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Question Number 32334 by abdo imad last updated on 23/Mar/18

p is apolynomial having n roots  (x_i ) with x_i ≠xj for  i≠j  calculate  Σ_(k=1) ^n    (1/(1−x_k )) .

$${p}\:{is}\:{apolynomial}\:{having}\:{n}\:{roots}\:\:\left({x}_{{i}} \right)\:{with}\:{x}_{{i}} \neq{xj}\:{for} \\ $$$${i}\neq{j}\:\:{calculate}\:\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\frac{\mathrm{1}}{\mathrm{1}−{x}_{{k}} }\:. \\ $$

Commented by abdo imad last updated on 01/Apr/18

p(x)=λ Π_(k=1) ^n  (x−x_i ) ⇒p^′ (x)=λ Σ_(k=1) ^n  Π_(p=1_(p≠k) ) ^n  (x−x_p )  ⇒ ((p^′ (x))/(p(x))) = Σ_(k=1) ^n   (1/(x−x_k )) ⇒ Σ_(k=1) ^n   (1/(1−x_k )) =((p^′ (1))/(p(1)))  with p(1)≠0

$${p}\left({x}\right)=\lambda\:\prod_{{k}=\mathrm{1}} ^{{n}} \:\left({x}−{x}_{{i}} \right)\:\Rightarrow{p}^{'} \left({x}\right)=\lambda\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\prod_{{p}=\mathrm{1}_{{p}\neq{k}} } ^{{n}} \:\left({x}−{x}_{{p}} \right) \\ $$$$\Rightarrow\:\frac{{p}^{'} \left({x}\right)}{{p}\left({x}\right)}\:=\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{{x}−{x}_{{k}} }\:\Rightarrow\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{1}−{x}_{{k}} }\:=\frac{{p}^{'} \left(\mathrm{1}\right)}{{p}\left(\mathrm{1}\right)}\:\:{with}\:{p}\left(\mathrm{1}\right)\neq\mathrm{0} \\ $$$$ \\ $$

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