Question Number 30599 by abdo imad last updated on 23/Feb/18
$${decompose}\:{inside}\:{C}\left({x}\right)\:\:{F}=\:\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)\left({x}^{{n}} \:−\mathrm{1}\right)}\:. \\ $$
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)](Q30791.png)
$${the}\:{roots}\:{of}\:{z}^{{n}} \:−\mathrm{1}=\mathrm{0}\:{are}\:{the}\:{complex}\:{z}_{{k}} =\:{e}^{{i}\frac{\mathrm{2}{k}\pi}{{k}}} \:\:{with} \\ $$$${k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right]\:\Rightarrow{F}\:{will}\:{be}\:{decomposed}\:{inside}\:{C}\left({x}\right)\:{at} \\ $$$${form}\:{F}\left({x}\right)=\:\frac{{a}}{{x}−\mathrm{1}}\:+\frac{{b}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:+\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\frac{\lambda_{{k}} }{{x}−{z}_{{k}} }\:{we}\:{have} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}^{{n}} \:−\mathrm{1}\right)={x}^{{n}+\mathrm{1}} −{x}\:−{x}^{{n}} \:+\mathrm{1}\:\Rightarrow \\ $$$$\frac{{d}}{{dx}}\left(\:\left({x}−\mathrm{1}\right)\left({x}^{{n}} −\mathrm{1}\right)\right)=\left({n}+\mathrm{1}\right){x}^{{n}} \:−{nx}^{{n}−\mathrm{1}} \:−\mathrm{1}\Rightarrow \\ $$$$\lambda_{{k}} =\:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right){z}_{{k}} ^{{n}} \:−{n}\:{z}_{{k}} ^{{n}−\mathrm{1}} \:−\mathrm{1}}\:=\frac{\mathrm{1}}{{n}+\mathrm{1}−\frac{{n}}{{z}_{{k}} }−\mathrm{1}}=\:\frac{{z}_{{k}} }{{nz}_{{k}} −{n}} \\ $$$${the}\:{ch}.{x}−\mathrm{1}={y}\:{give}\:{F}\left({x}\right)=\:\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{1}+{x}\:+{x}^{\mathrm{2}} \:+...+{x}^{{n}−\mathrm{1}} \right)} \\ $$$$={g}\left({y}\right)=\:\frac{\mathrm{1}}{{y}^{\mathrm{2}} \left(\mathrm{1}+\left(\mathrm{1}+{y}\right)\:+\left(\mathrm{1}+{y}\right)^{\mathrm{2}} \:+....\left(\mathrm{1}+{y}\right)^{{n}−\mathrm{1}} \right)}\:{after} \\ $$$${divide}\:\mathrm{1}\:{per}\:\mathrm{2}+{y}\:+\left(\mathrm{1}+{y}\right)^{\mathrm{2}} \:+...\left(\mathrm{1}+{y}\right)^{{n}−\mathrm{1}} \:{folowing}\:{the} \\ $$$${increasing}\:{powers}\:{after}\:{all}\:{calculus}\:{we}\:{find} \\ $$$${a}=\frac{\mathrm{1}−{n}}{\mathrm{2}{n}}\:{and}\:{b}=\:\frac{\mathrm{1}}{{n}}\:{look}\:{also}\:{that} \\ $$$${b}={lim}_{{x}\rightarrow\mathrm{1}} \left({x}−\mathrm{1}\right)^{\mathrm{2}} {F}\left({x}\right)={lim}_{{x}\rightarrow\mathrm{1}} \:\:\frac{\mathrm{1}}{\mathrm{1}+{x}\:+{x}^{\mathrm{2}} \:+...+{x}^{{n}−\mathrm{1}} }\:=\frac{\mathrm{1}}{{n}} \\ $$