p-is-a-polynom-having-n-roots-simples-with-x-i-1-caculate-k-1-n-1-1-x-i-and-k-1-n-1-1-x-i-2-

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

p i s a p o l y n o m h a v i n g n r o o t s s i m p l e s w i t h x_{i} \neq \bar{+} 1

c a c u l a t e \sum_{k = 1}^{n} \frac{1}{1 - x_{i}} a n d \sum_{k = 1}^{n} \frac{1}{1 - x_{i}^{2}}

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

p (x) = λ \prod_{i = 1}^{n} (x - x_{i}) \Rightarrow \frac{1}{p (x)} = \frac{1}{λ \prod_{i = 1}^{n} (x - x_{i})}

= \sum_{i = 1}^{n} \frac{α_{i}}{x - x_{i}} w i t h α_{i} = \frac{1}{p^{^{'}} (x_{i})} \Rightarrow

\frac{1}{p (x)} = \frac{1}{p^{^{'}} (x_{i})} \sum_{i = 1}^{n} \frac{1}{x - x_{i}} \Rightarrow \sum_{i = 1}^{n} \frac{1}{x - x_{i}} = \frac{p^{^{'}} (x_{i})}{p (x)}

x = 1 \Rightarrow \sum_{i = 1}^{n} \frac{1}{1 - x_{i}} = \frac{p^{^{'}} (x_{i})}{p (1)}

p (1) = λ \prod_{i = 1}^{n} (1 - x_{i}) .