p-is-a-polynomial-having-n-roots-x-i-with-x-i-x-j-for-i-j-prove-that-i-1-n-p-x-i-p-x-i-0-

Tinku Tara June 4, 2023 Algebra 0 Comments

Question Number 29832 by abdo imad last updated on 12/Feb/18

p is a polynomial having n roots x_i with x_i ≠x_j for i≠j prove that Σ_(i=1) ^n ((p^(′′) (x_i ))/(p^′ (x_i )))=0

$${p}\:{is}\:{a}\:{polynomial}\:{having}\:{n}\:{roots}\:{x}_{{i}} \:\:{with}\:{x}_{{i}} \neq{x}_{{j}} \:{for}\:{i}\neq{j} \\ $$$${prove}\:{that}\:\sum_{{i}=\mathrm{1}} ^{{n}} \:\:\frac{{p}^{''} \left({x}_{{i}} \right)}{{p}^{'} \left({x}_{{i}} \right)}=\mathrm{0} \\ $$

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