Question Number 28544 by abdo imad last updated on 26/Jan/18
$${if}\:\:{a}_{\mathrm{1}} \:,{a}_{\mathrm{2}} ,…{a}_{\mathrm{14}\:} {are}\:{roots}\:{of}\:{the}\:{polynomial} \\ $$$${p}\left({x}\right)={x}^{\mathrm{14}} +{x}^{\mathrm{8}} \:\mathrm{2}{x}+\mathrm{1}\:\:\:{calculate}\:\:\sum_{{i}=\mathrm{1}} ^{\mathrm{14}} \:\:\frac{\mathrm{1}}{\left({a}_{{i}} −\mathrm{1}\right)^{\mathrm{2}} }\:\:. \\ $$
Commented by abdo imad last updated on 26/Jan/18
$${p}\left({x}\right)={x}^{\mathrm{14}} \:+{x}^{\mathrm{8}} \:+\mathrm{2}{x}+\mathrm{1}\:. \\ $$
Commented by abdo imad last updated on 28/Jan/18
$${we}\:{know}\:{that}\:\:\frac{{p}^{'} \left({x}\right)}{{p}\left({x}\right)}\:=\:\sum_{{i}=\mathrm{1}} ^{\mathrm{14}\:} \:\frac{\mathrm{1}}{{x}−{x}_{{i}} } \\ $$$$\:\frac{{p}^{''} \:{p}\left({x}\right)\:−{p}'^{\mathrm{2}} \left({x}\right)}{{p}^{\mathrm{2}} \left({x}\right)}=\:−\sum_{{i}=\mathrm{1}} ^{\mathrm{14}} \:\:\:\frac{\mathrm{1}}{\left({x}−{x}_{{i}} \right)^{\mathrm{2}} }\:\:\:{but}\:{we}\:{have} \\ $$$${p}^{'} \left({x}\right)=\:\mathrm{14}\:{x}^{\mathrm{13}} +\mathrm{8}{x}^{\mathrm{7}} \:+\mathrm{2}\:\:{and}\:\:{p}^{\left(\mathrm{2}\right)} \left({x}\right)=\:\mathrm{14}\:×\mathrm{13}\:{x}^{\mathrm{12}} \:+\mathrm{56}\:{x}^{\mathrm{6}} \\ $$$$\sum_{{i}=\mathrm{2}} ^{\mathrm{14}} \:\:\:\frac{\mathrm{1}}{\left({x}−{a}_{{i}} \right)\mathrm{2}\:}=−\frac{\left(\mathrm{14}×\mathrm{13}{x}^{\mathrm{12}} \:+\mathrm{56}\:{x}^{\mathrm{6}} \right)\left({x}^{\mathrm{14}} +{x}^{\mathrm{8}} +\mathrm{2}{x}+\mathrm{1}\right)−\left(\mathrm{14}{x}^{\mathrm{13}} \:+\mathrm{8}{x}^{\mathrm{7}} +\mathrm{2}\right)^{\mathrm{2}} }{\left({x}^{\mathrm{14}} \:+{x}^{\mathrm{8}} \:+\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${and}\:{for}\:{x}=\mathrm{1}\:?{we}\:{find} \\ $$$$\sum_{{i}=\mathrm{1}} ^{\mathrm{14}} \:\:\:\frac{\mathrm{1}}{\left({a}_{{i}} \:−\mathrm{1}\right)^{\mathrm{2}} }=−\frac{\left(\mathrm{14}×\mathrm{13}\:\:+\mathrm{56}\right)×\mathrm{5}−\left(\mathrm{24}\right)}{\mathrm{25}}… \\ $$