Question Number 57923 by maxmathsup by imad last updated on 14/Apr/19
$${decompose}\:{the}\:{fraction}\:{F}\left({x}\right)\:=\frac{{x}^{{n}} }{{x}^{\mathrm{2}{n}} −\mathrm{1}}\:{inside}\:{C}\left({x}\right)\:{and}\:{R}\left({x}\right) \\ $$
Commented by maxmathsup by imad last updated on 28/Apr/19
![roots of z^(2n) −1 =0 let z =re^(iθ) so z^(2n) =1 ⇒r^n =1 and2 nθ =2kπ ⇒ θ_k =((kπ)/n) ⇒ the roots areZ_k =e^((ik)/n) and k∈[[0,2n−1]] ⇒ F(x) =(x^n /(Π_(k=0) ^(2n−1) (x−Z_k ))) =Σ_(k=0) ^(2n−1) (λ_k /(x−Z_k )) and λ_k =(Z_k ^n /(2n Z_k ^(2n−1) )) =(1/(2n)) Z_k ^(n+1) ⇒ F(x) =(1/(2n))Σ_(k=0) ^(2n−1) (Z_k ^(n+1) /(x−Z_k )) is the decomposition inside C(x) .](https://www.tinkutara.com/question/Q58721.png)
$${roots}\:{of}\:{z}^{\mathrm{2}{n}} −\mathrm{1}\:=\mathrm{0}\:\:\:\:\:\:{let}\:{z}\:={re}^{{i}\theta} \:\:\:{so}\:{z}^{\mathrm{2}{n}} =\mathrm{1}\:\Rightarrow{r}^{{n}} =\mathrm{1}\:{and}\mathrm{2}\:{n}\theta\:=\mathrm{2}{k}\pi\:\:\Rightarrow \\ $$$$\theta_{{k}} =\frac{{k}\pi}{{n}}\:\:\Rightarrow\:{the}\:{roots}\:{areZ}_{{k}} ={e}^{\frac{{ik}}{{n}}} \:\:\:{and}\:{k}\in\left[\left[\mathrm{0},\mathrm{2}{n}−\mathrm{1}\right]\right]\:\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\frac{{x}^{{n}} }{\prod_{{k}=\mathrm{0}} ^{\mathrm{2}{n}−\mathrm{1}} \left({x}−{Z}_{{k}} \right)}\:=\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}−\mathrm{1}} \:\:\:\frac{\lambda_{{k}} }{{x}−{Z}_{{k}} }\:\:\:\:{and}\:\:\lambda_{{k}} =\frac{{Z}_{{k}} ^{{n}} }{\mathrm{2}{n}\:{Z}_{{k}} ^{\mathrm{2}{n}−\mathrm{1}} }\:=\frac{\mathrm{1}}{\mathrm{2}{n}}\:{Z}_{{k}} ^{{n}+\mathrm{1}} \:\Rightarrow \\ $$$${F}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{n}}\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}−\mathrm{1}} \:\:\:\:\frac{{Z}_{{k}} ^{{n}+\mathrm{1}} }{{x}−{Z}_{{k}} }\:\:\:{is}\:{the}\:{decomposition}\:{inside}\:{C}\left({x}\right)\:. \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 28/Apr/19
![we have Z_k =e^((ikπ)/n) with k∈[[0,2n−1]] Z_0 =1 , Z_1 =e^((iπ)/n) , Z_(2 ) = e^((i2π)/n) , Z_(n−1) =e^(i(((n−1)π)/n)) ,Z_n =−1 ,Z_(n+1) =e^(i(((n+1)π)/n)) , ....Z_(2n−1) =e^(i(((2n−1)π)/n)) ⇒Z_(2n−1) =Z_1 ^− ,Z_(2n−2) =Z_2 ^− .... ⇒ F(x) =(1/(2n)) { (1/(Z−1)) +(((−1)^(n+1) )/(x+1)) +Σ_(k=1) ^(n−1) ( (Z_k ^(n+1) /(x−Z_k )) +(((Z_k ^− )^(n+1) )/(x−Z_k ^− )))} ⇒ F(x)= (1/(2n(x−1))) +(((−1)^(n+1) )/(2n(x+1))) +(1/(2n)) +Σ_(k=0) ^(n−1) ((Z_k ^(n+1) (x−Z_k ^− )+(x−Z_k )(Z_k ^− )^(n+1) )/(x^2 −2xRe(Z_k )+1)) F(x) =(1/(2n(x−1))) +(((−1)^(n+1) )/(2n(x+1))) +Σ_(k=0) ^(n−1) (((Z_k ^(n+1) +(Z_k ^− )^(n+1) )x −Z_k ^(n+1) Z_k ^− −Z_k (Z_k ^− )^(n+1) )/(x^2 −2cos(((kπ)/n))x +1)) is the decomposition inside R(x) with Z_k ^(n+1) +(Z_k ^− )^(n+1) = 2 Re(Z_k ^(n+1) ) =2cos(k(((n+1)π)/n)) −Z_k ^(n+1) Z_k ^− −Z_k (Z_k ^− )^(n+1) =−( Z_k ^n +(Z_k ^− )^n ) =−2cos(kπ) =−2(−1)^k .](https://www.tinkutara.com/question/Q58726.png)
