+2Σ_(1≤i<j≤n) sin(((iπ)/n))sin(((jπ)/n)) ⇒Σ_(1≤i<j≤j) sin(((iπ)/n))sin(((jπ)/n)) =(1/2){ (Σ_(i=1) ^n sin(((iπ)/n)))^2 −Σ_(i=1) ^n sin^2 (((iπ)/n))} let A_n =Σ_(k=1) ^n sin(((kπ)/n)) ⇒A_n =Im(Σ_(k=0) ^n e^((ikπ)/n) ) and Σ_(i=0) ^n e^((ikπ)/n) =Σ_(i.0) ^n (e^(i(π/n)) )^k =((1−(e^(i(π/n)) )^(n+1) )/(1−e^((iπ)/n) )) =((1−e^(i(((n+1)π)/n)) )/(1−e^((iπ)/n) )) =((1−cos((((n+1)π)/n))−isin((((n+1)π)/n)))/(1−cos((π/n))−isin((π/n)))) =((2sin^2 ((((n+1)π)/(2n)))−2isin((((n+1)π)/(2n)))cos((((n+1)π)/(2n))))/(2sin^2 ((π/(2n)))−2isin((π/(2n)))cos((π/(2n))))) =((−isin((((n+1)π)/(2n)))e^(i(((n+1)π)/(2n))) )/(−isin((π/(2n)))e^((iπ)/(2n)) )) =((sin((((n+1)π)/(2n))))/(sin((π/(2n)))))(i) ⇒ A_n =((sin((((n+1)π)/(2n))))/(sin((π/(2n))))) let calculate Σ_(k=1) ^n sin^2 (((kπ)/n))...be continued...$
$w e h a v e {(\sum_{i = 1}^{n} s i n (x_{i}))}^{2} = \sum_{i = 1}^{n} {s i n}^{2} (x_{i}) + 2 \sum_{1 ⩽ i < j ⩽ n} s i n (x_{i}) s i n (x_{j})$ $l e t x_{i} = \frac{i π}{n} \Rightarrow (\sum_{i = 1}^{n} s i n {(\frac{i π}{n})}^{2}) = \sum_{i = 1}^{n} {s i n}^{2} (\frac{i π}{n}) + 2 \sum_{1 ⩽ i < j ⩽ n} s i n (\frac{i π}{n}) s i n (\frac{j π}{n})$ $\Rightarrow \sum_{1 ⩽ i < j ⩽ j} s i n (\frac{i π}{n}) s i n (\frac{j π}{n})$ $= \frac{1}{2} {{(\sum_{i = 1}^{n} s i n (\frac{i π}{n}))}^{2} - \sum_{i = 1}^{n} {s i n}^{2} (\frac{i π}{n})}$ $l e t A_{n} = \sum_{k = 1}^{n} s i n (\frac{k π}{n}) \Rightarrow A_{n} = I m (\sum_{k = 0}^{n} e^{\frac{i k π}{n}}) a n d$ $\sum_{i = 0}^{n} e^{\frac{i k π}{n}} = \sum_{i . 0}^{n} {(e^{i \frac{π}{n}})}^{k} = \frac{1 - {(e^{i \frac{π}{n}})}^{n + 1}}{1 - e^{\frac{i π}{n}}} = \frac{1 - e^{i \frac{(n + 1) π}{n}}}{1 - e^{\frac{i π}{n}}}$ $= \frac{1 - c o s (\frac{(n + 1) π}{n}) - i s i n (\frac{(n + 1) π}{n})}{1 - c o s (\frac{π}{n}) - i s i n (\frac{π}{n})}$ $= \frac{2 {s i n}^{2} (\frac{(n + 1) π}{2 n}) - 2 i s i n (\frac{(n + 1) π}{2 n}) c o s (\frac{(n + 1) π}{2 n})}{2 {s i n}^{2} (\frac{π}{2 n}) - 2 i s i n (\frac{π}{2 n}) c o s (\frac{π}{2 n})}$ $= \frac{- i s i n (\frac{(n + 1) π}{2 n}) e^{i \frac{(n + 1) π}{2 n}}}{- i s i n (\frac{π}{2 n}) e^{\frac{i π}{2 n}}} = \frac{s i n (\frac{(n + 1) π}{2 n})}{s i n (\frac{π}{2 n})} (i) \Rightarrow$ $A_{n} = \frac{s i n (\frac{(n + 1) π}{2 n})}{s i n (\frac{π}{2 n})} l e t c a l c u l a t e \sum_{k = 1}^{n} {s i n}^{2} (\frac{k π}{n}) . . . b e c o n t i n u e d . . .$
