calculate-interms-of-n-U-n-1-i-lt-j-n-sin-ipi-n-sin-jpi-n-

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 by mathmax by abdo last updated on 26/Oct/19

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}) \dots b e c o n t i n u e d \dots

Commented by mathmax by abdo last updated on 26/Oct/19

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}}