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Question Number 30599 by abdo imad last updated on 23/Feb/18

decompose inside C(x) F= (1/((x−1)(x^n −1))) .

decomposeinsideC(x)F=1(x−1)(xn−1).

Commented by abdo imad last updated on 25/Feb/18

the roots of z^n −1=0 are the complex z_k = e^(i((2kπ)/k)) with k∈[[0,n−1]] ⇒F will be decomposed inside C(x) at form F(x)= (a/(x−1)) +(b/((x−1)^2 )) +Σ_(k=1) ^(n−1) (λ_k /(x−z_k )) we have (x−1)(x^n −1)=x^(n+1) −x −x^n +1 ⇒ (d/dx)( (x−1)(x^n −1))=(n+1)x^n −nx^(n−1) −1⇒ λ_k = (1/((n+1)z_k ^n −n z_k ^(n−1) −1)) =(1/(n+1−(n/z_k )−1))= (z_k /(nz_k −n)) the ch.x−1=y give F(x)= (1/((x−1)^2 (1+x +x^2 +...+x^(n−1) ))) =g(y)= (1/(y^2 (1+(1+y) +(1+y)^2 +....(1+y)^(n−1) ))) after divide 1 per 2+y +(1+y)^2 +...(1+y)^(n−1) folowing the increasing powers after all calculus we find a=((1−n)/(2n)) and b= (1/n) look also that b=lim_(x→1) (x−1)^2 F(x)=lim_(x→1) (1/(1+x +x^2 +...+x^(n−1) )) =(1/n)

therootsofzn−1=0arethecomplexzk=ei2kπkwithk∈[[0,n−1]]⇒FwillbedecomposedinsideC(x)atformF(x)=ax−1+b(x−1)2+∑k=1n−1λkx−zkwehave(x−1)(xn−1)=xn+1−x−xn+1⇒ddx((x−1)(xn−1))=(n+1)xn−nxn−1−1⇒λk=1(n+1)zkn−nzkn−1−1=1n+1−nzk−1=zknzk−nthech.x−1=ygiveF(x)=1(x−1)2(1+x+x2+...+xn−1)=g(y)=1y2(1+(1+y)+(1+y)2+....(1+y)n−1)afterdivide1per2+y+(1+y)2+...(1+y)n−1folowingtheincreasingpowersafterallcalculuswefinda=1−n2nandb=1nlookalsothatb=limx→1(x−1)2F(x)=limx→111+x+x2+...+xn−1=1n

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