Commented bymathmax by abdo last updated on 26/Oct/19
$we have Σ_(k=1) ^n sin^2 (((kπ)/n))=Σ_(k=1) ^n ((1−cos(((2kπ)/n)))/2) =(n/2)−(1/2)Σ_(k=1) ^n cos(((2kπ)/n)) and Σ_(k=1) ^n cos(((2kπ)/n))=Re(Σ_(k=0) ^n e^((i2kπ)/n) −1) Σ_(k=0) ^n e^((i2kπ)/n) =((1−(e^((i2π)/n) )^(n+1) )/(1−e^((i2π)/n) )) =((1−e^((i(2π)(n+1))/n) )/(1−e^((i2π)/n) )) =((1−cos(((2(n+1)π)/n))−isin(((2(n+1)π)/n)))/(1−cos(((2π)/n))−isin(((2π)/n)))) =((2sin^2 ((((n+1)π)/n))−2isin((((n+1)π)/n))cos((((n+1)π)/n)))/(2sin^2 ((π/n))−2isin((π/n))cos((π/n)))) =((−isin((((n+1)π)/n))e^(i(((n+1)π)/n)) )/(−isin((π/n))e^((iπ)/n) )) =((sin((((n+1)π)/n)))/(sin((π/n))))×(−1) ⇒ Σ_(k=1) ^n cos(((2kπ)/n))=−1−((sin((((n+1)π)/n)))/(sin((π/n)))) =−1−((sin(π+(π/n)))/(sin((π/n)))) =−1+1 =0 ⇒Σ_(k=1) ^n sin^2 (((kπ)/n))=(n/2) ⇒ Σ_(1≤i<j≤n) sin(((iπ)/n))sin(((jπ)/n)) =(1/2){ ((sin^2 ((((n+1)π)/(2n))))/(sin^2 ((π/(2n))))) −(n^2 /4)}$
$w e h a v e \sum_{k = 1}^{n} {s i n}^{2} (\frac{k π}{n}) = \sum_{k = 1}^{n} \frac{1 - c o s (\frac{2 k π}{n})}{2}$ $= \frac{n}{2} - \frac{1}{2} \sum_{k = 1}^{n} c o s (\frac{2 k π}{n}) a n d$ $\sum_{k = 1}^{n} c o s (\frac{2 k π}{n}) = R e (\sum_{k = 0}^{n} e^{\frac{i 2 k π}{n}} - 1)$ $\sum_{k = 0}^{n} e^{\frac{i 2 k π}{n}} = \frac{1 - {(e^{\frac{i 2 π}{n}})}^{n + 1}}{1 - e^{\frac{i 2 π}{n}}} = \frac{1 - e^{\frac{i (2 π) (n + 1)}{n}}}{1 - e^{\frac{i 2 π}{n}}}$ $= \frac{1 - c o s (\frac{2 (n + 1) π}{n}) - i s i n (\frac{2 (n + 1) π}{n})}{1 - c o s (\frac{2 π}{n}) - i s i n (\frac{2 π}{n})}$ $= \frac{2 {s i n}^{2} (\frac{(n + 1) π}{n}) - 2 i s i n (\frac{(n + 1) π}{n}) c o s (\frac{(n + 1) π}{n})}{2 {s i n}^{2} (\frac{π}{n}) - 2 i s i n (\frac{π}{n}) c o s (\frac{π}{n})}$ $= \frac{- i s i n (\frac{(n + 1) π}{n}) e^{i \frac{(n + 1) π}{n}}}{- i s i n (\frac{π}{n}) e^{\frac{i π}{n}}} = \frac{s i n (\frac{(n + 1) π}{n})}{s i n (\frac{π}{n})} \times (- 1) \Rightarrow$ $\sum_{k = 1}^{n} c o s (\frac{2 k π}{n}) = - 1 - \frac{s i n (\frac{(n + 1) π}{n})}{s i n (\frac{π}{n})} = - 1 - \frac{s i n (π + \frac{π}{n})}{s i n (\frac{π}{n})}$ $= - 1 + 1 = 0 \Rightarrow \sum_{k = 1}^{n} {s i n}^{2} (\frac{k π}{n}) = \frac{n}{2} \Rightarrow$ $\sum_{1 ⩽ i < j ⩽ n} s i n (\frac{i π}{n}) s i n (\frac{j π}{n})$ $= \frac{1}{2} {\frac{{s i n}^{2} (\frac{(n + 1) π}{2 n})}{{s i n}^{2} (\frac{π}{2 n})} - \frac{n^{2}}{4}}$
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