Question Number 28432 by abdo imad last updated on 25/Jan/18
$${let}\:{put}\:{w}={e}^{{i}\frac{\mathrm{2}\pi}{{n}}} \:{calculate}\:\:{S}_{{n}} =\:\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\:\frac{\mathrm{1}}{{x}−{w}^{{k}} }\:\:{and} \\ $$$${W}_{{n}} =\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\frac{\mathrm{1}}{\left({x}−{w}^{{k}} \right)^{\mathrm{2}} }\:. \\ $$
Commented by abdo imad last updated on 27/Jan/18
![let introduce the polynomial p(x)= x^n −1 the roots of p(x)are the complex z_k =e^(i((2kπ)/n)) and k∈[[0,n−1]] we know that ((p^′ (x))/(p(x))) = Σ_(k=0) ^(n−1) (1/(x−z_k )) =Σ_(k=0) ^(n−1) (1/(x−w^k )) so S_n = ((nx^(n−1) )/(x^n −1)) also we have by derivation (d/dx)( ((p^′ (x))/(p(x))))=−Σ_(k=0) ^(n−1) (1/((x−w^k )^2 )) ⇒((p^(′′) (x)p(x) −(p^′ (x))^2 )/((p(x))^2 ))=−Σ_(k=0) ^(n−1) (1/((x−w^k )^2 )) W_n =−((n(n−1)x^(n−2) ( x^n −1) −(nx^(n−1) )^2 )/((x^n −1)^2 )) =−((n(n−1) x^(2n−2) −n(n−1)x^(n−2) −n^2 x^(2n−2) )/((x^n −1)^2 )) =−((−n x^(2n−2) −n(n−1)x^(n−2) )/((x^n −1)^2 ))=((n x^(2n−2) −n(n−1)x^(n−2) )/((x^n −1)^2 )) for n≥2 .](Q28579.png)
$${let}\:{introduce}\:{the}\:{polynomial}\:{p}\left({x}\right)=\:{x}^{{n}} −\mathrm{1}\:\:{the}\:{roots}\:{of} \\ $$$${p}\left({x}\right){are}\:{the}\:{complex}\:{z}_{{k}} \:={e}^{{i}\frac{\mathrm{2}{k}\pi}{{n}}} \:\:\:{and}\:{k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right] \\ $$$${we}\:{know}\:{that}\:\:\frac{{p}^{'} \left({x}\right)}{{p}\left({x}\right)}\:=\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\:\frac{\mathrm{1}}{{x}−{z}_{{k}} }\:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\frac{\mathrm{1}}{{x}−{w}^{{k}} }\:{so} \\ $$$${S}_{{n}} \:=\:\frac{{nx}^{{n}−\mathrm{1}} }{{x}^{{n}} −\mathrm{1}}\:\:{also}\:{we}\:{have}\:{by}\:{derivation} \\ $$$$\frac{{d}}{{dx}}\left(\:\frac{{p}^{'} \left({x}\right)}{{p}\left({x}\right)}\right)=−\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\frac{\mathrm{1}}{\left({x}−{w}^{{k}} \right)^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{p}^{''} \left({x}\right){p}\left({x}\right)\:−\left({p}^{'} \left({x}\right)\right)^{\mathrm{2}} }{\left({p}\left({x}\right)\right)^{\mathrm{2}} }=−\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\:\:\frac{\mathrm{1}}{\left({x}−{w}^{{k}} \right)^{\mathrm{2}} } \\ $$$${W}_{{n}} \:=−\frac{{n}\left({n}−\mathrm{1}\right){x}^{{n}−\mathrm{2}} \left(\:{x}^{{n}} −\mathrm{1}\right)\:−\left({nx}^{{n}−\mathrm{1}} \right)^{\mathrm{2}} }{\left({x}^{{n}} \:−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=−\frac{{n}\left({n}−\mathrm{1}\right)\:{x}^{\mathrm{2}{n}−\mathrm{2}} \:−{n}\left({n}−\mathrm{1}\right){x}^{{n}−\mathrm{2}} \:−{n}^{\mathrm{2}} {x}^{\mathrm{2}{n}−\mathrm{2}} }{\left({x}^{{n}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=−\frac{−{n}\:{x}^{\mathrm{2}{n}−\mathrm{2}} \:−{n}\left({n}−\mathrm{1}\right){x}^{{n}−\mathrm{2}} }{\left({x}^{{n}} −\mathrm{1}\right)^{\mathrm{2}} }=\frac{{n}\:{x}^{\mathrm{2}{n}−\mathrm{2}} \:−{n}\left({n}−\mathrm{1}\right){x}^{{n}−\mathrm{2}} }{\left({x}^{{n}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${for}\:{n}\geqslant\mathrm{2}\:. \\ $$