Question Number 36911 by prof Abdo imad last updated on 07/Jun/18
$${p}\:{is}\:{a}\:{polynome}\:{having}\:{nroots}\:{simples} \\ $$$${x}_{{i}} \:\left(\mathrm{1}\leqslant{x}_{{i}} \leqslant{n}\:\right)\:{with}\:{x}_{{i}} ^{\mathrm{2}} \:\neq\mathrm{1}\:\:{calculste} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{1}−{x}_{{k}} }\:. \\ $$
Commented by math khazana by abdo last updated on 09/Jun/18
$${we}\:{have}\:\:{p}\left({x}\right)=\:\lambda\:\prod_{{i}=\mathrm{1}} ^{{n}} \:\left({x}−{x}_{{i}} \right)\:{and}\: \\ $$$${p}^{'} \left({x}\right)\:=\:\lambda\:\sum_{{i}=\mathrm{1}} ^{{n}} \:\prod_{{k}=\mathrm{1}\:{and}\:{k}\neq{i}} ^{{n}} \left({x}−{x}_{{k}} \right)\:\Rightarrow \\ $$$$\frac{{p}^{'} \left({x}\right)}{{p}\left({x}\right)}\:=\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{{x}−{x}_{{k}} }\:\Rightarrow\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{1}−{x}_{{k}} }\:=\:\frac{{p}^{'} \left(\mathrm{1}\right)}{{p}\left(\mathrm{1}\right)}\:\:. \\ $$