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

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

$${calculate}\:{interms}\:{of}\:{n}\:\:{U}_{{n}} =\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \:{sin}\left(\frac{{i}\pi}{{n}}\right){sin}\left(\frac{{j}\pi}{{n}}\right) \\ $$

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...

$${we}\:{have}\:\left(\sum_{{i}=\mathrm{1}} ^{{n}} {sin}\left({x}_{{i}} \right)\right)^{\mathrm{2}} =\sum_{{i}=\mathrm{1}} ^{{n}} \:{sin}^{\mathrm{2}} \left({x}_{{i}} \right)\:+\mathrm{2}\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \:{sin}\left(\:\:{x}_{{i}} \right){sin}\left({x}_{{j}} \right) \\ $$ $${let}\:{x}_{{i}} =\frac{{i}\pi}{{n}}\:\Rightarrow\left(\sum_{{i}=\mathrm{1}} ^{{n}} {sin}\left(\frac{{i}\pi}{{n}}\right)^{\mathrm{2}} \right)=\sum_{{i}=\mathrm{1}} ^{{n}} {sin}^{\mathrm{2}} \left(\frac{{i}\pi}{{n}}\right)+\mathrm{2}\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} {sin}\left(\frac{{i}\pi}{{n}}\right){sin}\left(\frac{{j}\pi}{{n}}\right) \\ $$ $$\Rightarrow\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{j}} {sin}\left(\frac{{i}\pi}{{n}}\right){sin}\left(\frac{{j}\pi}{{n}}\right) \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\left(\sum_{{i}=\mathrm{1}} ^{{n}} \:{sin}\left(\frac{{i}\pi}{{n}}\right)\right)^{\mathrm{2}} −\sum_{{i}=\mathrm{1}} ^{{n}} \:{sin}^{\mathrm{2}} \left(\frac{{i}\pi}{{n}}\right)\right\} \\ $$ $${let}\:{A}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} {sin}\left(\frac{{k}\pi}{{n}}\right)\:\Rightarrow{A}_{{n}} ={Im}\left(\sum_{{k}=\mathrm{0}} ^{{n}} \:{e}^{\frac{{ik}\pi}{{n}}} \right)\:{and} \\ $$ $$\sum_{{i}=\mathrm{0}} ^{{n}} \:{e}^{\frac{{ik}\pi}{{n}}} \:=\sum_{{i}.\mathrm{0}} ^{{n}} \:\left({e}^{{i}\frac{\pi}{{n}}} \right)^{{k}} \:=\frac{\mathrm{1}−\left({e}^{{i}\frac{\pi}{{n}}} \right)^{{n}+\mathrm{1}} }{\mathrm{1}−{e}^{\frac{{i}\pi}{{n}}} }\:=\frac{\mathrm{1}−{e}^{{i}\frac{\left({n}+\mathrm{1}\right)\pi}{{n}}} }{\mathrm{1}−{e}^{\frac{{i}\pi}{{n}}} } \\ $$ $$=\frac{\mathrm{1}−{cos}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{{n}}\right)−{isin}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{{n}}\right)}{\mathrm{1}−{cos}\left(\frac{\pi}{{n}}\right)−{isin}\left(\frac{\pi}{{n}}\right)} \\ $$ $$=\frac{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right)−\mathrm{2}{isin}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right){cos}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right)}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}{n}}\right)−\mathrm{2}{isin}\left(\frac{\pi}{\mathrm{2}{n}}\right){cos}\left(\frac{\pi}{\mathrm{2}{n}}\right)}\: \\ $$ $$=\frac{−{isin}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right){e}^{{i}\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}} }{−{isin}\left(\frac{\pi}{\mathrm{2}{n}}\right){e}^{\frac{{i}\pi}{\mathrm{2}{n}}} }\:=\frac{{sin}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right)}{{sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)}\left({i}\right)\:\Rightarrow \\ $$ $${A}_{{n}} =\frac{{sin}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right)}{{sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)}\:\:{let}\:\:{calculate}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{sin}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)...{be}\:{continued}... \\ $$ $$ \\ $$

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)}

$${we}\:{have}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{sin}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)=\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\mathrm{1}−{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)}{\mathrm{2}} \\ $$ $$=\frac{{n}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{1}} ^{{n}} \:{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)\:{and} \\ $$ $$\sum_{{k}=\mathrm{1}} ^{{n}} \:{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)={Re}\left(\sum_{{k}=\mathrm{0}} ^{{n}} \:{e}^{\frac{{i}\mathrm{2}{k}\pi}{{n}}} −\mathrm{1}\right) \\ $$ $$\sum_{{k}=\mathrm{0}} ^{{n}} \:{e}^{\frac{{i}\mathrm{2}{k}\pi}{{n}}} \:=\frac{\mathrm{1}−\left({e}^{\frac{{i}\mathrm{2}\pi}{{n}}} \right)^{{n}+\mathrm{1}} }{\mathrm{1}−{e}^{\frac{{i}\mathrm{2}\pi}{{n}}} }\:=\frac{\mathrm{1}−{e}^{\frac{{i}\left(\mathrm{2}\pi\right)\left({n}+\mathrm{1}\right)}{{n}}} }{\mathrm{1}−{e}^{\frac{{i}\mathrm{2}\pi}{{n}}} } \\ $$ $$=\frac{\mathrm{1}−{cos}\left(\frac{\mathrm{2}\left({n}+\mathrm{1}\right)\pi}{{n}}\right)−{isin}\left(\frac{\mathrm{2}\left({n}+\mathrm{1}\right)\pi}{{n}}\right)}{\mathrm{1}−{cos}\left(\frac{\mathrm{2}\pi}{{n}}\right)−{isin}\left(\frac{\mathrm{2}\pi}{{n}}\right)} \\ $$ $$=\frac{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\left({n}+\mathrm{1}\right)\pi}{{n}}\right)−\mathrm{2}{isin}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{{n}}\right){cos}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{{n}}\right)}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\pi}{{n}}\right)−\mathrm{2}{isin}\left(\frac{\pi}{{n}}\right){cos}\left(\frac{\pi}{{n}}\right)} \\ $$ $$=\frac{−{isin}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{{n}}\right){e}^{{i}\frac{\left({n}+\mathrm{1}\right)\pi}{{n}}} }{−{isin}\left(\frac{\pi}{{n}}\right){e}^{\frac{{i}\pi}{{n}}} }\:=\frac{{sin}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{{n}}\right)}{{sin}\left(\frac{\pi}{{n}}\right)}×\left(−\mathrm{1}\right)\:\Rightarrow \\ $$ $$\sum_{{k}=\mathrm{1}} ^{{n}} \:{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)=−\mathrm{1}−\frac{{sin}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{{n}}\right)}{{sin}\left(\frac{\pi}{{n}}\right)}\:=−\mathrm{1}−\frac{{sin}\left(\pi+\frac{\pi}{{n}}\right)}{{sin}\left(\frac{\pi}{{n}}\right)} \\ $$ $$=−\mathrm{1}+\mathrm{1}\:=\mathrm{0}\:\Rightarrow\sum_{{k}=\mathrm{1}} ^{{n}} \:{sin}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)=\frac{{n}}{\mathrm{2}}\:\Rightarrow \\ $$ $$\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \:\:\:{sin}\left(\frac{{i}\pi}{{n}}\right){sin}\left(\frac{{j}\pi}{{n}}\right) \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\:\frac{{sin}^{\mathrm{2}} \left(\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right)}{{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}{n}}\right)}\:−\frac{{n}^{\mathrm{2}} }{\mathrm{4}}\right\} \\ $$

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