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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

findthedecompositioninC[x]thenR[x]fortherationslfractionF(x)=1x2n−1.withnintegernot0

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].

letfindthepolesofFletputz=rcosθz2n=1⇔2nθ=2kπandr=1sothepolesofarezk=eikπnwithkfrom[[0,2n−1]]F(x)=∑k=02n−1αkx−zkandαk=12nzk2n−1=12nzk⇒F(x)=12n∑k=02n−1zkx−zkisthedecompositionofF(x)inC[x].

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].

wehaveF(x)=12n∑k=02n−1zkx−xkbutz0=1,z1=eiπn,z2=ei2πn,zn−1=ein−1nπzn=−1zn+1=ei(n+1)πn=z1−,zn+2=ei(n+2)πn=z2−,zn−1=z1−⇒F(x)=12n(z0x−z0+znx−zn+∑k=1k=n−1(zkx−zk+zk−x−zk−))=12n(1x−1−1x+1)+12n∑k=1n−1(zk+zk−)x−2x2−2cos(kπn)x+1F(x)=12n(1x−1−1x+1)+1n∑k=1k=n−1cos(kπn)x−1x2−2cos(kπn)x+1isthedecompositionofF(x)insideR[x].

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