Previous in Relation and Functions Next in Relation and Functions
Question Number 50385 by Abdo msup. last updated on 16/Dec/18
$${p}\:{is}\:{a}\:{polynom}\:{having}\:{n}\:{roots}\:{simples}\:{with}\:{x}_{{i}} \neq\overset{−} {+}\mathrm{1} \\ $$$${caculate}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{1}−{x}_{{i}} }\:\:{and}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{1}−{x}_{{i}} ^{\mathrm{2}} } \\ $$$$ \\ $$
Commented by Abdo msup. last updated on 19/Dec/18
$${p}\left({x}\right)=\lambda\:\prod_{{i}=\mathrm{1}} ^{{n}} \:\left({x}−{x}_{{i}} \right)\:\Rightarrow\frac{\mathrm{1}}{{p}\left({x}\right)}\:=\frac{\mathrm{1}}{\lambda\:\prod_{{i}=\mathrm{1}} ^{{n}} \left({x}−{x}_{{i}} \right)} \\ $$$$=\sum_{{i}=\mathrm{1}} ^{{n}} \:\:\frac{\alpha_{{i}} }{{x}−{x}_{{i}} }\:\:{with}\:\alpha_{{i}} =\frac{\mathrm{1}}{{p}^{'} \left({x}_{{i}} \right)}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{{p}\left({x}\right)}\:=\frac{\mathrm{1}}{{p}^{'} \left({x}_{{i}} \right)}\:\sum_{{i}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{x}−{x}_{{i}} }\:\Rightarrow\sum_{{i}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{x}−{x}_{{i}} }\:=\frac{{p}^{'} \left({x}_{{i}} \right)}{{p}\left({x}\right)} \\ $$$${x}\:=\mathrm{1}\:\Rightarrow\sum_{{i}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{1}−{x}_{{i}} }\:=\frac{{p}^{'} \left({x}_{{i}} \right)}{{p}\left(\mathrm{1}\right)} \\ $$$${p}\left(\mathrm{1}\right)\:=\lambda\:\prod_{{i}=\mathrm{1}} ^{{n}} \left(\mathrm{1}−{x}_{{i}} \right)\:. \\ $$