$${we}\:{have}\:{Z}_{{k}} ={e}^{\frac{{ik}\pi}{{n}}} \:\:\:{with}\:\:{k}\in\left[\left[\mathrm{0},\mathrm{2}{n}−\mathrm{1}\right]\right] \\ $$$${Z}_{\mathrm{0}} =\mathrm{1}\:\:,\:\:{Z}_{\mathrm{1}} ={e}^{\frac{{i}\pi}{{n}}} \:\:,\:{Z}_{\mathrm{2}\:} =\:{e}^{\frac{{i}\mathrm{2}\pi}{{n}}} \:\:,\:\:\:{Z}_{{n}−\mathrm{1}} ={e}^{{i}\frac{\left({n}−\mathrm{1}\right)\pi}{{n}}} \:\:,{Z}_{{n}} =−\mathrm{1}\:\:,{Z}_{{n}+\mathrm{1}} \:={e}^{{i}\frac{\left({n}+\mathrm{1}\right)\pi}{{n}}} \:, \\ $$$$….{Z}_{\mathrm{2}{n}−\mathrm{1}} ={e}^{{i}\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)\pi}{{n}}} \:\Rightarrow{Z}_{\mathrm{2}{n}−\mathrm{1}} =\overset{−} {{Z}}_{\mathrm{1}} \:\:,{Z}_{\mathrm{2}{n}−\mathrm{2}} =\overset{−} {{Z}}_{\mathrm{2}} \:….\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{n}}\:\left\{\:\:\frac{\mathrm{1}}{{Z}−\mathrm{1}}\:\:+\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{x}+\mathrm{1}}\:+\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\:\left(\:\frac{{Z}_{{k}} ^{{n}+\mathrm{1}} }{{x}−{Z}_{{k}} }\:+\frac{\left(\overset{−} {{Z}}_{{k}} \right)^{{n}+\mathrm{1}} }{{x}−\overset{−} {{Z}}_{{k}} }\right)\right\}\:\Rightarrow \\ $$$${F}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}{n}\left({x}−\mathrm{1}\right)}\:+\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{2}{n}\left({x}+\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{2}{n}}\:+\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\frac{{Z}_{{k}} ^{{n}+\mathrm{1}} \left({x}−\overset{−} {{Z}}_{{k}} \right)+\left({x}−{Z}_{{k}} \right)\left(\overset{−} {{Z}}_{{k}} \right)^{{n}+\mathrm{1}} }{{x}^{\mathrm{2}} −\mathrm{2}{xRe}\left({Z}_{{k}} \right)+\mathrm{1}} \\ $$$${F}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{n}\left({x}−\mathrm{1}\right)}\:+\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{2}{n}\left({x}+\mathrm{1}\right)}\:+\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\:\frac{\left({Z}_{{k}} ^{{n}+\mathrm{1}} +\left(\overset{−} {{Z}}_{{k}} \right)^{{n}+\mathrm{1}} \right){x}\:−{Z}_{{k}} ^{{n}+\mathrm{1}} \:\overset{−} {{Z}}_{{k}} \:−{Z}_{{k}} \left(\overset{−} {{Z}}_{{k}} \right)^{{n}+\mathrm{1}} }{{x}^{\mathrm{2}} \:−\mathrm{2}{cos}\left(\frac{{k}\pi}{{n}}\right){x}\:+\mathrm{1}}\:{is}\:{the}\:{decomposition}\:{inside}\:{R}\left({x}\right) \\ $$$$\:{with}\:\:{Z}_{{k}} ^{{n}+\mathrm{1}} \:+\left(\overset{−} {{Z}}_{{k}} \right)^{{n}+\mathrm{1}} \:=\:\mathrm{2}\:{Re}\left({Z}_{{k}} ^{{n}+\mathrm{1}} \right)\:=\mathrm{2}{cos}\left({k}\frac{\left({n}+\mathrm{1}\right)\pi}{{n}}\right) \\ $$$$−{Z}_{{k}} ^{{n}+\mathrm{1}} \:\overset{−} {{Z}}_{{k}} \:−{Z}_{{k}} \left(\overset{−} {{Z}}_{{k}} \right)^{{n}+\mathrm{1}} \:=−\left(\:{Z}_{{k}} ^{{n}} \:+\left(\overset{−} {{Z}}_{{k}} \right)^{{n}} \right)\:=−\mathrm{2}{cos}\left({k}\pi\right)\:=−\mathrm{2}\left(−\mathrm{1}\right)^{{k}} \:. \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 29/Apr/19
$${F}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{n}\left({x}−\mathrm{1}\right)}\:+\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{2}{n}\left({x}+\mathrm{1}\right)}\:+\frac{\mathrm{1}}{{n}}\:\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\frac{{cos}\left({k}\frac{{n}+\mathrm{1}}{{n}}\pi\right){x}\:−\left(−\mathrm{1}\right)^{{k}} }{{x}^{\mathrm{2}} \:−\mathrm{2}{cos}\left(\frac{{k}\pi}{{n}}\right){x}\:+\mathrm{1}}\:\:. \\ $$