Question Number 166754 by mathlove last updated on 27/Feb/22
Answered by mindispower last updated on 28/Feb/22
![S=Σ_(n=1) ^m cot^2 (((nπ)/(2m+1))),Σ_1 ^(2m) cot^2 (((nπ)/(2m+1)))=2S put z_k =e^((ikπ)/(2m+1)) cot^2 (((kπ)/(2m+1)))=(i((z_k ^2 +1)/(z_k ^2 −1)))^2 =−(1+(2/(z_k ^2 −1)))^2 =−(1+(1/((z_k +1)^2 ))+(1/((z_k −1)^2 ))+(1/(z_k −1))−(1/(z_k +1))) Σ_(k=1) ^(2m) cot^2 (((kπ)/(2m+1)))=2S⇔ −2S=Σ_(k=1) ^(2m) (1+(1/(z_k −1))−(1/(z_k +1))+(1/((z_k −1)^2 ))+(1/((z_k +1)^2 )))q Put P(x)=Σ_(k=0) ^(2m) x^k ,P(x)=0⇒x∈{x_k }_(k∈[1,2m]) ,x_k =e^((2ikπ)/(2m+1)) We Have ((P′(x))/(p(x)))=Σ_(k=1) ^(2m) (1/(X−x_k ))⇒((P′^2 (x)−p(x)p′′(x))/(p^2 (x)))=Σ_(k=1) ^(2m) (1/((x−x_k )^2 )) Σ_(k=1) ^(2m) (1/(z_k −1))=((−P′(1))/(P(1)))=−m −Σ_(k=1) ^(2m) (1/(z_k +1))=((P′(−1))/(P(1)))=−m p′′(1)=((2(2m−1)(2m^2 +m))/3) p′′(−1)=2m^2 Σ_(k=1) ^(2m) (1/((z_k −1)^2 ))=((−m(m−2))/3) Σ_(k=1) ^(2m) (1/((z_k +1)^2 ))=−m^2 ⇒−2S=2m−m−m−((m(m−2))/3)−m^2 −2S=((−4m^2 +2m)/3)⇒S=((m(2m−1))/3)](Q166852.png)
$${S}=\underset{{n}=\mathrm{1}} {\overset{{m}} {\sum}}{cot}^{\mathrm{2}} \left(\frac{{n}\pi}{\mathrm{2}{m}+\mathrm{1}}\right),\underset{\mathrm{1}} {\overset{\mathrm{2}{m}} {\sum}}{cot}^{\mathrm{2}} \left(\frac{{n}\pi}{\mathrm{2}{m}+\mathrm{1}}\right)=\mathrm{2}{S} \\ $$$${put}\:{z}_{{k}} ={e}^{\frac{{ik}\pi}{\mathrm{2}{m}+\mathrm{1}}} \\ $$$${cot}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{2}{m}+\mathrm{1}}\right)=\left({i}\frac{{z}_{{k}} ^{\mathrm{2}} +\mathrm{1}}{{z}_{{k}} ^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} =−\left(\mathrm{1}+\frac{\mathrm{2}}{{z}_{{k}} ^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} \\ $$$$=−\left(\mathrm{1}+\frac{\mathrm{1}}{\left({z}_{{k}} +\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({z}_{{k}} −\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{{z}_{{k}} −\mathrm{1}}−\frac{\mathrm{1}}{{z}_{{k}} +\mathrm{1}}\right) \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{m}} {\sum}}{cot}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{2}{m}+\mathrm{1}}\right)=\mathrm{2}{S}\Leftrightarrow \\ $$$$−\mathrm{2}{S}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{m}} {\sum}}\left(\mathrm{1}+\frac{\mathrm{1}}{{z}_{{k}} −\mathrm{1}}−\frac{\mathrm{1}}{{z}_{{k}} +\mathrm{1}}+\frac{\mathrm{1}}{\left({z}_{{k}} −\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({z}_{{k}} +\mathrm{1}\right)^{\mathrm{2}} }\right){q} \\ $$$${Put}\:{P}\left({x}\right)=\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{m}} {\sum}}{x}^{{k}} ,{P}\left({x}\right)=\mathrm{0}\Rightarrow{x}\in\left\{{x}_{{k}} \right\}_{{k}\in\left[\mathrm{1},\mathrm{2}{m}\right]} ,{x}_{{k}} ={e}^{\frac{\mathrm{2}{ik}\pi}{\mathrm{2}{m}+\mathrm{1}}} \\ $$$${We}\:{Have} \\ $$$$\frac{{P}'\left({x}\right)}{{p}\left({x}\right)}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{m}} {\sum}}\frac{\mathrm{1}}{{X}−{x}_{{k}} }\Rightarrow\frac{{P}'^{\mathrm{2}} \left({x}\right)−{p}\left({x}\right){p}''\left({x}\right)}{{p}^{\mathrm{2}} \left({x}\right)}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{m}} {\sum}}\frac{\mathrm{1}}{\left({x}−{x}_{{k}} \right)^{\mathrm{2}} } \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{m}} {\sum}}\frac{\mathrm{1}}{{z}_{{k}} −\mathrm{1}}=\frac{−{P}'\left(\mathrm{1}\right)}{{P}\left(\mathrm{1}\right)}=−{m} \\ $$$$−\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{m}} {\sum}}\frac{\mathrm{1}}{{z}_{{k}} +\mathrm{1}}=\frac{{P}'\left(−\mathrm{1}\right)}{{P}\left(\mathrm{1}\right)}=−{m} \\ $$$${p}''\left(\mathrm{1}\right)=\frac{\mathrm{2}\left(\mathrm{2}{m}−\mathrm{1}\right)\left(\mathrm{2}{m}^{\mathrm{2}} +{m}\right)}{\mathrm{3}} \\ $$$${p}''\left(−\mathrm{1}\right)=\mathrm{2}{m}^{\mathrm{2}} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{m}} {\sum}}\frac{\mathrm{1}}{\left({z}_{{k}} −\mathrm{1}\right)^{\mathrm{2}} }=\frac{−{m}\left({m}−\mathrm{2}\right)}{\mathrm{3}} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{m}} {\sum}}\frac{\mathrm{1}}{\left({z}_{{k}} +\mathrm{1}\right)^{\mathrm{2}} }=−{m}^{\mathrm{2}} \\ $$$$\Rightarrow−\mathrm{2}{S}=\mathrm{2}{m}−{m}−{m}−\frac{{m}\left({m}−\mathrm{2}\right)}{\mathrm{3}}−{m}^{\mathrm{2}} \\ $$$$−\mathrm{2}{S}=\frac{−\mathrm{4}{m}^{\mathrm{2}} +\mathrm{2}{m}}{\mathrm{3}}\Rightarrow{S}=\frac{{m}\left(\mathrm{2}{m}−\mathrm{1}\right)}{\mathrm{3}} \\ $$$$ \\ $$