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Question Number 50385 by Abdo msup. last updated on 16/Dec/18

$$pisapolynomhavingnrootssimpleswith{x}_i\ne \stackrel{-}{+}1$$$$caculate\sum _{k=1}^n\frac{1}{1-x_i}and\sum _{k=1}^n\frac{1}{1-x_i^2}$$$$$$

Commented by Abdo msup. last updated on 19/Dec/18

$$p\left(x\right)=\lambda \prod _{i=1}^n(x-x_i)\Rightarrow \frac{1}{p\left(x\right)}=\frac{1}{\lambda \prod _{i=1}^n(x-x_i)}$$$$=\sum _{i=1}^n\frac{{\alpha }_i}{x-x_i}with{\alpha }_i=\frac{1}{p^{{}^{\prime }}\left(x_i\right)}\Rightarrow$$$$\frac{1}{p\left(x\right)}=\frac{1}{p^{{}^{\prime }}\left(x_i\right)}\sum _{i=1}^n\frac{1}{x-x_i}\Rightarrow \sum _{i=1}^n\frac{1}{x-x_i}=\frac{p^{{}^{\prime }}\left(x_i\right)}{p\left(x\right)}$$$$x=1\Rightarrow \sum _{i=1}^n\frac{1}{1-x_i}=\frac{p^{{}^{\prime }}\left(x_i\right)}{p\left(1\right)}$$$$p\left(1\right)=\lambda \prod _{i=1}^n(1-x_i).$$

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