Question Number 26583 by abdo imad last updated on 27/Dec/17
![find the decomposition in C[x] then R[x] for the rationsl fraction F(x)= ((1 )/(x^(2n) −1)) .with n integer not 0](Q26583.png)
$${find}\:{the}\:{decomposition}\:{in}\:\mathbb{C}\left[{x}\right]\:{then}\:\mathbb{R}\left[{x}\right] \\ $$$${for}\:{the}\:{rationsl}\:{fraction} \\ $$$${F}\left({x}\right)=\:\:\frac{\mathrm{1}\:}{{x}^{\mathrm{2}{n}} −\mathrm{1}}\:\:.{with}\:{n}\:{integer}\:{not}\:\mathrm{0} \\ $$
Commented by abdo imad last updated on 28/Dec/17
![let find the poles of F let put z=rcosθ z^(2n) =1⇔ 2nθ =2kπ and r=1 so the poles of are z_k = e^((ikπ)/n) with k from [[0,2n−1]] F(x) =Σ_(k=0) ^(2n−1) (α_k /(x−z_k )) and α_k = (1/(2n z_k ^(2n−1) )) = (1/(2n)) z_k ⇒ F(x)= (1/(2n)) Σ_(k=0) ^(2n−1) (z_k /(x−z_k )) is the decomposition of F(x) in C[x].](Q26720.png)
$${let}\:{find}\:{the}\:{poles}\:{of}\:{F}\:\:\:{let}\:{put}\:{z}={rcos}\theta \\ $$$${z}^{\mathrm{2}{n}} =\mathrm{1}\Leftrightarrow\:\:\:\mathrm{2}{n}\theta\:\:=\mathrm{2}{k}\pi\:{and}\:\:{r}=\mathrm{1}\:\:{so}\:{the}\:{poles}\:{of}\:{are}\: \\ $$$${z}_{{k}} =\:{e}^{\frac{{ik}\pi}{{n}}} \:\:{with}\:\:{k}\:{from}\:\left[\left[\mathrm{0},\mathrm{2}{n}−\mathrm{1}\right]\right] \\ $$$${F}\left({x}\right)\:\:=\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}−\mathrm{1}} \:\:\frac{\alpha_{{k}} }{{x}−{z}_{{k}} }\:\:\:\:{and}\:\:\alpha_{{k}} =\:\:\:\frac{\mathrm{1}}{\mathrm{2}{n}\:{z}_{{k}} ^{\mathrm{2}{n}−\mathrm{1}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}{n}}\:{z}_{{k}} \\ $$$$\Rightarrow\:\:\:\:\:{F}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}{n}}\:\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}−\mathrm{1}} \:\frac{{z}_{{k}} }{{x}−{z}_{{k}} }\:\:{is}\:{the}\:{decomposition}\:{of}\:{F}\left({x}\right) \\ $$$${in}\:\mathbb{C}\left[{x}\right]. \\ $$
Commented by abdo imad last updated on 28/Dec/17
![we have F(x)= (1/(2n)) Σ_(k=0) ^(2n−1) ((zk)/(x−x_k )) but z_0 =1, z_1 =e^(i(π/n)) , z_2 =e^(i((2π)/n)) , z_(n−1) =e^(i((n−1)/n)π) z_n =−1 z_(n+1) =e^(i(((n+1)π)/n)) =z_1 ^− , z_(n+2) = e^(i(((n+2)π)/n)) =z_2 ^− , z_(n−1) =z_1 ^− ⇒ F(x)= (1/(2n))( (z_0 /(x−z_0 )) + (z_n /(x−z_n )) + Σ_(k=1) ^(k=n−1) ( (z_k /(x−z_k )) + (z_k ^− /(x−z_k ^− )))) = (1/(2n))( (1/(x−1)) − (1/(x+1))) + (1/(2n))Σ_(k=1) ^(n−1) (((z_k +z_k ^− )x −2)/(x^2 −2 cos(((kπ)/n))x+1)) F(x)= (1/(2n)) ( (1/(x−1)) − (1/(x+1))) + (1/n) Σ_(k=1) ^(k=n−1) (( cos(((kπ)/n))x −1)/(x^(2 ) −2cos(((kπ)/n))x+1)) is the decomposition of F(x) inside R[x].](Q26740.png)
$${we}\:{have}\:{F}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}{n}}\:\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}−\mathrm{1}} \:\frac{{zk}}{{x}−{x}_{{k}} }\:\:{but} \\ $$$${z}_{\mathrm{0}} =\mathrm{1},\:\:\:\:{z}_{\mathrm{1}} ={e}^{{i}\frac{\pi}{{n}}} \:\:\:,\:\:\:{z}_{\mathrm{2}} ={e}^{{i}\frac{\mathrm{2}\pi}{{n}}} \:\:\:,\:{z}_{{n}−\mathrm{1}} ={e}^{{i}\frac{{n}−\mathrm{1}}{{n}}\pi} \:\:\:{z}_{{n}} \:\:=−\mathrm{1} \\ $$$${z}_{{n}+\mathrm{1}} ={e}^{{i}\frac{\left({n}+\mathrm{1}\right)\pi}{{n}}} \:\:={z}_{\mathrm{1}} ^{−} \:\:\:\:\:\:,\:\:{z}_{{n}+\mathrm{2}} =\:{e}^{{i}\frac{\left({n}+\mathrm{2}\right)\pi}{{n}}} \:\:={z}_{\mathrm{2}} ^{−} \:\:\:,\:{z}_{{n}−\mathrm{1}} \:\:={z}_{\mathrm{1}} ^{−} \\ $$$$\Rightarrow\:\:\:{F}\left({x}\right)=\:\:\frac{\mathrm{1}}{\mathrm{2}{n}}\left(\:\:\:\:\frac{{z}_{\mathrm{0}} }{{x}−{z}_{\mathrm{0}} }\:+\:\frac{{z}_{{n}} }{{x}−{z}_{{n}} }\:+\:\:\:\sum_{{k}=\mathrm{1}} ^{{k}={n}−\mathrm{1}} \left(\:\:\frac{{z}_{{k}} }{{x}−{z}_{{k}} }\:+\:\frac{{z}_{{k}} ^{−} }{{x}−{z}_{{k}} ^{−} }\right)\right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}{n}}\left(\:\:\frac{\mathrm{1}}{{x}−\mathrm{1}}\:−\:\:\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)\:\:+\:\:\frac{\mathrm{1}}{\mathrm{2}{n}}\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \frac{\left({z}_{{k}} \:+{z}_{{k}} ^{−} \right){x}\:−\mathrm{2}}{{x}^{\mathrm{2}} \:−\mathrm{2}\:{cos}\left(\frac{{k}\pi}{{n}}\right){x}+\mathrm{1}} \\ $$$${F}\left({x}\right)=\:\:\frac{\mathrm{1}}{\mathrm{2}{n}}\:\left(\:\frac{\mathrm{1}}{{x}−\mathrm{1}}\:−\:\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)\:+\:\frac{\mathrm{1}}{{n}}\:\sum_{{k}=\mathrm{1}} ^{{k}={n}−\mathrm{1}} \frac{\:{cos}\left(\frac{{k}\pi}{{n}}\right){x}\:−\mathrm{1}}{{x}^{\mathrm{2}\:} −\mathrm{2}{cos}\left(\frac{{k}\pi}{{n}}\right){x}+\mathrm{1}} \\ $$$${is}\:{the}\:{decomposition}\:{of}\:{F}\left({x}\right)\:{inside}\:\mathbb{R}\left[{x}\right]. \\ $$