Question Number 26582 by abdo imad last updated on 27/Dec/17
$${p}\:{is}\:{a}\:{polynomial}\:{having}\:{the}\:{roots}\:{x}_{\mathrm{1}} ,{x}_{\mathrm{2}} ,...{x}_{{n}} \\ $$$${with}\:{x}_{{i}} \neq\:{x}_{{j}} \:{fori}\neq{j}\:{give}\:{the}\:{decomposition} \\ $$$${of}\:{the}\:{fravtion}\:{F}\left({x}\right)=\:\frac{{p}^{'} \left({x}\right)}{{p}\left({x}\right)} \\ $$
Commented by abdo imad last updated on 28/Dec/17
$${the}\:{roots}\:{of}\:\:{p}\left({x}\right)\:{are}\:{simple}\:{so}\:{p}\left({x}\right)=\alpha\:\prod_{{k}=\mathrm{1}} ^{{k}={n}} \left({x}−{x}_{{k}} \:\right) \\ $$$$\Rightarrow\:{p}^{,} \left({x}\right)=\:\:\alpha\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\prod_{{p}=\mathrm{1}_{{p}\neq{k}} } ^{{p}={n}} \:\left({x}−{x}_{{p}} \:\right)\:{and} \\ $$$$\frac{{p}^{,} \left({x}\right)}{{p}\left({x}\right)}\:=\:\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\:\:\frac{\mathrm{1}}{{x}−{x}_{{k}} }\:. \\ $$